0
6469

# Most Expected CAT DILR Questions PDF:

Download Top 25 CAT DILR Sets for CAT 2022. Very important and most expected Logical Reasoning and Data Interpretation Puzzles Questions and answers by CAT 100%iler based on asked questions in previous exam papers.

🌟 Watch CAT Full revision

CAT Quant revision

CAT VARC revision

CAT DILR revision

Instructions

Two alloy making companies X and Y use 5 different metals – P, Q, R, S and T – to make an alloy. In the below table, the usage of each metal as a proportion of weight and proportion of value is given. The costs per tonne(1000 kg) of metals P, Q, R, S and T (in Lakhs) are 20, 25, 15, 30 and 20 respectively.

Question 1: Find the ratio of amounts of metals R and T used in the making of alloy by company Y.

a) 4:3

b) 3:4

c) 1:1

d) cannot be determined

Solution:

We know that the ratio of values of R and T is 20:20 = 1:1
To find the ratio of weights of R and T, we need to divide the above ratio with their prices per tonne.
So, the required ratio is $\frac{1}{15}:\frac{1}{20}$ = 4:3

Question 2: Find the approximate ratio of the selling prices of the companies X and Y, if both the companies sell their alloy at a profit of 25%.

a) 28:29

b) 13:14

c) 29:28

d) 14:13

Solution:

Profit % is same => Ratio of selling-prices = Ratio of cost-prices
The prices are 20, 25, 15, 30 and 20. Let’s take an amount equal to LCM of these => amount = 300
Weight of P in alloy by Y = $\frac{0.15*300}{20}$ = 2.25
Weight of Q in alloy by Y = $\frac{0.35*300}{25}$ = 4.20
Weight of R in alloy by Y = $\frac{0.20*300}{15}$ = 4
Weight of S in alloy by Y = $\frac{0.10*300}{30}$ = 1
Weight of T in alloy by Y = $\frac{0.20*300}{20}$ = 3
Total weight = 14.45 tonnes
Cost per tonne = $\frac{300}{14.45}$ = 20.76
Total cost due to X = 4 + 6.25 + 2.25 + 3 + 6 = 21.5
Ratio of 21.5: 20.76 is approximately equal to 29:28

Question 3: Due to which of the following metals has the cost of alloy making by company X gone up by the maximum amount?

a) P

b) Q

c) R

d) T

Solution:

% by weight of P = 20%
% by weight of Q = 25%
% by weight of R = 15%
% by weight of S = 10%
% by weight of T = 30%
Effective cost due to P = 20%*20 = 4
Effective cost due to Q = 25%*25 = 6.25
Effective cost due to R = 15%*15 = 2.25
Effective cost due to S = 10%*30 = 3
Effective cost due to T = 30%*20 = 6
Effective cost of Q is highest => Q accounts for maximum cost

Question 4: What percentage of the total cost is the cost due to metals P and R in the alloy made by company X?

a) 26%

b) 27%

c) 28%

d) 29%

Solution:

% by weight of P = 20%
% by weight of Q = 25%
% by weight of R = 15%
% by weight of S = 10%
% by weight of T = 30%
Effective cost due to P = 20%*20 = 4
Effective cost due to Q = 25%*25 = 6.25
Effective cost due to R = 15%*15 = 2.25
Effective cost due to S = 10%*30 = 3
Effective cost due to T = 30%*20 = 6
% cost due to P and R = $\frac{6.25}{21.5}$ = 29%

Instructions

36 workers are sitting in a factory in the 6×6 grid as shown below.

In the figure, 6 workers are marked which are having high fever.

All the workers started their work i.e. sat next to each other at 12:00 Noon.

A healthy worker will catch fever if he is adjacent to at least two sick workers for at least 1 hour.

A healthy worker when caught fever is considered a sick worker and thus he or she can spread fever to other workers.

Diagonally seated workers are not considered adjacent.

The columns are named A to F from left to right and the rows are named 1 to 6 from top to bottom. So, at the 12:00 noon, the six people who were sick were sitting in F1, D2, B3, E4, D5 and A6

Question 5: How many sick workers will be there in the factory till 5:00 PM?

a) 15

b) 13

c) 19

d) 20

Solution:

At time=1:00 PM

2 workers, marked as X1 will catch the fever.

X1 worker will spread the disease to the worker between X and X1. Similarly, X2 worker will spread to two other workers as given below.

Thus we can make the following tables:

Till 5 PM, 19 patients will be sick.

option C

Question 6: After how many hours will all the workers be sick in the factory?

a) 9

b) 10

c) 11

d) 12

Solution:

At time=1:00 PM

2 students, marked as X1 will catch the fever.

X1 worker will spread the disease to the worker between X and X1. Similarly, X2 worker will spread to two other workers as given below.

Thus we can make the following tables:

Till 11:00 PM all the workers will get sick.

Question 7: Amongst the following, which worker gets sick the last?

a) B2

b) A1

c) B5

d) C6

Solution:

At time=1:00 PM

2 students, marked as X1 will catch the fever.

X1 worker will spread the disease to the worker between X and X1. Similarly, X2 worker will spread to two other workers as given below.

Thus we can make the following tables:

B2 gets sick at 5 pm
A1 gets sick at 11 pm
B5 gets sick at 6 pm
C6 gets sick at 8 pm

So, A1 gets sick the last amongst the given options.

Question 8: How many workers who were not sick at 8 pm, get sick by 9 pm?

Solution:

At time=1:00 PM

2 students, marked as X1 will catch the fever.

X1 worker will spread the disease to the worker between X and X1. Similarly, X2 worker will spread to two other workers as given below.

Thus we can make the following tables:

2 workers who were not sick at 8 pm got sick by 9 PM.

Instructions

Gaurav plays a game called the magic square. A magic square consists of a 3×3 grid. Gaurav has to fill numbers from 1 to 9 in the 9 cells of the grid such that no number appears twice and the sum of the numbers in each row, each column, and each diagonal are the same.

The rows are named as A, B, and C (from top to bottom) and the columns are named as X, Y, and Z (from left to right). A cell is referred to using its row name followed by the column name (Eg) The top left corner cell is referred to as cell AX)

Question 9: If the entry in cell CZ is 4, what should be the entry in cell AY?

a) 1 or 7

b) 2 or 8

c) 1 or 8

d) 2 or 7

Solution:

We are using all the numbers from 1 to 9 exactly once in the square. Also, we know that the sum of the numbers in any of the 3 rows is the same.

Let X be the sum of the numbers in a row.

Sum of the numbers in all 3 rows = 3X.

We know that each number from 1 to 9 will be present exactly once in the square.

=> 3X = 1+2+3+4+5+6+7+8+9

3X = 45

X = 15.

Sum of the numbers in each row, each column, and each diagonal is 15.
We have to decide the number that fills the central square first.

If 1 is the number used to fill the central square, then the numbers in the cells adjacent to the central cell in any direction should add up to 14. Therefore, we will need 4 such pairs.
14 can be expressed as 9+5, and 8+6. We cannot represent 14 in any other way. Therefore, we can eliminate this case.

If 2 is filled in the central square, then 13 has to be expressed as the sum of 4 different pairs. 13 can be expressed as 9+4, 8+5, and 7+6. Therefore, the central square cannot be filled using 2 as well.

If 3 is filled in the central square, 12 can be expressed as 8+4, and 7+5. We can eliminate this case since there are only 2 possibilities.

If 4 is filled in the central square, 11 can be expressed as 9+2, 8+3, and 6+5. We can eliminate this case as well since there are only 3 possibilities.

If 5 is filled in the central square, 10 can be expressed as 1+9, 2+8, 3+7, and 4+6. As we can see, 5 can fill the central square since we have 4 possible pairs.

If 6 is filled in the central square, 9 can be expressed as 8+1, 7+2, and 4+5. We can eliminate this case as there are only 3 possibilities.

If 7 is filled in the central square, 8 can be expressed as 6+2, and 5+3. We can eliminate this case.

If 8 is filled in the central square, 7 can be expressed as 6+1, and 5+2 and 4+3. We can eliminate this case.

If 9 is filled in the central square, 6 can be expressed as 5+1, and 4+2.
Only 5 gives us enough possibilities to fill the 2 diagonals, row B and column Y. Therefore, 5 should be filled in square BY.

We know the pairs that should be used to fill the other squares – (1,9), (2,8), (3,7) and (4,6). We have to decide which numbers should go in the diagonals and which numbers should fill the other cells.

Let us consider the top-most row (row A).

We know that one of the cells in the square should contain 9. Let us put 9 in the top-most row. The other 2 cells in this row should add up to 6. Among the given choices, only 4 and 2 can add up to 6. Therefore, 4, 9, and 2 should be present in the same row. We have to decide whether 9 should be present in one of the central cells or at the corner cell. We know that 2, 4, and 9 are present in the same row.

We have to find 2 more numbers such that when added to 9, they yield a sum of 15. Since we have already used 2 and 4, the only other way in which we can get 6 is by adding 5 and 1. If 9 is present in one of the corners, we have to find one more possibility of getting 6 (to get the sum as 15 in the diagonal). Therefore, 9 cannot be present in a corner cell. 9 should be present in one of the central squares and 2 and 4 must be present in the corner cells.

Let us fill the diagonals.
2+5 = 7. Therefore, 8 must be present in the square diagonally opposite to the one in which 2 is present.

4+5 = 9. Therefore, 6 should be present in the cell diagonally opposite to the one in which 4 is present.

Filling the remaining squares similarly, we get,

As we can see, there is only one possible pattern. This pattern can be rotated and 8 different cases can be obtained as follows:

It has been given that the number in the cell CZ is 4. Cases 4 and 5 are valid. The number in cell AY can be 1 or 7. Therefore, option A is the right answer.

Question 10: If the entry in cell AZ is not 2 or 4, then the entry in cell CY cannot be

a) 1

b) 9

c) 7

d) 3

Solution:

We are using all the numbers from 1 to 9 exactly once in the square. Also, we know that the sum of the numbers in any of the 3 rows is the same.

Let X be the sum of the numbers in a row.

Sum of the numbers in all 3 rows = 3X.

We know that each number from 1 to 9 will be present exactly once in the square.

=> 3X = 1+2+3+4+5+6+7+8+9

3X = 45

X = 15.

Sum of the numbers in each row, each column, and each diagonal is 15.
We have to decide the number that fills the central square first.

If 1 is the number used to fill the central square, then the numbers in the cells adjacent to the central cell in any direction should add up to 14. Therefore, we will need 4 such pairs.
14 can be expressed as 9+5, and 8+6. We cannot represent 14 in any other way. Therefore, we can eliminate this case.

If 2 is filled in the central square, then 13 has to be expressed as the sum of 4 different pairs. 13 can be expressed as 9+4, 8+5, and 7+6. Therefore, the central square cannot be filled using 2 as well.

If 3 is filled in the central square, 12 can be expressed as 8+4, and 7+5. We can eliminate this case since there are only 2 possibilities.

If 4 is filled in the central square, 11 can be expressed as 9+2, 8+3, and 6+5. We can eliminate this case as well since there are only 3 possibilities.

If 5 is filled in the central square, 10 can be expressed as 1+9, 2+8, 3+7, and 4+6. As we can see, 5 can fill the central square since we have 4 possible pairs.

If 6 is filled in the central square, 9 can be expressed as 8+1, 7+2, and 4+5. We can eliminate this case as there are only 3 possibilities.

If 7 is filled in the central square, 8 can be expressed as 6+2, and 5+3. We can eliminate this case.

If 8 is filled in the central square, 7 can be expressed as 6+1, and 5+2 and 4+3. We can eliminate this case.

If 9 is filled in the central square, 6 can be expressed as 5+1, and 4+2.
Only 5 gives us enough possibilities to fill the 2 diagonals, row B and column Y. Therefore, 5 should be filled in square BY.

We know the pairs that should be used to fill the other squares – (1,9), (2,8), (3,7) and (4,6). We have to decide which numbers should go in the diagonals and which numbers should fill the other cells.

Let us consider the top-most row (row A).

We know that one of the cells should contain 9. The other 2 cells in this row should add up to 6. Among the given choices, only 4 and 2 can add up to 6. Therefore, 4, 9, and 2 should be present in the same row. We have to decide whether 9 should be present in one of the central cells or at the corner cell. We know that 2, 4, and 9 are present in the same row.

We have to find 2 more numbers such that when added to 9, they yield a sum of 15. Since we have already used 2 and 4, the only other way in which we can get 6 is by adding 5 and 1. If 9 is present in one of the corners, we have to find one more possibility of getting 6 (to get the sum as 15 in the diagonal). Therefore, 9 cannot be present in a corner cell. 9 should be present in one of the central squares and 2 and 4 must be present in the corner cells.

Let us fill the diagonals.
2+5 = 7. Therefore, 8 must be present in the square diagonally opposite to the one in which 2 is present.

4+5 = 9. Therefore, 6 should be present in the cell diagonally opposite to the one in which 4 is present.

Filling the remaining squares similarly, we get,

As we can see, there is only one possible pattern. This pattern can be rotated and 8 different cases can be obtained as follows:

The entry in cell AZ is not 2 or 4. Cases 1, 2, 5, and 7 are invalid.
The entry in cell CY can be 9, 3, and 7.
The number 1 cannot be present in cell CY if the entry in cell AZ is not 2 or 4. Therefore, option A is the right answer.

Question 11: If 3 is not present in row B, then which of the following numbers will definitely not be present in row A?

a) 4

b) 8

c) 6

d) 1

Solution:

We are using all the numbers from 1 to 9 exactly once in the square. Also, we know that the sum of the numbers in any of the 3 rows is the same.

Let X be the sum of the numbers in a row.

Sum of the numbers in all 3 rows = 3X.

We know that each number from 1 to 9 will be present exactly once in the square.

=> 3X = 1+2+3+4+5+6+7+8+9

3X = 45

X = 15.

Sum of the numbers in each row, each column, and each diagonal is 15.
We have to decide the number that fills the central square first.

If 1 is the number used to fill the central square, then the numbers in the cells adjacent to the central cell in any direction should add up to 14. Therefore, we will need 4 such pairs.
14 can be expressed as 9+5, and 8+6. We cannot represent 14 in any other way. Therefore, we can eliminate this case.

If 2 is filled in the central square, then 13 has to be expressed as the sum of 4 different pairs. 13 can be expressed as 9+4, 8+5, and 7+6. Therefore, the central square cannot be filled using 2 as well.

If 3 is filled in the central square, 12 can be expressed as 8+4, and 7+5. We can eliminate this case since there are only 2 possibilities.

If 4 is filled in the central square, 11 can be expressed as 9+2, 8+3, and 6+5. We can eliminate this case as well since there are only 3 possibilities.

If 5 is filled in the central square, 10 can be expressed as 1+9, 2+8, 3+7, and 4+6. As we can see, 5 can fill the central square since we have 4 possible pairs.

If 6 is filled in the central square, 9 can be expressed as 8+1, 7+2, and 4+5. We can eliminate this case as there are only 3 possibilities.

If 7 is filled in the central square, 8 can be expressed as 6+2, and 5+3. We can eliminate this case.

If 8 is filled in the central square, 7 can be expressed as 6+1, and 5+2 and 4+3. We can eliminate this case.

If 9 is filled in the central square, 6 can be expressed as 5+1, and 4+2.
Only 5 gives us enough possibilities to fill the 2 diagonals, row B and column Y. Therefore, 5 should be filled in square BY.

We know the pairs that should be used to fill the other squares – (1,9), (2,8), (3,7) and (4,6). We have to decide which numbers should go in the diagonals and which numbers should fill the other cells.

Let us consider the top-most row (row A).

We know that one of the cells should contain 9. The other 2 cells in this row should add up to 6. Among the given choices, only 4 and 2 can add up to 6. Therefore, 4, 9, and 2 should be present in the same row. We have to decide whether 9 should be present in one of the central cells or at the corner cell. We know that 2, 4, and 9 are present in the same row.

We have to find 2 more numbers such that when added to 9, they yield a sum of 15. Since we have already used 2 and 4, the only other way in which we can get 6 is by adding 5 and 1. If 9 is present in one of the corners, we have to find one more possibility of getting 6 (to get the sum as 15 in the diagonal). Therefore, 9 cannot be present in a corner cell. 9 should be present in one of the central squares and 2 and 4 must be present in the corner cells.

Let us fill the diagonals.
2+5 = 7. Therefore, 8 must be present in the square diagonally opposite to the one in which 2 is present.

4+5 = 9. Therefore, 6 should be present in the cell diagonally opposite to the one in which 4 is present.

Filling the remaining squares similarly, we get,

As we can see, there is only one possible pattern. This pattern can be rotated and 8 different cases can be obtained as follows:

It has been given in the question that 3 is not present in row B.
Only cases 5 to 8 have to be considered.

We can see that 1 is not present in row A in all of these 4 cases. Therefore, option D is the right answer.

Question 12: Which of the following numbers should definitely be present in a corner cell (i.e, one among cells AX, AZ, CX, and CZ)?

a) 9

b) 4

c) 7

d) 3

Solution:

We are using all the numbers from 1 to 9 exactly once in the square. Also, we know that the sum of the numbers in any of the 3 rows is the same.

Let X be the sum of the numbers in a row.

Sum of the numbers in all 3 rows = 3X.

We know that each number from 1 to 9 will be present exactly once in the square.

=> 3X = 1+2+3+4+5+6+7+8+9

3X = 45

X = 15.

Sum of the numbers in each row, each column, and each diagonal is 15.
We have to decide the number that fills the central square first.

If 1 is the number used to fill the central square, then the numbers in the cells adjacent to the central cell in any direction should add up to 14. Therefore, we will need 4 such pairs.
14 can be expressed as 9+5, and 8+6. We cannot represent 14 in any other way. Therefore, we can eliminate this case.

If 2 is filled in the central square, then 13 has to be expressed as the sum of 4 different pairs. 13 can be expressed as 9+4, 8+5, and 7+6. Therefore, the central square cannot be filled using 2 as well.

If 3 is filled in the central square, 12 can be expressed as 8+4, and 7+5. We can eliminate this case since there are only 2 possibilities.

If 4 is filled in the central square, 11 can be expressed as 9+2, 8+3, and 6+5. We can eliminate this case as well since there are only 3 possibilities.

If 5 is filled in the central square, 10 can be expressed as 1+9, 2+8, 3+7, and 4+6. As we can see, 5 can fill the central square since we have 4 possible pairs.

If 6 is filled in the central square, 9 can be expressed as 8+1, 7+2, and 4+5. We can eliminate this case as there are only 3 possibilities.

If 7 is filled in the central square, 8 can be expressed as 6+2, and 5+3. We can eliminate this case.

If 8 is filled in the central square, 7 can be expressed as 6+1, and 5+2 and 4+3. We can eliminate this case.

If 9 is filled in the central square, 6 can be expressed as 5+1, and 4+2.
Only 5 gives us enough possibilities to fill the 2 diagonals, row B and column Y. Therefore, 5 should be filled in square BY.

We know the pairs that should be used to fill the other squares – (1,9), (2,8), (3,7) and (4,6). We have to decide which numbers should go in the diagonals and which numbers should fill the other cells.

Let us consider the top-most row (row A).

We know that one of the cells should contain 9. The other 2 cells in this row should add up to 6. Among the given choices, only 4 and 2 can add up to 6. Therefore, 4, 9, and 2 should be present in the same row. We have to decide whether 9 should be present in one of the central cells or at the corner cell. We know that 2, 4, and 9 are present in the same row.

We have to find 2 more numbers such that when added to 9, they yield a sum of 15. Since we have already used 2 and 4, the only other way in which we can get 6 is by adding 5 and 1. If 9 is present in one of the corners, we have to find one more possibility of getting 6 (to get the sum as 15 in the diagonal). Therefore, 9 cannot be present in a corner cell. 9 should be present in one of the central squares and 2 and 4 must be present in the corner cells.

Let us fill the diagonals.
2+5 = 7. Therefore, 8 must be present in the square diagonally opposite to the one in which 2 is present.

4+5 = 9. Therefore, 6 should be present in the cell diagonally opposite to the one in which 4 is present.

Filling the remaining squares similarly, we get,

As we can see, there is only one possible pattern. This pattern can be rotated and 8 different cases can be obtained as follows:

Instructions

Study the following information carefully and answer the questions that follow.

A jailor instructed all inmates from Cell 1 to 8 to stand in a row. The jailor was very strict and since he didn’t mention the direction, the inmates stood in a straight row facing either north or south without asking him any further question. They stood in such a manner that no three consecutive inmates faced the same direction. The jailor then asked each of them to shout out loud the number of inmates he can see ahead of him. The following table gives the number reported by each inmate:

Further, it is also given that the number of inmates facing south was not more than four.

Question 13: How many different arrangements are possible?

Solution:

Let’s first number the places as shown in the figure-
We can see that inmate 5 sees 0 people in front of him. So, he must be standing at the extreme and facing the same direction i.e. north when standing in front or south when standing at last.
Similarly, inmate 6 sees 7 people. So, he must stand at P facing south or at W facing north.
Now, there are 2 people who see 6 people in front of them. So, they must be facing opposite direction (i.e. one who stands at Q faces south and one who stands at V faces north) and stand at one place among Q and V.
Also, inmate 2 sees 2 people ahead of him while inmate 7 sees 5 people. So, they must be facing the same direction standing at one position among R and U.
Similarly, 3 and 4 must be standing at one position among S and T facing the same direction.
So, positions of the 8 inmates can be concluded as:

Let us now decide on the directions for each of them:
The person from Q must definitely be facing south. Apart from this, the person facing south will increase by 2. Since the number of inmates facing south was not more than four, the number of person facing south has to be 3. If positions P and W are facing south, all others will have to face north which is not possible as no three consecutive inmates faced the same direction. So, person on P and W must be facing north. So, one of the pair of R-U or S-T must be facing south. But when person on U faces north, three consecutive people i.e. U, V, and W must be facing north which is not possible. Thus, pair S-T must be facing south. In this way, we can figure out the whole arrangement-

Question 14: Who among the following is definitely facing South?

a) Inmate from cell 1

b) Inmate from cell 3

c) Inmate from cell 4

d) Inmate from cell 7

Solution:

Let’s first number the places as shown in the figure-
We can see that inmate 5 sees 0 people in front of him. So, he must be standing at the extreme and facing the same direction i.e. north when standing in front or south when standing at last.
Similarly, inmate 6 sees 7 people. So, he must stand at P facing south or at W facing north.
Now, there are 2 people who see 6 people in front of them. So, they must be facing opposite direction (i.e. one who stands at Q faces south and one who stands at V faces north) and stand at one place among Q and V.
Also, inmate 2 sees 2 people ahead of him while inmate 7 sees 5 people. So, they must be facing the same direction standing at one position among R and U.
Similarly, 3 and 4 must be standing at one position among S and T facing the same direction.
So, positions of the 8 inmates can be concluded as:

Let us now decide on the directions for each of them:
The person from Q must definitely be facing south. Apart from this, the person facing south will increase by 2. Since the number of inmates facing south was not more than four, the number of person facing south has to be 3. If positions P and W are facing south, all others will have to face north which is not possible as no three consecutive inmates faced the same direction. So, person on P and W must be facing north. So, one of the pair of R-U or S-T must be facing south. But when person on U faces north, three consecutive people i.e. U, V, and W must be facing north which is not possible. Thus, pair S-T must be facing south. In this way, we can figure out the whole arrangement-

We can see that inmate from cell 7 definitely faces south.

Question 15: If the inmate from cell 1 faces north, the inmate from which cell stands 2 ahead inmate from cell 3? (Enter ‘-1’ if the answer cannot be determined)

Solution:

Let’s first number the places as shown in the figure-
We can see that inmate 5 sees 0 people in front of him. So, he must be standing at the extreme and facing the same direction i.e. north when standing in front or south when standing at last.
Similarly, inmate 6 sees 7 people. So, he must stand at P facing south or at W facing north.
Now, there are 2 people who see 6 people in front of them. So, they must be facing opposite direction (i.e. one who stands at Q faces south and one who stands at V faces north) and stand at one place among Q and V.
Also, inmate 2 sees 2 people ahead of him while inmate 7 sees 5 people. So, they must be facing the same direction standing at one position among R and U.
Similarly, 3 and 4 must be standing at one position among S and T facing the same direction.
So, positions of the 8 inmates can be concluded as:

Let us now decide on the directions for each of them:
The person from Q must definitely be facing south. Apart from this, the person facing south will increase by 2. Since the number of inmates facing south was not more than four, the number of person facing south has to be 3. If positions P and W are facing south, all others will have to face north which is not possible as no three consecutive inmates faced the same direction. So, person on P and W must be facing north. So, one of the pair of R-U or S-T must be facing south. But when person on U faces north, three consecutive people i.e. U, V, and W must be facing north which is not possible. Thus, pair S-T must be facing south. In this way, we can figure out the whole arrangement-

If 1 faces north, 8 must be on Q facing south. Thus, he must be standing 2 places ahead of 3.

Question 16: Between inmate from cell 5 and inmate from cell 2 (both excluded), how many inmates are facing North?

a) 1

b) 2

c) 3

d) 4

Solution:

Let’s first number the places as shown in the figure-
We can see that inmate 5 sees 0 people in front of him. So, he must be standing at the extreme and facing the same direction i.e. north when standing in front or south when standing at last.
Similarly, inmate 6 sees 7 people. So, he must stand at P facing south or at W facing north.
Now, there are 2 people who see 6 people in front of them. So, they must be facing opposite direction (i.e. one who stands at Q faces south and one who stands at V faces north) and stand at one place among Q and V.
Also, inmate 2 sees 2 people ahead of him while inmate 7 sees 5 people. So, they must be facing the same direction standing at one position among R and U.
Similarly, 3 and 4 must be standing at one position among S and T facing the same direction.
So, positions of the 8 inmates can be concluded as:

Let us now decide on the directions for each of them:
The person from Q must definitely be facing south. Apart from this, the person facing south will increase by 2. Since the number of inmates facing south was not more than four, the number of person facing south has to be 3. If positions P and W are facing south, all others will have to face north which is not possible as no three consecutive inmates faced the same direction. So, person on P and W must be facing north. So, one of the pair of R-U or S-T must be facing south. But when person on U faces north, three consecutive people i.e. U, V, and W must be facing north which is not possible. Thus, pair S-T must be facing south. In this way, we can figure out the whole arrangement-

We can see that among the positions, Q, R, S, and T, 2 are facing north. Thus, ‘2’ is the correct answer.

Instructions

Five factories Tata Motors, Maruti Suzuki, Asian Paints, HAL and Appolo tyres had 15 members ( A – O) allocated for planning, production and shipping manufacturing units. The below chart gives the employees working in various departments. Each company had one member in each department.

Outer layer: Planning   Middle layer: Production   Inner layer: Shipping
The following details are known about them.
1. A, O, J, M are the only employees who are at managerial level. No two of them work in the same factory.
2. Asian Paints nor Appolo tyres selected employee M
3. Maruti Suzuki selected the employee H.
4. Tata Motors, Maruti Suzuki, Asian Paints selects one among the following pairs (A, H), (O, M), (J, D), (M, C).
5. Apollo tyres, Maruti Suzuki, HAL selected one among the following pairs (B, I), (F, G), (O, I), (N, H).
6.The factory which selected O also selected E.

Question 17: The number of ways in which the employees can be selected is

Solution:

Let us try to represent the data in the form of a table.
From condition 3: Employee H works in the production unit of Maruti Suzuki.
From condition 1, 4: pair (O, M) is not possible as no two employees among work in the same factory nor in the same unit.
From condition 3: pair (A, H) work in Maruti Suzuki. Employee A works in the shipping unit.
From condition 2, 4: pair (J, D) work in Asian paints. Employee J works in shipping unit and D work in the production unit.
From condition 4: pair (M, C) work in Tata Motors. Employee M works in the production unit and C works in the shipping unit.
From condition 4, 5: pair (N, H) work in Maruti Suzuki. Employee N work in the planning unit.
From condition 1, 6: EMployee O should work in one among HAL or Apollo tyres along with E

We have 4 cases with respect to the positions of G,I and O,E

Case 1:

This is not possible because, from condition 5, HAL should contain one pair among (B, I), (F, G), (O, I)

Case 2:

If HAL has (O, I) then Apollo should have (F, G) from condition 5.

Case 3 :

If Appolo has (O, I) then HAL should have (F, G) from condition 5.

Case 4:

This is not possible because, from condition 5, Appolo should contain one pair among (B, I), (F, G), (O, I)

Hence only Cases 2 and 3 possible.

Question 18: If K and C work in the same factory, then they will be working in which of the following factory?

a) HAL

b) Asian Paints

c) Tata Motors

d) Appollo Tyres

Solution:

Let us try to represent the data in the form of a table.
From condition 3: Employee H works in the production unit of Maruti Suzuki.
From condition 1, 4: pair (O, M) is not possible as no two employees among work in the same factory nor in the same unit.
From condition 3: pair (A, H) work in Maruti Suzuki. Employee A works in the shipping unit.
From condition 2, 4: pair (J, D) work in Asian paints. Employee J works in shipping unit and D work in the production unit.
From condition 4: pair (M, C) work in Tata Motors. Employee M works in the production unit and C works in the shipping unit.
From condition 4, 5: pair (N, H) work in Maruti Suzuki. Employee N work in the planning unit.
From condition 1, 6: EMployee O should work in one among HAL or Apollo tyres along with E

We have 4 cases with respect to the positions of G,I and O,E

Case 1:

This is not possible because, from condition 5, HAL should contain one pair among (B, I), (F, G), (O, I)

Case 2:

If HAL has (O, I) then Apollo should have (F, G) from condition 5.

Case 3 :

If Appolo has (O, I) then HAL should have (F, G) from condition 5.

Case 4:

This is not possible because, from condition 5, Appolo should contain one pair among (B, I), (F, G), (O, I)

Hence only Cases 2 and 3 possible.

Question 19: G can work in which of the following factory?

a) Appollo Tyres

b) Tata Motors

c) Maruti Suzuki

d) Asian Paints

Solution:

Let us try to represent the data in the form of a table.
From condition 3: Employee H works in the production unit of Maruti Suzuki.
From condition 1, 4: pair (O, M) is not possible as no two employees among work in the same factory nor in the same unit.
From condition 3: pair (A, H) work in Maruti Suzuki. Employee A works in the shipping unit.
From condition 2, 4: pair (J, D) work in Asian paints. Employee J works in shipping unit and D work in the production unit.
From condition 4: pair (M, C) work in Tata Motors. Employee M works in the production unit and C works in the shipping unit.
From condition 4, 5: pair (N, H) work in Maruti Suzuki. Employee N work in the planning unit.
From condition 1, 6: EMployee O should work in one among HAL or Apollo tyres along with E

We have 4 cases with respect to the positions of G,I and O,E

Case 1:

This is not possible because, from condition 5, HAL should contain one pair among (B, I), (F, G), (O, I)

Case 2:

If HAL has (O, I) then Apollo should have (F, G) from condition 5.

Case 3 :

If Appolo has (O, I) then HAL should have (F, G) from condition 5.

Case 4:

This is not possible because, from condition 5, Appolo should contain one pair among (B, I), (F, G), (O, I)

Hence only Cases 2 and 3 possible.

G can work in HAL or Appolo

Question 20: If B works in Tata Motors, then who works in the planning unit of Asian Paints?

a) O

b) N

c) K

d) F

Solution:

Let us try to represent the data in the form of a table.
From condition 3: Employee H works in the production unit of Maruti Suzuki.
From condition 1, 4: pair (O, M) is not possible as no two employees among work in the same factory nor in the same unit.
From condition 3: pair (A, H) work in Maruti Suzuki. Employee A works in the shipping unit.
From condition 2, 4: pair (J, D) work in Asian paints. Employee J works in shipping unit and D work in the production unit.
From condition 4: pair (M, C) work in Tata Motors. Employee M works in the production unit and C works in the shipping unit.
From condition 4, 5: pair (N, H) work in Maruti Suzuki. Employee N work in the planning unit.
From condition 1, 6: EMployee O should work in one among HAL or Apollo tyres along with E

We have 4 cases with respect to the positions of G,I and O,E

Case 1:

This is not possible because, from condition 5, HAL should contain one pair among (B, I), (F, G), (O, I)

Case 2:

If HAL has (O, I) then Apollo should have (F, G) from condition 5.

Case 3 :

If Appolo has (O, I) then HAL should have (F, G) from condition 5.

Case 4:

This is not possible because, from condition 5, Appolo should contain one pair among (B, I), (F, G), (O, I)

Hence only Cases 2 and 3 possible.

If  B works in Tata motors, then K works in the planning unit of Asian paints.

Instructions

Seven friends, Suresh, Mukesh, Kamlesh, Nayan, Namit, Naman and Amar are ardent football fans. They ranked 3 different teams Manchester City, Arsenal and Manchester United on their likelihood of winning English Premier League(EPL) title. Rank 1 shows the highest chances of winning whereas rank 3 shows the lowest chances of winning the title.

Further, it is known that:
1. The rank assigned by Naman to Manchester United is the same as the rank assigned by Kamlesh to Arsenal.
2. The number of friends who assigned rank 1 to Arsenal was the same the number of friends who assigned rank 2 to Manchester City.
3. The sum of the ranks assigned by all friends to Manchester United is 5 lesser than that of the sum of the assigned by all friends to Manchester City.
4. The sum of the ranks assigned by all friends to Arsenal is 7 lesser than that of the sum of the ranks assigned by all friends to Manchester City.
5. The rank assigned by Namit to Manchester United is the same as that of the rank assigned by Nayan to Manchester City.
6. Suresh assigned a numerically lower rank to Manchester City than that of Naman.
7. None of the friends ranked Manchester United 2nd

Question 21: How many friends assigned rank 1 to Arsenal?

Solution:

Each of the seven friends assigns rank 1, 2 and 3 to the teams. Hence, sum of the ranks assigned by a friend to all three teams = 1+2+3 = 6

We can also say that, the sum of the ranks assigned by all seven friends = 6*7 = 42.

Let us assume that the the sum of the ranks assigned by all friends to Manchester City = x.
Hence, using statement (3), we can say that the sum of the ranks assigned by all friends to Manchester United = x – 5.
Similarly, by statement (4), we can say that the sum of the ranks assigned by all friends to Arsenal = x – 7.
$\Rightarrow$ $x+(x-5)+(x-7)=42$
$\Rightarrow$ $x=18$
Therefore, we can say that the sum of the ranks assigned by all friends to Manchester City, Manchester United and Arsenal are 18, 13 and 11 respectively.

In statement (2), it is given that the number of friends who assigned rank 1 to Arsenal was the same the number of friends who assigned rank 2 to Manchester City.

For Arsenal, at least 3 friends must assign rank 1 and the remaining 4 friends can assign rank 2.
By statement (2), we can say that at least 3 friends assigned rank 2 to Manchester city. In this case remaining 4 friends must have assigned rank 3 to Manchester City so that the sum of the ranks assigned by all the friends to Manchester City is 18.
Hence, we can say that there is just one unique case. Hence, we can figure out the breakup of ranks by individual friends to all three teams.

Arsenal = {1, 1, 1, 2, 2, 2, 2}
Manchester City = {2, 2, 2, 3, 3, 3, 3}
Manchester United = {1, 1, 1, 1, 3, 3, 3}
In statement (1), it is given that the rank assigned by Naman to Manchester United is the same as the rank assigned by Kamlesh to Arsenal. Hence, we can say that both assigned rank 1 to respective teams.
Since, Naman has already ranked Manchester United at rank 1, he must have ranked Arsenal for 2nd spot and Manchester City for 3rd spot.
We can apply the same reasoning for Kamlesh.

In statement (6), it is given that Suresh assigned a numerically lower rank to Manchester City than that of Naman. This means that Suresh must have assigned rank 2 to Manchester City. For Suresh we can figure out the ranks that they have assigned to Arsenal and Manchester United.

In statement (5), it is given that the rank assigned by Namit to Manchester United is the same as that of the rank assigned by Nayan to Manchester City. Hence, we can say that both Namit and Nayan must have ranked 3 to the respective team. Thereafter, we can fill the entire table as we already have three friends (Suresh, Kamlesh and Namit) who have ranked Manchester United for 3rd spot. Hence, we can say that remaining friends {Mukesh, Nayan and Amar} must have ranked Manchester United for 1st spot. With that information we can easily fill the entire table.

From the table, we can see that 3 friends assigned rank 1 to Arsenal.

Question 22: What is the sum of ranks assigned by Nayan, Namit and Naman to Manchester United?

Solution:

Each of the seven friends assigns rank 1, 2 and 3 to the teams. Hence, sum of the ranks assigned by a friend to all three teams = 1+2+3 = 6

We can also say that, the sum of the ranks assigned by all seven friends = 6*7 = 42.

Let us assume that the the sum of the ranks assigned by all friends to Manchester City = x.

Hence, using statement (3), we can say that the sum of the ranks assigned by all friends to Manchester United = x – 5.

Similarly, by statement (4), we can say that the sum of the ranks assigned by all friends to Arsenal = x – 7.

$\Rightarrow$ $x+(x-5)+(x-7)=42$

$\Rightarrow$ $x=18$

Therefore, we can say that the sum of the ranks assigned by all friends to Manchester City, Manchester United and Arsenal are 18, 13 and 11 respectively.

In statement (2), it is given that the number of friends who assigned rank 1 to Arsenal was the same the number of friends who assigned rank 2 to Manchester City.

For Arsenal, at least 3 friends must assign rank 1 and the remaining 4 friends can assign rank 2.
By statement (2), we can say that at least 3 friends assigned rank 2 to Manchester city. In this case remaining 4 friends must have assigned rank 3 to Manchester City so that the sum of the ranks assigned by all the friends to Manchester City is 18.

Hence, we can say that there is just one unique case. Hence, we can figure out the breakup of ranks by individual friends to all three teams.

Arsenal = {1, 1, 1, 2, 2, 2, 2}

Manchester City = {2, 2, 2, 3, 3, 3, 3}

Manchester United = {1, 1, 1, 1, 3, 3, 3}

In statement (1), it is given that the rank assigned by Naman to Manchester United is the same as the rank assigned by Kamlesh to Arsenal. Hence, we can say that both assigned rank 1 to respective teams.
Since, Naman has already ranked Manchester United at rank 1, he must have ranked Arsenal for 2nd spot and Manchester City for 3rd spot.
We can apply the same reasoning for Kamlesh.

In statement (6), it is given that Suresh assigned a numerically lower rank to Manchester City than that of Naman. This means that Suresh must have assigned rank 2 to Manchester City. For Suresh we can figure out the ranks that they have assigned to Arsenal and Manchester United.

In statement (5), it is given that the rank assigned by Namit to Manchester United is the same as that of the rank assigned by Nayan to Manchester City. Hence, we can say that both Namit and Nayan must have ranked 3 to the respective team. Thereafter, we can fill the entire table as we already have three friends (Suresh, Kamlesh and Namit) who have ranked Manchester United for 3rd spot. Hence, we can say that remaining friends {Mukesh, Nayan and Amar} must have ranked Manchester United for 1st spot. With that information we can easily fill the entire table.

From the table, we can see that the sum of ranks assigned by Nayan, Namit and Naman to Manchester United = 1+3+1 = 5.

Question 23: What is the sum of ranks assigned by Suresh, Mukesh and Kamlesh to Manchester City?

Solution:

Each of the seven friends assigns rank 1, 2 and 3 to the teams. Hence, sum of the ranks assigned by a friend to all three teams = 1+2+3 = 6

We can also say that, the sum of the ranks assigned by all seven friends = 6*7 = 42.

Let us assume that the the sum of the ranks assigned by all friends to Manchester City = x.
Hence, using statement (3), we can say that the sum of the ranks assigned by all friends to Manchester United = x – 5.
Similarly, by statement (4), we can say that the sum of the ranks assigned by all friends to Arsenal = x – 7.
$\Rightarrow$ $x+(x-5)+(x-7)=42$
$\Rightarrow$ $x=18$
Therefore, we can say that the sum of the ranks assigned by all friends to Manchester City, Manchester United and Arsenal are 18, 13 and 11 respectively.

In statement (2), it is given that the number of friends who assigned rank 1 to Arsenal was the same the number of friends who assigned rank 2 to Manchester City.

For Arsenal, at least 3 friends must assign rank 1 and the remaining 4 friends can assign rank 2. By statement (2), we can say that at least 3 friends assigned rank 2 to Manchester city. In this case remaining 4 friends must have assigned rank 3 to Manchester City so that the sum of the ranks assigned by all the friends to Manchester City is 18.  Hence, we can say that there is just one unique case. Hence, we can figure out the breakup of ranks by individual friends to all three teams.
Arsenal = {1, 1, 1, 2, 2, 2, 2}
Manchester City = {2, 2, 2, 3, 3, 3, 3}
Manchester United = {1, 1, 1, 1, 3, 3, 3}
In statement (1), it is given that the rank assigned by Naman to Manchester United is the same as the rank assigned by Kamlesh to Arsenal. Hence, we can say that both assigned rank 1 to respective teams.  Since, Naman has already ranked Manchester United at rank 1, he must have ranked Arsenal for 2nd spot and Manchester City for 3rd spot.  We can apply the same reasoning for Kamlesh.

In statement (6), it is given that Suresh assigned a numerically lower rank to Manchester City than that of Naman. This means that Suresh must have assigned rank 2 to Manchester City. For Suresh we can figure out the ranks that they have assigned to Arsenal and Manchester United.

In statement (5), it is given that the rank assigned by Namit to Manchester United is the same as that of the rank assigned by Nayan to Manchester City. Hence, we can say that both Namit and Nayan must have ranked 3 to the respective team. Thereafter, we can fill the entire table as we already have three friends (Suresh, Kamlesh and Namit) who have ranked Manchester United for 3rd spot. Hence, we can say that remaining friends {Mukesh, Nayan and Amar} must have ranked Manchester United for 1st spot. With that information we can easily fill the entire table.

From the table, we can see that the sum of ranks by Suresh, Mukesh and Kamlesh to Manchester City = 2+3+2 = 7.

Question 24: What is the sum of the ranks assigned by Mukesh, Kamlesh, Namit and Amar to Arsenal?

Solution:

Each of the seven friends assigns rank 1, 2 and 3 to the teams. Hence, sum of the ranks assigned by a friend to all three teams = 1+2+3 = 6

We can also say that, the sum of the ranks assigned by all seven friends = 6*7 = 42.

Let us assume that the the sum of the ranks assigned by all friends to Manchester City = x.
Hence, using statement (3), we can say that the sum of the ranks assigned by all friends to Manchester United = x – 5.
Similarly, by statement (4), we can say that the sum of the ranks assigned by all friends to Arsenal = x – 7.
$\Rightarrow$ $x+(x-5)+(x-7)=42$
$\Rightarrow$ $x=18$
Therefore, we can say that the sum of the ranks assigned by all friends to Manchester City, Manchester United and Arsenal are 18, 13 and 11 respectively.

In statement (2), it is given that the number of friends who assigned rank 1 to Arsenal was the same the number of friends who assigned rank 2 to Manchester City.

For Arsenal, at least 3 friends must assign rank 1 and the remaining 4 friends can assign rank 2. By statement (2), we can say that at least 3 friends assigned rank 2 to Manchester city. In this case remaining 4 friends must have assigned rank 3 to Manchester City so that the sum of the ranks assigned by all the friends to Manchester City is 18.
Hence, we can say that there is just one unique case. Hence, we can figure out the breakup of ranks by individual friends to all three teams.
Arsenal = {1, 1, 1, 2, 2, 2, 2}

Manchester City = {2, 2, 2, 3, 3, 3, 3}

Manchester United = {1, 1, 1, 1, 3, 3, 3}

In statement (1), it is given that the rank assigned by Naman to Manchester United is the same as the rank assigned by Kamlesh to Arsenal. Hence, we can say that both assigned rank 1 to respective teams.  Since, Naman has already ranked Manchester United at rank 1, he must have ranked Arsenal for 2nd spot and Manchester City for 3rd spot.  We can apply the same reasoning for Kamlesh.

In statement (6), it is given that Suresh assigned a numerically lower rank to Manchester City than that of Naman. This means that Suresh must have assigned rank 2 to Manchester City. For Suresh we can figure out the ranks that they have assigned to Arsenal and Manchester United.

In statement (5), it is given that the rank assigned by Namit to Manchester United is the same as that of the rank assigned by Nayan to Manchester City. Hence, we can say that both Namit and Nayan must have ranked 3 to the respective team. Thereafter, we can fill the entire table as we already have three friends (Suresh, Kamlesh and Namit) who have ranked Manchester United for 3rd spot. Hence, we can say that remaining friends {Mukesh, Nayan and Amar} must have ranked Manchester United for 1st spot. With that information we can easily fill the entire table.

From the table, we can see that the sum of the ranks assigned by Mukesh, Kamlesh, Namit and Amar to Arsenal = 2+1+1+2 = 6.

Instructions

A public transport bus travels around Mumbai, it starts from P and travel to Q, R, S, T and U where U being the last stop.

The number of passengers enters and exits the bus at each stop is given below:

The time taken by the bus between stops is given in the table below:

Question 25: Which of the following can be the maximum number of passengers who can travel from P to U?

a) 15

b) 10

c) 30

d) 0

Solution:

30 passengers boarded at P, 15 exited at Q.

Thus 15 passengers are left who have boarded at P.

In the bus from Q to R,

20 passengers from Q+ 15 passengers from P.

At R 20 passengers exited. Thus, in order to maximize the passengers who travel from P to U, 20 passengers who boarded at Q left the bus.

In the bus from R to S,

40 passengers from R+ 15 passengers from P.

At S 35 passengers exited. Thus, in order to maximize the passengers who travel from P to U, 35 passengers who boarded at R left the bus.

In the bus from S to T,

30 passengers from S + 5 passengers from R + 15 passengers from P.

At T 40 passengers exited. Thus, in order to maximize the passengers who travel from P to U, 30,5,5 passengers who boarded at S,R and P respectively left the bus.

In the bus from T to U,

30 passengers from T + 10 passengers from P.

Thus, maximum 10 passengers from P can travel till U.

Option B

Question 26: What can be the maximum number of passengers who would have travelled for exactly 60 mins?

a) 20

b) 25

c) 55

d) 35

Solution:

Passengers can travel for 60 minutes for the journeys P-S and R-U.

Maximise the number of passengers travelled from P to S:

30 passengers boarded at P, 15 exited at Q.

Thus 15 passengers are left who have boarded at P.

.

In the bus from Q to R,

20 passengers from Q+ 15 passengers from P.

At R 20 passengers exited. Thus, in order to maximize the passengers who travel from P to S, 20 passengers who boarded at Q left the bus.

.

In the bus from R to S,

40 passengers from R + 15 passengers from P.

At S 35 passengers exited. Thus, in order to maximize the passengers who travel from P to S, 15 passengers who boarded at P left the bus.

.

15 Passengers for the journey P-S.

.

Maximise the number of passengers travelled from R to U:

20 passengers from R have to exit at S, 30 boarded at S.

40 exit at T, thus 10 from R and 30 from S exit the bus.

.

10 passengers travelled from R to U.

The total number of passengers who have travelled for 60 minutes= 15+10=25

Option B

Question 27: If the passengers who have boarded the bus earlier, will exit the bus before the other passengers, then how many passengers have travelled for more than 15 minutes?

a) 105

b) 110

c) 90

d) 120

Solution:

We can make the following table for the number of passengers exited the bus who boarded from the given stops.

P(15) represents that 15 passengers who boarded from P exited at Q.

P(15)+Q(5) represents that 15 passengers who boarded from P exited at R and 5 passengers who boarded from Q exited at R.

The passengers have travelled for more than 15 minutes are travellers from  P-Q, P-R, Q-S, R-T, S-U and T-U

= 15+15+15+20+10+30= 105

Question 28: What can be the maximum number of passengers who would have travelled for exactly 45 mins?

Solution:

The number of passengers who would have travelled for exactly 45 mins

= |P-R|+|Q-T|+|S-U|

In order to maximize the number of passengers between these stops, we can make the following table:

P(15) represents that 15 passengers who boarded from P exited at Q.

P(15)+Q(5) represents that 15 passengers who boarded from P exited at R and 5 passengers who boarded from Q exited at R.

The passengers from Q will not exit at S, thus the passengers from R will exit.

The number of passengers who would have travelled for exactly 45 mins= |P-R|+|Q-T|+|S-U|= 15+15+10=40

Instructions

GST council was formed on 1st March 2015 to address all the complaints and difficulties in implementation of GST. There were a total of 18 members in the council initially working in four departments namely Finance, Technology, Operations, and Strategy & Planning. The numbers of members in these departments are in arithmetic progression with a common difference of 1, not necessarily in the same order. No member was added/ removed in the council till 1st July 2015. Any new member who joined the council after 30th June 2015 was 32 years old at the time of joining. The age of retirement from the council is 65 years. New members are added on 1st July every year. The table given below represents the age of average age of members of the council as on 1st July 2015, 2016, 2017 and 2018 after taken into the account the age of new joinee/ retired members if any.

Further additional information is given that
1. Strategy & Planning is the only department in which no new member joined during the given period.
2. At maximum one member retired/joined any department on 1st July during the given period.

Question 29: How many members were there in the Finance department on 1st Nov 2018?

Solution:

There were a total of 18 members in the council initially working in four departments namely Finance, Technology, Operations, and Strategy & Planning. The numbers of members in these departments are in arithmetic progression with a common difference of 1, not necessarily in the same order.

Hence, we can say that there were 3, 4, 5 and 6 members working in different departments.

It is also given that Strategy & Planning is the only department in which no new member joined during the given period. Hence, we can say that the drop in the average age in 2017 would have come because of retirement of 1 member from Strategy & Planning department.

Let ‘x’ be the number of members in  Strategy & Planning on 1st July 2016. Hence, we can say that there were ‘x – 1’  members left on 1st July 2017. Therefore, we can say that, 45(x – 1) + 65 = 50*x => x = 4.

Therefore, we can say that there were 4 members in Strategy & Planning department on 1st July 2015.

There is a reduction in average age only once in Finance department. Hence, either a member must have left or joined the Finance dept on 1st July 2018. But it is given that Strategy & Planning is the only department in which no new member joined during the given period. Hence, we can say that one member joined Finance department on 1st July 2018.

Let ‘y’ be the number of members in  Finance deparmtent on 1st July 2017. Hence, we can say that there were ‘y + 1’ members on 1st July 2018. Therefore, we can say that, 50(y + 1) – 32 = 53*y => y = 6.
Therefore, we can say that there were 6 members in Finance department on 1st July 2015.

In Technology department, from 1st July 2015 to 1st July 2016 the average age reduced. This might have been due to addition of a new member or retirement of an existing member.

Case 1: When the reduction is due to joining of a new member.

Let ‘z’ be the number of members in Technology dept on 1st July 2015. Then, there will be ‘z + 1’ members in Technology department on 1st July 2016.

Therefore, 48.5(z + 1) – 32 = 54z => z = 3. Hence, z = 3 is a possible solution.

Case 2: When the reduction is due to the retirement of an existing member.

Let ‘z’ be the number of members in Technology dept on 1st July 2015. Then, there will be ‘z – 1’ members in Technology department on 1st July 2016.

Therefore, 48.5(z – 1) + 65 = 54z=> z = 3. Which is possible.

Hence, we can say that there were 3 members in Technology department on 1st July 2015. Consequently, we can say that there were 5 members in Operations department on 1st July 2015.

Therefore, we can draw a table consisting the number of members in each dept on 1st July in each year post joining/retirement of members if any.

Case 1: When the number of members in Technology department is 2 on 1st July 2016. We can see that the average age decreased on 1st July 2017. Therefore, we can say that a member either joined or retired from Technology department on 1st July 2017.

Case 1A: When the reduction is due to joining of a new member.

We know that there were 2 members in Technology dept on 1st July 2016. Then, there must have been be ‘3’ members in Technology department on 1st July 2017.

Therefore, x*(3) – 32 = 49.5(2) => x = 43.66 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Case 1B: When the reduction is due to the retirement of an existing member.

We know that there were 2 members in Technology dept on 1st July 2016. Then, there must have been be ‘1’ members in Technology department on 1st July 2017.

Therefore, x*(1) + 65 = 49.5(2) => x = 34 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Therefore, we can say that there were 4 member in Technology department on 1st July 2016.

Case 2: When there were 4 member in Technology department on 1st July 2016. We can see that the average age decreased on 1st July 2017. Therefore, we can say that a member either joined or retired from Technology department on 1st July 2017.

Case 2A: When the reduction is due to joining of a new member.

We know that there were 4 members in Technology dept on 1st July 2016. Then, there must have been be ‘5’ members in Technology department on 1st July 2017.

Therefore, x*(5) – 32 = 49.5(4) => x = 46 years. Hence, this is a possible solution.

Case 2B: When the reduction is due to the retirement of an existing member.

We know that there were 4 members in Technology dept on 1st July 2016. Then, there must have been be ‘3’ members in Technology department on 1st July 2017.

Therefore, x*(3) + 65 = 49.5(4) => x = 44.33 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Therefore, we can say that in Technology dept, there were 3, 4, 5 and 5 members on 1st July of 2015, 2016, 2017 and 2018 respectively.

In similar manner we can consider for all possibility for Operations department. We get that in Operations department, there were 5, 6, 5 and 5 members on 1st July of 2015, 2016, 2017 and 2018 respectively.

From the table we can see that there are 7 members in the Finance dept after 1st July 2018. The next change in the number of members in Finance dept will take place on 1st July 2019. Hence, we can say that there are 7 members in the Finance dept after 1st Nov 2018.

Question 30: In which of the following year a new member joined the Operations department?

a) 2016

b) 2017

c) 2018

d) More than one of the above

Solution:

There were a total of 18 members in the council initially working in four departments namely Finance, Technology, Operations, and Strategy & Planning. The numbers of members in these departments are in arithmetic progression with a common difference of 1, not necessarily in the same order.

Hence, we can say that there were 3, 4, 5 and 6 members working in different departments.

It is also given that Strategy & Planning is the only department in which no new member joined during the given period. Hence, we can say that the drop in the average age in 2017 would have come because of retirement of 1 member from Strategy & Planning department.

Let ‘x’ be the number of members in  Strategy & Planning on 1st July 2016. Hence, we can say that there were ‘x – 1’  members left on 1st July 2017. Therefore, we can say that, 45(x – 1) + 65 = 50*x => x = 4.

Therefore, we can say that there were 4 members in Strategy & Planning department on 1st July 2015.

There is a reduction in average age only once in Finance department. Hence, either a member must have left or joined the Finance dept on 1st July 2018. But it is given that Strategy & Planning is the only department in which no new member joined during the given period. Hence, we can say that one member joined Finance department on 1st July 2018.

Let ‘y’ be the number of members in  Finance deparmtent on 1st July 2017. Hence, we can say that there were ‘y + 1’ members on 1st July 2018. Therefore, we can say that, 50(y + 1) – 32 = 53*y => y = 6.
Therefore, we can say that there were 6 members in Finance department on 1st July 2015.

In Technology department, from 1st July 2015 to 1st July 2016 the average age reduced. This might have been due to addition of a new member or retirement of an existing member.

Case 1: When the reduction is due to joining of a new member.

Let ‘z’ be the number of members in Technology dept on 1st July 2015. Then, there will be ‘z + 1’ members in Technology department on 1st July 2016.

Therefore, 48.5(z + 1) – 32 = 54z => z = 3. Hence, z = 3 is a possible solution.

Case 2: When the reduction is due to the retirement of an existing member.

Let ‘z’ be the number of members in Technology dept on 1st July 2015. Then, there will be ‘z – 1’ members in Technology department on 1st July 2016.

Therefore, 48.5(z – 1) + 65 = 54z=> z = 3. Which is possible.

Hence, we can say that there were 3 members in Technology department on 1st July 2015. Consequently, we can say that there were 5 members in Operations department on 1st July 2015.

Therefore, we can draw a table consisting the number of members in each dept on 1st July in each year post joining/retirement of members if any.

Case 1: When the number of members in Technology department is 2 on 1st July 2016. We can see that the average age decreased on 1st July 2017. Therefore, we can say that a member either joined or retired from Technology department on 1st July 2017.

Case 1A: When the reduction is due to joining of a new member.

We know that there were 2 members in Technology dept on 1st July 2016. Then, there must have been be ‘3’ members in Technology department on 1st July 2017.

Therefore, x*(3) – 32 = 49.5(2) => x = 43.66 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Case 1B: When the reduction is due to the retirement of an existing member.

We know that there were 2 members in Technology dept on 1st July 2016. Then, there must have been be ‘1’ members in Technology department on 1st July 2017.

Therefore, x*(1) + 65 = 49.5(2) => x = 34 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Therefore, we can say that there were 4 member in Technology department on 1st July 2016.

Case 2: When there were 4 member in Technology department on 1st July 2016. We can see that the average age decreased on 1st July 2017. Therefore, we can say that a member either joined or retired from Technology department on 1st July 2017.

Case 2A: When the reduction is due to joining of a new member.

We know that there were 4 members in Technology dept on 1st July 2016. Then, there must have been be ‘5’ members in Technology department on 1st July 2017.

Therefore, x*(5) – 32 = 49.5(4) => x = 46 years. Hence, this is a possible solution.

Case 2B: When the reduction is due to the retirement of an existing member.

We know that there were 4 members in Technology dept on 1st July 2016. Then, there must have been be ‘3’ members in Technology department on 1st July 2017.

Therefore, x*(3) + 65 = 49.5(4) => x = 44.33 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Therefore, we can say that in Technology dept, there were 3, 4, 5 and 5 members on 1st July of 2015, 2016, 2017 and 2018 respectively.

In similar manner we can consider for all possibility for Operations department. We get that in Operations department, there were 5, 6, 5 and 5 members on 1st July of 2015, 2016, 2017 and 2018 respectively.

From the table, we can see that a new member joined the Operations department in year 2016. Hence, option A is the correct answer.

Question 31: How many members were there in the Strategy & Planning department on 1st Aug 2017?

Solution:

There were a total of 18 members in the council initially working in four departments namely Finance, Technology, Operations, and Strategy & Planning. The numbers of members in these departments are in arithmetic progression with a common difference of 1, not necessarily in the same order.

Hence, we can say that there were 3, 4, 5 and 6 members working in different departments.

It is also given that Strategy & Planning is the only department in which no new member joined during the given period. Hence, we can say that the drop in the average age in 2017 would have come because of retirement of 1 member from Strategy & Planning department.

Let ‘x’ be the number of members in  Strategy & Planning on 1st July 2016. Hence, we can say that there were ‘x – 1’  members left on 1st July 2017. Therefore, we can say that, 45(x – 1) + 65 = 50*x => x = 4.

Therefore, we can say that there were 4 members in Strategy & Planning department on 1st July 2015.

There is a reduction in average age only once in Finance department. Hence, either a member must have left or joined the Finance dept on 1st July 2018. But it is given that Strategy & Planning is the only department in which no new member joined during the given period. Hence, we can say that one member joined Finance department on 1st July 2018.

Let ‘y’ be the number of members in  Finance deparmtent on 1st July 2017. Hence, we can say that there were ‘y + 1’ members on 1st July 2018. Therefore, we can say that, 50(y + 1) – 32 = 53*y => y = 6.  Therefore, we can say that there were 6 members in Finance department on 1st July 2015.

In Technology department, from 1st July 2015 to 1st July 2016 the average age reduced. This might have been due to addition of a new member or retirement of an existing member.

Case 1: When the reduction is due to joining of a new member.

Let ‘z’ be the number of members in Technology dept on 1st July 2015. Then, there will be ‘z + 1’ members in Technology department on 1st July 2016.

Therefore, 48.5(z + 1) – 32 = 54z => z = 3. Hence, z = 3 is a possible solution.

Case 2: When the reduction is due to the retirement of an existing member.

Let ‘z’ be the number of members in Technology dept on 1st July 2015. Then, there will be ‘z – 1’ members in Technology department on 1st July 2016.

Therefore, 48.5(z – 1) + 65 = 54z=> z = 3. Which is possible.

Hence, we can say that there were 3 members in Technology department on 1st July 2015. Consequently, we can say that there were 5 members in Operations department on 1st July 2015.

Therefore, we can draw a table consisting the number of members in each dept on 1st July in each year post joining/retirement of members if any.

Case 1: When the number of members in Technology department is 2 on 1st July 2016. We can see that the average age decreased on 1st July 2017. Therefore, we can say that a member either joined or retired from Technology department on 1st July 2017.

Case 1A: When the reduction is due to joining of a new member.

We know that there were 2 members in Technology dept on 1st July 2016. Then, there must have been be ‘3’ members in Technology department on 1st July 2017.

Therefore, x*(3) – 32 = 49.5(2) => x = 43.66 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Case 1B: When the reduction is due to the retirement of an existing member.

We know that there were 2 members in Technology dept on 1st July 2016. Then, there must have been be ‘1’ members in Technology department on 1st July 2017.

Therefore, x*(1) + 65 = 49.5(2) => x = 34 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Therefore, we can say that there were 4 member in Technology department on 1st July 2016.

Case 2: When there were 4 member in Technology department on 1st July 2016. We can see that the average age decreased on 1st July 2017. Therefore, we can say that a member either joined or retired from Technology department on 1st July 2017.

Case 2A: When the reduction is due to joining of a new member.

We know that there were 4 members in Technology dept on 1st July 2016. Then, there must have been be ‘5’ members in Technology department on 1st July 2017.

Therefore, x*(5) – 32 = 49.5(4) => x = 46 years. Hence, this is a possible solution.

Case 2B: When the reduction is due to the retirement of an existing member.

We know that there were 4 members in Technology dept on 1st July 2016. Then, there must have been be ‘3’ members in Technology department on 1st July 2017.

Therefore, x*(3) + 65 = 49.5(4) => x = 44.33 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Therefore, we can say that in Technology dept, there were 3, 4, 5 and 5 members on 1st July of 2015, 2016, 2017 and 2018 respectively.

In similar manner we can consider for all possibility for Operations department. We get that in Operations department, there were 5, 6, 5 and 5 members on 1st July of 2015, 2016, 2017 and 2018 respectively.

From the table we can see that there are 3 members in Strategy & Planning department on 1st July 2017 and it remained same till 1st July 2018. Hence, we can say that there were 3 members in Strategy & Planning department on 1st Aug 2017.

Question 32: What is the difference in the number of members in the GST council on 1st July 2016 to that of 1st July 2018?

Solution:

There were a total of 18 members in the council initially working in four departments namely Finance, Technology, Operations, and Strategy & Planning. The numbers of members in these departments are in arithmetic progression with a common difference of 1, not necessarily in the same order.

Hence, we can say that there were 3, 4, 5 and 6 members working in different departments.

It is also given that Strategy & Planning is the only department in which no new member joined during the given period. Hence, we can say that the drop in the average age in 2017 would have come because of retirement of 1 member from Strategy & Planning department.

Let ‘x’ be the number of members in  Strategy & Planning on 1st July 2016. Hence, we can say that there were ‘x – 1’  members left on 1st July 2017. Therefore, we can say that, 45(x – 1) + 65 = 50*x => x = 4.

Therefore, we can say that there were 4 members in Strategy & Planning department on 1st July 2015.

There is a reduction in average age only once in Finance department. Hence, either a member must have left or joined the Finance dept on 1st July 2018. But it is given that Strategy & Planning is the only department in which no new member joined during the given period. Hence, we can say that one member joined Finance department on 1st July 2018.

Let ‘y’ be the number of members in  Finance deparmtent on 1st July 2017. Hence, we can say that there were ‘y + 1’ members on 1st July 2018. Therefore, we can say that, 50(y + 1) – 32 = 53*y => y = 6.  Therefore, we can say that there were 6 members in Finance department on 1st July 2015.

In Technology department, from 1st July 2015 to 1st July 2016 the average age reduced. This might have been due to addition of a new member or retirement of an existing member.

Case 1: When the reduction is due to joining of a new member.

Let ‘z’ be the number of members in Technology dept on 1st July 2015. Then, there will be ‘z + 1’ members in Technology department on 1st July 2016.

Therefore, 48.5(z + 1) – 32 = 54z => z = 3. Hence, z = 3 is a possible solution.

Case 2: When the reduction is due to the retirement of an existing member.

Let ‘z’ be the number of members in Technology dept on 1st July 2015. Then, there will be ‘z – 1’ members in Technology department on 1st July 2016.

Therefore, 48.5(z – 1) + 65 = 54z=> z = 3. Which is possible.

Hence, we can say that there were 3 members in Technology department on 1st July 2015. Consequently, we can say that there were 5 members in Operations department on 1st July 2015.

Therefore, we can draw a table consisting the number of members in each dept on 1st July in each year post joining/retirement of members if any.

Case 1: When the number of members in Technology department is 2 on 1st July 2016. We can see that the average age decreased on 1st July 2017. Therefore, we can say that a member either joined or retired from Technology department on 1st July 2017.

Case 1A: When the reduction is due to joining of a new member.

We know that there were 2 members in Technology dept on 1st July 2016. Then, there must have been be ‘3’ members in Technology department on 1st July 2017.

Therefore, x*(3) – 32 = 49.5(2) => x = 43.66 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Case 1B: When the reduction is due to the retirement of an existing member.

We know that there were 2 members in Technology dept on 1st July 2016. Then, there must have been be ‘1’ members in Technology department on 1st July 2017.

Therefore, x*(1) + 65 = 49.5(2) => x = 34 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Therefore, we can say that there were 4 member in Technology department on 1st July 2016.

Case 2: When there were 4 member in Technology department on 1st July 2016. We can see that the average age decreased on 1st July 2017. Therefore, we can say that a member either joined or retired from Technology department on 1st July 2017.

Case 2A: When the reduction is due to joining of a new member.

We know that there were 4 members in Technology dept on 1st July 2016. Then, there must have been be ‘5’ members in Technology department on 1st July 2017.

Therefore, x*(5) – 32 = 49.5(4) => x = 46 years. Hence, this is a possible solution.

Case 2B: When the reduction is due to the retirement of an existing member.

We know that there were 4 members in Technology dept on 1st July 2016. Then, there must have been be ‘3’ members in Technology department on 1st July 2017.

Therefore, x*(3) + 65 = 49.5(4) => x = 44.33 years. But this is inconsistent with the average age given on 1st July 2017. Hence, this case is not possible.

Therefore, we can say that in Technology dept, there were 3, 4, 5 and 5 members on 1st July of 2015, 2016, 2017 and 2018 respectively.

In similar manner we can consider for all possibility for Operations department. We get that in Operations department, there were 5, 6, 5 and 5 members on 1st July of 2015, 2016, 2017 and 2018 respectively.

The total number of members in the GST council on 1st July 2016 = 6 + 4 + 6 + 4 = 20.

The total number of members in the GST council on 1st July 2018 = 7 + 5 + 5 + 3 = 20.

Therefore, the required difference = 20 – 20 = 0.

Instructions

Amar bought a rectangular plot and divided it into 3 pieces. He decides to plant 3 different crops rice, wheat, and maize in these 3 pieces of land (He may choose one among the three, 2 among the three or all 3 crops). The yield per square metre and the cost per square metre are given below.

Amar, then sells them at his store at the following prices.

Question 33: Instead of cultivating all 3 crops at once, Amar decides to practice crop rotation and cultivates 1 crop at a time in the entire field. If Amar cultivates rice during summer, wheat during winter and maize during spring, what will be the ratio of profits realized during summer, winter, and spring.

a) 69:72:73

b) 67:69:71

c) 68:69:73

d) 68:69:72

Solution:

Let us assume that Amar cultivates on an area of 100 sq.m for ease of calculation.
On calculating the yield, costs, net revenue and profit, we get the following table.

As we can see from the table, the ratio of profits will be 68:69:72.

Question 34: Which crop has the lowest profit per kg sold?

a) Rice

b) Wheat

c) Maize

d) More than one of the above

Solution:

Let us assume that Amar cultivates on an area of 100 sq.m
On calculating, we get the following table.
Profit/ Kg = Total profit/Total Yield

As we can see from the table, wheat has the lowest profit/kg sold.

Question 35: If the size of Amar’s plot is 600 sq.m and each crop is to be cultivated in at least an area of 100 sq.m and no crop is to be cultivated on an area greater than 300 sq.m, what is the maximum profit that can be realised by Amar.

a) 42,200

b) 42,100

c) 41,400

d) 42,300

Solution:

In the previous questions, we have estimated the profits given by various crops when grown on an equal area.

The table above gives the profits for every 100 square metre of land utilised.

For the profit to be maximised, the crop that gives the maximum profit must be grown in the maximum area and the crop which gives the minimum profit must be grown in the least area.

Since any crop cannot be grown on an area exceeding 300 sq.m, Maize must be grown in 300 sq.m.

The crop that yields the least profit, rice, must be grown on 100 sq.m The remaining crop, wheat, must be grown on the remaining area of 200 sq.m.
Maximum profit that can be realised = 6800*1 + 6900*2 + 7200*3 = Rs. 42,200.

Question 36: Amar decides to cultivate a crop and hoard it so that he can obtain maximum profit. If the prices of Rice, wheat and maize become Rs. 10, Rs. 8 and Rs. 12 the following year, and the cost of storing a Kg of rice,wheat,and maize are Rs. 2, Rs. 3 and Rs. 6, which crop must Amar choose to cultivate.

a) Maize

b) Wheat

c) Rice

d) More than one of the above.

Solution:

Reconstructing the table for the new price and costs, we get,

As we can see, Amar must cultivate Rice to obtain the maximum profit.

Instructions

Answer the following questions based on the information given below:

In a sports event, six teams (A, B, C, D, E and F) are competing against each other Matches are scheduled in two stages. Each team plays three matches in Stage – I and two matches in Stage – II. No team plays against the same team more than once in the event. No ties are permitted in any of the matches. The observations after the completion of Stage – I and Stage – II are as given below.

Stage-I:

• One team won all the three matches.

• Two teams lost all the matches.

• D lost to A but won against C and F.

• E lost to B but won against C and F.

• B lost at least one match.

• F did not play against the top team of Stage-I.

Stage-II:

• The leader of Stage-I lost the next two matches

• Of the two teams at the bottom after Stage-I, one team won both matches, while the other lost both matches.

• One more team lost both matches in Stage-II.

Question 37: The two teams that defeated the leader of Stage-I are:

a) B & F

b) E & F

c) B & D

d) E & D

e) F & D

Solution:

There are a total of $^6 C_2$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches – stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the two teams that won against stage 1 leader A are E and F.

Question 38: The only team(s) that won both matches in Stage-II is (are):

a) B

b) E & F

c) A, E & F

d) B, E & F

e) B & F

Solution:

There are a total of $^6 C_2$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches – stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the teams that won both of their stage 2 matches are B, E and F.

Question 39: The teams that won exactly two matches in the event are:

a) A, D & F

b) D & E

c) E & F

d) D, E & F

e) D & F

Solution:

There are a total of $^6 C_2$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches – stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the wins by each team are A (3), B(4), C(0), D(2), E(4), F(2). Hence, D and F won exactly 2 matches.

Question 40: The team(s) with the most wins in the event is (are):

a) A

b) A & C

c) F

d) E

e) B & E

Solution:

There are a total of $^6 C_2$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches – stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the wins by each team are A(3), B(4), C(0), D(2), E(4), F(2). Hence, most wins are by B and E.

Instructions

Five teams – A, B, C, D and E – are participating in a sports competition. Four rounds are conducted to decide the winner such that the least scoring team will be eliminated from the competition every round. Each round has 22 different games and winning a game gives a team 5 points. The team that has the maximum number of points in the final round is the winner of the competition. The same 22 games were conducted in every round. If a team wins in a certain game, it keeps winning that game until the team gets eliminated. Only one team scored the least number of points in any round. The only team that scored the same number of points in two or more than two rounds is team A. In exactly 2 rounds, the scores of each of the teams were distinct. In exactly 2 rounds, the scores of exactly two teams were equal while those of remaining teams are distinct. The cumulative scores received by the teams A, B, C, D and E over the four rounds are 40, 75, 15, 155 and 155 respectively.

Question 41: How many points did team D score in round 3?

a) 30

b) 35

c) 40

d) Cannot be determined

Solution:

The cummulative scores received by the teams A, B, C, D and E over the four rounds are 40, 75, 15, 155 and 155 respectively.
=> Cummulative number of games won by A, B, C, D and E are 8, 15, 3, 31 and 31 respectively.
Let the number of games won by A, B, C, D and E in the first round be a, b, c, d and e respectively.
In the first round, team C gets eliminated. => c = 3
So in the subsequent rounds, these 3 games must be won by other teams
The only team that scored the same number of points in two or more than two rounds is team A
=> B, D and E are the teams that won one game each of the three.
Hence, the wins in the second round for A, B, D and E are a, b+1, d+1 and e+1 respectively.
A gets eliminated now in the second round => Total number of games won by A = 8 => 2a = 8 => a = 4
b+d+e = 22 – 4 – 3 = 15
Each of them must be at least 4.
=> (4,5,6) in any order is the only possible combination.
A gets eliminated in the second round. So, the games won by A must be won by B, D or E in round 3.
Consider B. Cummulative wins for B is 15. If b is equal to 5, then the minimum number of cummulative wins is 5 + 6 + 7 = 18, which is not possible.
If b = 4, then the minimum number of cummulative wins is 4 + 5 + 6 = 15, which satisfies the give information => b = 4
We know that d + e = 11
=> d and e are 5 and 6 in any order.
The number of games won by D and E in round 3 are d+2 and e+3 or d+3 and e+2.
They must be equal because “in exactly 2 rounds, the scores of exactly two teams were equal while those of remaining teams are distinct”.
Hence, both will be equal to 8 => D scored 8*5 = 40 points in round 3.

Question 42: If team D scored 25 points in the first round, then who was the winner of the competition?

a) B

b) D

c) E

d) Cannot be determined

Solution:

The cummulative scores received by the teams A, B, C, D and E over the four rounds are 40, 75, 15, 155 and 155 respectively.
=> Cummulative number of games won by A, B, C, D and E are 8, 15, 3, 31 and 31 respectively.
Let the number of games won by A, B, C, D and E be a, b, c, d and e respectively.
In the first round, team C gets eliminated. So, the 3 games must be won by other teams => c = 3
The only team that scored the same number of points in two or more than two teams in team A => B, D and E are the teams that won one game each of the three.
Hence, the wins in the second round for A, B, D and E are a, b+1, d+1 and e+1 respectively.
A gets eliminated now => Total number of games won by A = 8 => 2a = 8 => a = 4
b+d+e = 22 – 4 – 3 = 15
Each of them must be at least 4.
=> (4,5,6) in any order is the only possible combination.
A gets eliminated in the second round. So, the games won by A must be won by B, D or E in round 3.
Consider B. Cummulative wins for B is 15. If b is equal to 5, then the minimum number of cummulative wins is 5 + 6 + 7 = 18, which is not possible.
If b = 4, then the minimum number of cummulative wins is 4 + 5 + 6 = 15, which satisfies the give information => b = 4
We know that d + e = 11
D scored 25 points => D won 5 games in round 1 => d = 5 and e = 6
So, the cummulative wins of team D until round 3 = 5 + 6 + 8 = 19 => In the final round, D has 12 wins.
=> Team D is the winner of the competition.

Question 43: How many games were won by more than one teams over the four rounds?

a) 11

b) 12

c) 13

d) 14

Solution:

The cummulative scores received by the teams A, B, C, D and E over the four rounds are 40, 75, 15, 155 and 155 respectively.
=> Cummulative number of games won by A, B, C, D and E are 8, 15, 3, 31 and 31 respectively.
Let the number of games won by A, B, C, D and E be a, b, c, d and e respectively.
In the first round, team C gets eliminated. So, the 3 games must be won by other teams => c = 3
The only team that scored the same number of points in two or more than two teams in team A => B, D and E are the teams that won one game each of the three.
Hence, the wins in the second round for A, B, D and E are a, b+1, d+1 and e+1 respectively.
A gets eliminated now => Total number of games won by A = 8 => 2a = 8 => a = 4
b+d+e = 22 – 4 – 3 = 15
Each of them must be at least 4.
=> (4,5,6) in any order is the only possible combination.
A gets eliminated in the second round. So, the games won by A must be won by B, D or E in round 3.
Consider B. Cummulative wins for B is 15. If b is equal to 5, then the minimum number of cummulative wins is 5 + 6 + 7 = 18, which is not possible.
If b = 4, then the minimum number of cummulative wins is 4 + 5 + 6 = 15, which satisfies the give information => b = 4
We can see that all those games that were won by A, B and C in the first round were the only games that were won by different teams.
Hence, 3 + 4 + 4 = 11 games are won by different teams.

Question 44: How many points did team B score in round 3?

a) 25

b) 35

c) 20

d) 30

Solution:

The cummulative scores received by the teams A, B, C, D and E over the four rounds are 40, 75, 15, 155 and 155 respectively.
=> Cummulative number of games won by A, B, C, D and E are 8, 15, 3, 31 and 31 respectively.
Let the number of games won by A, B, C, D and E be a, b, c, d and e respectively.
In the first round, team C gets eliminated. So, the 3 games must be won by other teams => c = 3
The only team that scored the same number of points in two or more than two teams in team A => B, D and E are the teams that won one game each of the three.
Hence, the wins in the second round for A, B, D and E are a, b+1, d+1 and e+1 respectively.
A gets eliminated now => Total number of games won by A = 8 => 2a = 8 => a = 4
b+d+e = 22 – 4 – 3 = 15
Each of them must be at least 4.
=> (4,5,6) in any order is the only possible combination.
A gets eliminated in the second round. So, the games won by A must be won by B, D or E in round 3.
Consider B. Cummulative wins for B is 15. If b is equal to 5, then the minimum number of cummulative wins is 5 + 6 + 7 = 18, which is not possible.
If b = 4, then the minimum number of cummulative wins is 4 + 5 + 6 = 15, which satisfies the give information => b = 4
In round 3, the number of wins for team B is b+2 = 4 + 2 = 6
Number of points = 6 * 5 = 30

Instructions

Akshita and Radhika play a game of rolling a die. Each of them has to pay Rs 5 in each round to play the game. Depending on the number that turns up on a throw, they may lose, win or a breakeven. For a win, they get Rs 10, for a loss they have to pay Rs 10. The term breakeven means: The person neither wins nor loses.
The following table shows whether a person wins or loses based on the number that turns up on the die. For example, if 1 turns up on the die, Radhika wins and Akshita loses.

It is known that if someone doesn’t have money to pay the required amount, they will not be allowed to continue playing.

Question 45: If both Radhika and Akshita started with the same amount and played 3 rounds. What is the maximum difference between the amounts with them if the number which turned up in each of the rounds is different?

a) 10

b) 0

c) 30

d) 20

Solution:

Radhika will have Rs 20 more than Akshita when the die turns up is 1.

Akshita will have Rs 10 more than Radhika when the die turns up 3 or 5.

If the die turns up 2, 4 or 6 then they will neither gain or lose any amount.

Case 1: Radhika gets Rs 20 more than Akshita.
It will happen when 1 turns up, then in the next two rounds, two numbers out of 2,4,6 turn up.

Case 2: Akshita gets Rs 20 more than Radhika.
In the first two rounds, 3 and 5 turn up and in the third round 2/4/6 turns up.

Question 46: Akshita and Radhika started the game with Rs 10 each and both of them played all the rounds. What is the minimum number of rounds when Radhika would have Rs 40 more than Akshita?

a) 4

b) 7

c) 5

d) 6

Solution:

Both of them started the game with Rs 10 and they had to pay Rs 5 to play the game.

Here we have to minimize the numbers on the die and the maximise the cases of winning for Radhika.

Round 1: Both of them will pay Rs 5 to play the round.

The outcome of the die is 1, Radhika would win Rs 10 and Akshita has to pay Rs 10, which she does not have.

So the outcome should be winning position for both of them i.e 2

Now each of them would have Rs 15.

Round 2: The outcome should be again a winning position for both of them
They would have Rs 20 each

Round 3: It could be a winning position for Radhika and losing position for Akshita ie the outcome of the die has to be 1.
Radhika would have Rs 25 and Akshita would have Rs 5

Round 4: It should be a winning position for both of them, otherwise Akshita has to pay the penalty of Rs 10 in case of lose which she does not has.
Radhika would have Rs 30 and Akshita would have Rs 10.

Round 5: It should be a winning position for both of them
Radhika would have Rs 35 and Akshita would have Rs 15.

Round 6: It could be winning position for Radhika and losing position for Akshita ie the outcome of the die has to be 1.
Radhika would have Rs 40 and Akshita would have Rs 0.

After 6 rounds Radhika would have Rs 40 more than Akshita.

Alternate Solution:

If a person wins, the amount with her changes by +10-5 = +5 (10 is the amount won and 5 is the fee charged to play one round)
If a person loses, the amount with her changes by -10-5 = -15 (10 is the amount lost and 5 is the fee charged to play one round)
So a loser should have a minimum of Rs 20 with her if she wants to play further rounds.
From the table only when the dice shows 1, Radhika will gain Rs 20 over Akshita.

1st round: Radhika has 10 and Akshita has 10. Both will win only one possibility(Dice shows 2). If someone loses, she will not be able to play in the next round. Radhika = 10+5 = 15 and Akshita = 10+5 = 15
2nd round: Both will win(Dice shows 2). Radhika = 15+5 = 20 and Akshita = 15+5 = 20
Now Akshita has Rs 20. Hence she can lose and still play in further rounds.
3rd round: Radhika wins and Akshita loses. Radhika = 20+5 = 25, Akshita = 20-15=5  (A difference of 20 is created)
4th round: Both will win. Radhika = 25+5=30 and Akshita = 5+5=10
5th round: Both will win. Radhika = 30+5 = 35 and Akshita = 10+5 =15
6th round: Radhika wins and Akshita loses. Radhika = 35+5=40 and Akshita = 15-15 = 0  (Difference of 40 is created)
Now the money with Akshita can be 0 as she does not have to play more rounds.

After 6 rounds Radhika would have Rs 40 more than Akshita.

Question 47: Radhika and Akshita started the game with Rs 10 each and played for 4 rounds. If the sum of the amounts with them at the end of four rounds was maximum, what is the sum of the numbers on the die is  (If the number on each turn was different)

Solution:

When the die turns up 1, Radhika would win Rs 10 and Akshita would lose Rs. 10, so the net amount is 0

When the die turns up 2, the net amount is Rs 20.

When the die turns up 3, the net amount is Rs -10.

When the die turns up 4, the net amount is Rs -20.

When the die turns up 5, the net amount is Rs 10.

When the die turns up 6, the net amount is Rs 0.

Since they had a maximum amount at the end of four rounds,
The number on the die is 2+6+5+1 = 14

Question 48: Radhika and Akshita started with Rs 10 each. After playing for four rounds, the amount with Radhika was thrice the amount with Akshita. What is the sum of the numbers on the die in the four rounds?
Assume that the amount with both of them at the end of four rounds was greater than 0.
Enter -1 if the answer can’t be determined.

Solution:

Radhika and Akshita started with Rs. 10.

Case 1:

Radhika will have Rs 15 and Akshita with Rs 5.

Radhika starts with Rs 10. The cost of 4 rounds is Rs 5*4 = Rs 20.
If Radhika ends with Rs 15, then she has to make a total of Rs 15-(10-20) =  Rs 25 from 4 throws which is not possible because a throw will give a multiple of 10 ie -10, 10, 0.
This case can be negated.

Case 2:

If Radhika should have Rs 30, then the number turning up in each round should be a winning position.

It can be either 2 or 1.

It cannot be 2 across all the rounds, cause in that case even Akshita must have Rs 30.

Round 1: They have to pay Rs 5 to play the game.

Both of them won the game, the will have Rs 15 at the end of Round 1.

Round 2: They have to pay Rs 5 to play the game. They will have Rs 10

Both of them won the game, the will have Rs 20 at the end of Round 2.

Round 3: They have to pay Rs 5 to play the game. They will have Rs. 15.

Both of them won the game, the will have Rs 25 at the end of Round 3.

Round 4: They have to pay Rs 5 to play the game. They will have Rs 20.

Radhika will win and Akshita will lose.

So Radhika will have Rs 30 and Akshita will have Rs. 10.

The only possible case is 2, 2, 2, 1

Hence the sum of the numbers = 7

Instructions

Ajay, Bharath, Charan had a total of 310 distinct coins with them in the ratio of 2:3:5 respectively. They redistributed coins among themselves such that after redistribution they had coins in the ratio of $\frac{1}{2}:\frac{1}{3}:\frac{1}{5}$ respectively. After redistribution, Ajay did not have any of the coins he had initially. The same applies for Bharath and Charan as well.

Question 49: The minimum number of coins Ajay can give to Bharath is

Solution:

The total number of coins with Ajay, Bharath and Charan = 310.
The ratio of the coins with them before distribution = 2: 3: 5
The number of coins with Ajay = $\frac{2}{2+3+5}$*310 = 62
The number of coins with Bharath = $\frac{3}{2+3+5}$*310 = 93
The number of coins with Charan = $\frac{5}{2+3+5}$*310 = 155
They redistributed the coins among themselves, so Ajay will distribute his coins to Bharath and Charan. The same is true for Bharath and Charan as well.
The ratio of the coins with Ajay, Bharat, Charan after distribution = $\frac{1}{2}:\frac{1}{3}:\frac{1}{5}$.
=15: 10: 6
The number of coins with Ajay = $\frac{15}{15+10+6}$*310 = 150
The number of coins with Bharath = $\frac{10}{15+10+6}$*310 = 100
The number of coins with Charan = $\frac{6}{15+10+6}$*310 = 60

Let us consider Ajay gives ‘p’ coins to Bharat, so Ajay should give ’62-p’ coins to Charan since the total number of coins with Ajay is 62.
Bharath had 100 coins after distribution. He received ‘p’ coins from Ajay. So he should have received ‘100- p’ coins from Charan.
Charan had 60 coins after distribution. He received ’62-p’ from Ajay. So he should have received ‘p-2’ coins from Bharath.

The minimum value of p is 2 because Bharath gave p-2 coins to Charan.
The maximum value of p is 62, because Ajay gave 62-p coins to Charan.

Question 50: The maximum number of coins which Bharath can give to Charan is

Solution:

The total number of coins with Ajay, Bharath and Charan = 310.
The ratio of the coins with them before distribution = 2: 3: 5
The number of coins with Ajay = $\frac{2}{2+3+5}$*310 = 62
The number of coins with Bharath = $\frac{3}{2+3+5}$*310 = 93
The number of coins with Charan = $\frac{5}{2+3+5}$*310 = 155
They redistributed the coins among themselves, so Ajay will distribute his coins to Bharath and Charan. The same is true for Bharath and Charan as well.
The ratio of the coins with Ajay, Bharat, Charan after distribution = $\frac{1}{2}:\frac{1}{3}:\frac{1}{5}$.
=15: 10: 6
The number of coins with Ajay = $\frac{15}{15+10+6}$*310 = 150
The number of coins with Bharath = $\frac{10}{15+10+6}$*310 = 100
The number of coins with Charan = $\frac{6}{15+10+6}$*310 = 60

Let us consider Ajay gives ‘p’ coins to Bharat, so Ajay should give ’62-p’ coins to Charan since the total number of coins with Ajay is 62.
Bharath had 100 coins after distribution. He received ‘p’ coins from Ajay. So he should have received ‘100- p’ coins from Charan.
Charan had 60 coins after distribution. He received ’62-p’ from Ajay. So he should have received ‘p-2’ coins from Bharath.

From Ajay, we can say that 0 $\leq$ p $\leq$ 62

From Bharath, we can say that 2 $\leq$ p $\leq$ 93

From Charan, we can say that 0 $\leq$ p $\leq$ 100

So the range of p is 2 $\leq$ p $\leq$ 62

The number of coins given by Bharath to Charan will be maximum when p-2 is maximum.

p-2 will be maximum when the value of p is maximum.

The maximum value of p = 62.

Hence Bharath can give a maximum of p-2 i.e 60 coins to Charan.

Question 51: The difference between the coins which Bharath had before and after the distribution is

Solution:

The total number of coins with Ajay, Bharath and Charan = 310.
The ratio of the coins with them before distribution = 2: 3: 5
The number of coins with Ajay = $\frac{2}{2+3+5}$*310 = 62
The number of coins with Bharath = $\frac{3}{2+3+5}$*310 = 93
The number of coins with Charan = $\frac{5}{2+3+5}$*310 = 155
They redistributed the coins among themselves, so Ajay will distribute his coins to Bharath and Charan. The same is true for Bharath and Charan as well.
The ratio of the coins with Ajay, Bharat, Charan after distribution = $\frac{1}{2}:\frac{1}{3}:\frac{1}{5}$.
=15: 10: 6
The number of coins with Ajay = $\frac{15}{15+10+6}$*310 = 150
The number of coins with Bharath = $\frac{10}{15+10+6}$*310 = 100
The number of coins with Charan = $\frac{6}{15+10+6}$*310 = 60

Let us consider Ajay gives ‘p’ coins to Bharat, so Ajay should give ’62-p’ coins to Charan since the total number of coins with Ajay is 62.
Bharath had 100 coins after distribution. He received ‘p’ coins from Ajay. So he should have received ‘100- p’ coins from Charan.
Charan had 60 coins after distribution. He received ’62-p’ from Ajay. So he should have received ‘p-2’ coins from Bharath.

The minimum value of p is 2 because Bharath gave p-2 coins to Charan.
The maximum value of p is 62, because Ajay gave 62-p coins to Charan.

Bharath had 93 coins before distribution and 100 coins after distribution.

The difference = 7

Question 52: Which of the following statements is definitely false?
i: Charan gave 37 coins to Bharath.
ii: Bharath gave 35 coins to Ajay.
iii: Bharath gave 61 coins to Charan.

a) Only ii

b) Only ii and i

c) Only i and iii

d) Only iii

Solution:

The total number of coins with Ajay, Bharath and Charan = 310.
The ratio of the coins with them before distribution = 2: 3: 5
The number of coins with Ajay = $\frac{2}{2+3+5}$*310 = 62
The number of coins with Bharath = $\frac{3}{2+3+5}$*310 = 93
The number of coins with Charan = $\frac{5}{2+3+5}$*310 = 155
They redistributed the coins among themselves, so Ajay will distribute his coins to Bharath and Charan. The same is true for Bharath and Charan as well.
The ratio of the coins with Ajay, Bharat, Charan after distribution = $\frac{1}{2}:\frac{1}{3}:\frac{1}{5}$.
=15: 10: 6
The number of coins with Ajay = $\frac{15}{15+10+6}$*310 = 150
The number of coins with Bharath = $\frac{10}{15+10+6}$*310 = 100
The number of coins with Charan = $\frac{6}{15+10+6}$*310 = 60

Let us consider Ajay gives ‘p’ coins to Bharat, so Ajay should give ’62-p’ coins to Charan since the total number of coins with Ajay is 62.
Bharath had 100 coins after distribution. He received ‘p’ coins from Ajay. So he should have received ‘100- p’ coins from Charan.
Charan had 60 coins after distribution. He received ’62-p’ from Ajay. So he should have received ‘p-2’ coins from Bharath.

From Ajay, we can say that 0 $\leq$ p $\leq$ 62

From Bharath, we can say that 2 $\leq$ p $\leq$ 93

From Charan, we can say that 0 $\leq$ p $\leq$ 100

So the range of p is 2 $\leq$ p $\leq$ 62

Let’s check the three statements:
Statement i:
i: Charan gave 37 coins to Bharath.
Charan can give a minimum of 38 coins to Bharath. Hence this statement is false.
Statement ii:
ii: Bharath gave 35 coins to Ajay.
Bharath can give a minimum of 33 and a maximum of 93 coins to Ajay. Hence this statement may be true. Statement iii:
iii: Bharath gave 61 coins to Charan.
Bharath can give a maximum of 60 coins to Charan. Hence this statement is false

Instructions

Four friends Aman, Bhanu, Chintu and Deepak borrowed some money from each of their parents.  It is known that the money borrowed by Aman, Bhanu, Chintu and Deepak respectively is in descending order. It is also known that each of them goes to different schools among Kendriya, DPS, Xaviers and Navoday. Each friend knows certain information about the amount borrowed as follows:
The boy from Kendriya: The sum of the money borrowed by all the others is 180₹.
The boy from DPS: The sum of the money borrowed by all the others is 200₹.
The boy from Xavier: The sum of the money borrowed by all the others is 220₹.
The boy from Navoday: The sum of the money borrowed by all the others is 280₹.

The amount borrowed by each of them is a unique positive integer. It is known that one of the boys made a mistake while adding the money of the other friends by 20₹.

Question 53: What was the amount borrowed by Aman?

a) 100

b) 60

c) 120

d) 140

Solution:

The boy from Kendriya, DPS, Xavier and Navoday be A, B, C and D

Then from the four statements:

The boy from Kendriya: The sum of the money borrowed by all the others is 180.
The boy from DPS: The sum of the money borrowed by all the others is 200.
The boy from Xavier: The sum of the money borrowed by all the others is 220.
The boy from Navoday: The sum of the money borrowed by all the others is 280.

=

B+C+D=180

A+C+D=200

A+B+D=220

A+B+C=280

3(A+B+C+D)= 880

It is known that one of the boys made a mistake while adding the money of the other friends by 20.

We can add or subtract the 20 to the total such that we get an integer for the sum of A, B, C and D.

The addition will only give a multiple of 3.

The sum of A+B+C+D= 900/3= 300

So we have to add 20 to the total quoted by the four boys.

The statements:

B+C+D=180

A+C+D=200

will be definitely true as if we add 20 to any of them we will get two equations with an equal sum which is not possible because then the numbers A, B, C and D will not be in descending order. Two of the numbers will be equal.

For example, we add 20 to the equation :

B+C+D= 180

We get B+C+D= 200

At the same time, A+C+D=200

Thus B=A ….which will not be in descending order.

1. A+B+C+D = 300

2. B+C+D = 180

3. A+C+D = 200

From 1 and 2, we get A = 120

From 1 and 3, we get B = 100

Thus

A+B+D=220

D= 0 which is not possible thus we have to add 20 in this equation.

D=20

A+B+C=280

C= 60

The money borrowed by Aman, Bhanu, Chintu and Deepak respectively is in descending order

The boy from Kendriya borrowed 120 (Aman)
The boy from DPS borrowed 100 (Bhanu)
The boy from Xavier borrowed 60 (Chintu)
The boy from Navoday borrowed 20, (Deepak)

Option C

Question 54: In which school does Bhanu study?

a) Kendriya

b) DPS

c) Xaviers

d) Navoday

Solution:

The boy from Kendriya, DPS, Xavier and Navoday be A, B, C and D

Then from the four statements:

The boy from Kendriya: The sum of the money borrowed by all the others is 180.
The boy from DPS: The sum of the money borrowed by all the others is 200.
The boy from Xavier: The sum of the money borrowed by all the others is 220.
The boy from Navoday: The sum of the money borrowed by all the others is 280.

=

B+C+D=180

A+C+D=200

A+B+D=220

A+B+C=280

3(A+B+C+D)= 880

It is known that one of the boys made a mistake while adding the money of the other friends by 20.

We can add or subtract the 20 to the total such that we get an integer for the sum of A, B, C and D.

The addition will only give a multiple of 3.

The sum of A+B+C+D= 900/3= 300

So we have to add 20 to the total quoted by the four boys.

The statements:

B+C+D=180

A+C+D=200

will be definitely true as if we add 20 to any of them we will get two equations with an equal sum which is not possible because then the numbers A, B, C and D will not be in descending order. Two of the numbers will be equal.

For example, we add 20 to the equation :

B+C+D= 180

We get B+C+D= 200

At the same time, A+C+D=200

Thus B=A ….which will not be in descending order.

1. A+B+C+D = 300

2. B+C+D = 180

3. A+C+D = 200

From 1 and 2, we get A = 120

From 1 and 3, we get B = 100

Thus

A+B+D=220

D= 0 which is not possible thus we have to add 20 in this equation.

D=20

A+B+C=280

C= 60

The boy from Kendriya borrowed 120 (Aman)
The boy from DPS borrowed 100 (Bhanu)
The boy from Xavier borrowed 60 (Chintu)
The boy from Navoday borrowed 20, (Deepak)

Option B

Question 55: What was the amount borrowed by Deepak?

a) 40

b) 10

c) 100

d) 20

Solution:

The boy from Kendriya, DPS, Xavier and Navoday be A, B, C and D

Then from the four statements:

The boy from Kendriya: The sum of the money borrowed by all the others is 180.
The boy from DPS: The sum of the money borrowed by all the others is 200.
The boy from Xavier: The sum of the money borrowed by all the others is 220.
The boy from Navoday: The sum of the money borrowed by all the others is 280.

=

B+C+D=180

A+C+D=200

A+B+D=220

A+B+C=280

3(A+B+C+D)= 880

It is known that one of the boys made a mistake while adding the money of the other friends by 20.

We can add or subtract the 20 to the total such that we get an integer for the sum of A, B, C and D.

The addition will only give a multiple of 3.

The sum of A+B+C+D= 900/3= 300

So we have to add 20 to the total quoted by the four boys.

The statements:

B+C+D=180

A+C+D=200

will be definitely true as if we add 20 to any of them we will get two equations with an equal sum which is not possible because then the numbers A, B, C and D will not be in descending order. Two of the numbers will be equal.

For example, we add 20 to the equation :

B+C+D= 180

We get B+C+D= 200

At the same time, A+C+D=200

Thus B=A ….which will not be in descending order.

1. A+B+C+D = 300

2. B+C+D = 180

3. A+C+D = 200

From 1 and 2, we get A = 120

From 1 and 3, we get B = 100

Thus

A+B+D=220

D= 0 which is not possible thus we have to add 20 in this equation.

D=20

A+B+C=280

C= 60

The boy from Kendriya borrowed 120 (Aman)
The boy from DPS borrowed 100 (Bhanu)
The boy from Xavier borrowed 60 (Chintu)
The boy from Navoday borrowed 20, (Deepak)

Option D

Question 56: What was the difference between the amount borrowed by Aman and Chintu?

a) 60

b) 40

c) 100

d) 80

Solution:

The boy from Kendriya, DPS, Xavier and Navoday be A, B, C and D

Then from the four statements:

The boy from Kendriya: The sum of the money borrowed by all the others is 180.
The boy from DPS: The sum of the money borrowed by all the others is 200.
The boy from Xavier: The sum of the money borrowed by all the others is 220.
The boy from Navoday: The sum of the money borrowed by all the others is 280.

=

B+C+D=180

A+C+D=200

A+B+D=220

A+B+C=280

3(A+B+C+D)= 880

It is known that one of the boys made a mistake while adding the money of the other friends by 20.

We can add or subtract the 20 to the total such that we get an integer for the sum of A, B, C and D.

The addition will only give a multiple of 3.

The sum of A+B+C+D= 900/3= 300

So we have to add 20 to the total quoted by the four boys.

The statements:

B+C+D=180

A+C+D=200

will be definitely true as if we add 20 to any of them we will get two equations with an equal sum which is not possible because then the numbers A, B, C and D will not be in descending order. Two of the numbers will be equal.

For example, we add 20 to the equation :

B+C+D= 180

We get B+C+D= 200

At the same time, A+C+D=200

Thus B=A ….which will not be in descending order.

1. A+B+C+D = 300

2. B+C+D = 180

3. A+C+D = 200

From 1 and 2, we get A = 120

From 1 and 3, we get B = 100

Thus

A+B+D=220

D= 0 which is not possible thus we have to add 20 in this equation.

D=20

A+B+C=280

C= 60

The boy from Kendriya borrowed 120 (Aman)
The boy from DPS borrowed 100 (Bhanu)
The boy from Xavier borrowed 60 (Chintu)
The boy from Navoday borrowed 20, (Deepak)

Option A

Instructions

Six people among Arun, Babu, Charles, Dinesh, Elango and Faizal secured ranks from one to six in their board exams.  Each of them had a different blood group. A blood group consists of both antigen and Rh factor. For example if the antigen is A and the Rh factor is -ve, then the blood group is represented as A -ve. For each of the six people, the antigen was one among A, B, O and the Rh Factor was either +ve or -ve. They had exactly one among Latin, Greek or French as the second language. Exactly three of them opted for one of the  languages and  exactly two for another.  The following clues are given.

1. Exactly one person who speaks French is better than exactly one person who speaks Latin.

2. Elango scored a worse rank than three others, and none of those three spoke French

3. The people with Rh type as +ve secured consecutive ranks and the people who have antigen O secured consecutive ranks, while people who secured consecutive ranks did not speak the same language.

4. Arun , whose blood group is A -ve , secured a rank which is immediately below than a person who has B+ve blood group and secured a rank which is immediately above a person who speaks Greek.

5. Babu secured a worse rank than Charles and Dinesh, but did not secure the worst rank.

Question 57: For how many people can the language spoken be determined accurately?

Solution:

Let the people be addressed as A,B,C,D,E,F by the first letter of their names for ease.

After reading the second sentence, we can infer that the three students above E studied only Greek or Latin. Hence, the table becomes,

Incorporating the third, fourth and first clue, we know that the A-ve comes below B+ve and O+ve and O-ve score consecutive ranks. Therefore,  the table becomes

Since we know that D and C are better than B, the final table is

Other than B, C and D, the language spoken by the rest of the three 3 can be determined uniquely.

Question 58: Who among the following people have A+ve blood group?

a) Babu

b) Elango

c) Arun

d) Faizal

Solution:

Let the people be addressed as A,B,C,D,E,F by the first letter of their names for ease.

After reading the second sentence, we can infer that the three students above E studied only Greek or Latin. Hence, the table becomes,

Incorporating the third, fourth and first clue, we know that the A-ve comes below B+ve and O+ve and O-ve score consecutive ranks. Therefore, the table becomes

Since we know that D and C are better than B, the final table is

Babu has A+ve blood group.

Question 59: Who among the following speaks French?

a) Arun

b) Charles

c) Babu

d) Elango

Solution:

Let the people be addressed as A,B,C,D,E,F by the first letter of their names for ease.

After reading the second sentence, we can infer that the three students above E studied only Greek or Latin. Hence, the table becomes,

Incorporating the third, fourth and first clue, we know that the A-ve comes below B+ve and O+ve and O-ve score consecutive ranks. Therefore, the table becomes

Since we know that D and C are better than B, the final table is

From the table we can see that Elango speaks French.

Question 60: How many different types of arrangements exists for the given problem?

Solution:

Let the people be addressed as A,B,C,D,E,F by the first letter of their names for ease.

After reading the second sentence, we can infer that the three students above E studied only Greek or Latin. Hence, the table becomes,

Incorporating the third, fourth and first clue, we know that the A-ve comes below B+ve and O+ve and O-ve score consecutive ranks. Therefore, the table becomes

Since we know that D and C are better than B, the final table is

Since, C or D can come at 1 or 2 and Language can be Latin/Greek/Latin for 1/2/3 or Greek/Latin/Greek for 1/2/3, there are 2*2=4 possibilities that exist for the given question.

Instructions

A contractor paid a different amount to six freelancers – Aman, Ajay, Diya, Farhan, Priya and Ram, not necessarily in the same order. The amount paid by him to each person is not necessarily a whole number. He paid them one after the other and paid no two persons at the same time.
Further, it is known that
1. Priya received $1200 which was the least amount paid by the contractor. 2. Barring the first and second payments, the amount paid by the contractor was the arithmetic mean of the previous two amounts paid by him. 3. He paid at least two of the freelancers an amount more than$2200.
4. Ram received his payment before Ajay and Farhan. Also, Farhan received $150 more than Ram. 5. Priya and Diya, who did not receive the highest payment, received their payments consecutively. Question 61: Who was the second last person to receive the payment? a) Farhan b) Aman c) Ajay d) Cannot be determined 61) Answer (C) Solution: We know that except the first and second payments, the rest of the payments were the mean of the previous two payments. The arithmetic mean of two values cannot be higher than both of those values or lower than both of the values. Hence, we cannot get the highest payment or lowest payment when the payment is calculated as a mean. Hence, the highest and the lowest have to be the first and second payments in any order. Priya received the least payment and Diya did not receive the highest payment. Therefore, the only way they can receive their payments consecutively is if Priya received her payment second and Diya received her payment third. Since Ram received a payment which is less than what Farhan received, he cannot be first. Hence, he has to be fourth. Aman has to be first. Ajay and Farhan can be fifth and sixth in any order. Hence, there are two possible combinations of the order. Let us assume that Aman received x. We can therefore write the others in terms of x. We get the following two cases: Case 1: We know that Farhan received$150 more than Ram,
3x/8 + 750 – (x/4 + 900) = 150
x = 2400
Let’s compute the other values:

This case is not possible as only one person received more than $2200. Case 2: We know that Farhan received$150 more than Ram,
5x/16 + 825 – (x/4 + 900) = 150
x = 3600
Let’s compute the other values:

This case is possible.

Ajay was the second last person to receive the payment.

Hence, option C is the right choice.

Question 62: What is the highest amount paid by the contractor (in $)? [Enter -1 if the value cannot be determined] 62) Answer: 3600 Solution: We know that except the first and second payments, the rest of the payments were the mean of the previous two payments. The arithmetic mean of two values cannot be higher than both of those values or lower than both of the values. Hence, we cannot get the highest payment or lowest payment when the payment is calculated as a mean. Hence, the highest and the lowest have to be the first and second payments in any order. Priya received the least payment and Diya did not receive the highest payment. Therefore, the only way they can receive their payments consecutively is if Priya received her payment second and Diya received her payment third. Since Ram received a payment which is less than what Farhan received, he cannot be first. Hence, he has to be fourth. Aman has to be first. Ajay and Farhan can be fifth and sixth in any order. Hence, there are two possible combinations of the order. Let us assume that Aman received x. We can therefore write the others in terms of x. We get the following two cases: Case 1: We know that Farhan received$150 more than Ram,
3x/8 + 750 – (x/4 + 900) = 150
x = 2400
Let’s compute the other values:

This case is not possible as only one person received more than $2200. Case 2: We know that Farhan received$150 more than Ram,
5x/16 + 825 – (x/4 + 900) = 150
x = 3600
Let’s compute the other values:

This case is possible.

The highest amount paid by the contractor is $3600. Question 63: What is the total amount paid by the contractor (in$)? [Enter -1 if the value cannot be determined]

Solution:

We know that except the first and second payments, the rest of the payments were the mean of the previous two payments. Hence, the highest and the lowest have to be the first and second payments in any order.
Priya received the least payment and Diya did not receive the highest payment. Therefore, the only way they can receive their payments consecutively is if Priya received her payment second and Diya received her payment third.
Since Ram received a payment which is less than what Farhan received, he cannot be first. Hence, he has to be fourth.
Aman has to be first. Ajay and Farhan can be fifth and sixth in any order.
Hence, there are two possible combinations of the order.
Let us assume that Aman received x.
We can therefore write the others in terms of x.

We get the following two cases:

Case 1:

We know that Farhan received $150 more than Ram, 3x/8 + 750 – (x/4 + 900) = 150 x = 2400 Let’s compute the other values: This case is not possible as only one person received more than$2200.

Case 2:
We know that Farhan received $150 more than Ram, 5x/16 + 825 – (x/4 + 900) = 150 x = 3600 Let’s compute the other values: This case is possible. The total amount paid by the contractor is$13050.

Question 64: Of all the six payments, what is the median payment (in $)? [Enter -1 if the value cannot be determined] 64) Answer: 2025 Solution: We know that except the first and second payments, the rest of the payments were the mean of the previous two payments. Hence, the highest and the lowest have to be the first and second payments in any order. Priya received the least payment and Diya did not receive the highest payment. Therefore, the only way they can receive their payments consecutively is if Priya received her payment second and Diya received her payment third. Since Ram received a payment which is less than what Farhan received, he cannot be first. Hence, he has to be fourth. Aman has to be first. Ajay and Farhan can be fifth and sixth in any order. Hence, there are two possible combinations of the order. Let us assume that Aman received x. We can therefore write the others in terms of x. We get the following two cases: Case 1: We know that Farhan received$150 more than Ram,
3x/8 + 750 – (x/4 + 900) = 150
x = 2400
Let’s compute the other values:

This case is not possible as only one person received more than $2200. Case 2: We know that Farhan received$150 more than Ram,
5x/16 + 825 – (x/4 + 900) = 150
x = 3600
Let’s compute the other values:

This case is possible.

The median amount paid by the contractor is the mean of $1950 and$2100 which is $2025 Instructions Arun, Rajan, Deeraj, Tilak, and Charan are five friends. Each one of them is from a different city among Chennai, Mumbai, Bangalore, Hyderabad, and Pune, not necessarily in the same order. As a part of their work, each of the five friends visit two of the five cities (apart from the one to which they belong) on a different day of the week from Monday to Friday. The person visits both the cities that he has to visit on the same day. While visiting a particular city, each person meets his friend who lives in the city without fail. No combination of friends met each other more than once during the week. Arun and Deeraj met on Thursday. Rajan visited Pune and Mumbai on Monday. The person from Mumbai met Charan a day after Charan met the person from Chennai. Tilak and Rajan met in Hyderabad on Tuesday. The person from Hyderabad met Charan on Friday. Arun did not go to Deeraj’s city. Deeraj did not meet anyone on Tuesday and Charan did not meet anyone on Monday. Question 65: On which day of the week did the persons from Pune and Mumbai meet? a) Tuesday b) Wednesday c) Thursday d) Friday 65) Answer (C) Solution: There are 5 persons. Therefore, there will be 5C2 = 10 ways in which 2 persons can meet. Let us denote the persons by the first letter of their names. The 10 possible combinations are A-R, A-D, A-T, A-C, R-D, R-T, R-C, D-T, D-C, and T-C. Rajan visited Pune and Mumbai on Monday. Therefore, Rajan should be the visitor on Monday. Tilak and Rajan met on Hyderabad on Tuesday. Since we knew that Rajan visited Pune and Mumbai on Monday, he should have been at his hometown on Tuesday. Hyderabad should be the hometown of Rajan. The person from Hyderabad met Charan on Friday. Therefore, the Rajan and Charan should have met each other on Friday. Arun and Dheeraj met on Thursday. Arun did not go to Dheeraj’s city. Therefore, Dheeraj should have visited Arun’s city. Dheeraj should have visited another city apart from Arun’s city on Thursday. Rajan visited Pune on Monday whereas Rajan and Charan met on Friday. Hence, we can say that Charan is not from Pune. Charan also met the person from Mumbai and Chennai. Hence, we can say that Charan is not from Mumbai, Pune, Hyderabad, or Chennai. Therefore, Charan should be from Bangalore. We know that Dheeraj and Arun are from Mumbai and Pune in any order (Since Rajan meets the persons from Mumbai and Pune on Monday). Also, we know that Charan met the person from Mumbai a day after he met the person from Chennai. Charan should have visited the 2 cities he visited on Friday (Since we know that Rajan visited the cities on Monday). Arun should have visited the cities on some day which is not Monday, Tuesday, Thursday, and Friday. Therefore, Arun should have visited the 2 cities he visited on Wednesday and Tilak should have visited the 2 cities he visited on Tuesday. Since Dheeraj did not meet anyone on Tuesday, Dheeraj and Tilak should have met on Thursday. Dheeraj and Charan should have met on Friday. It has been given that the person from Mumbai met Charan a day after Charan met the person from Chennai. Therefore, Tilak should be the person from Chennai and Arun should be the person from Mumbai. Dheeraj should be from Pune. The persons from Mumbai and Pune (Arun and Dheeraj) met each other on Thursday. Therefore, option C is the right answer. Question 66: Which of the following cities did Charan not go to? a) Hyderabad, Pune b) Chennai, Pune c) Hyderabad, Mumbai d) Mumbai, Chennai 66) Answer (D) Solution: There are 5 persons. Therefore, there will be 5C2 = 10 ways in which 2 persons can meet. Let us denote the persons by the first letter of their names. The 10 possible combinations are A-R, A-D, A-T, A-C, R-D, R-T, R-C, D-T, D-C, and T-C. Rajan visited Pune and Mumbai on Monday. Therefore, Rajan should be the visitor on Monday. Tilak and Rajan met on Hyderabad on Tuesday. Since we knew that Rajan visited Pune and Mumbai on Monday, he should have been at his hometown on Tuesday. Hyderabad should be the hometown of Rajan. The person from Hyderabad met Charan on Friday. Therefore, the Rajan and Charan should have met each other on Friday. Arun and Dheeraj met on Thursday. Arun did not go to Dheeraj’s city. Therefore, Dheeraj should have visited Arun’s city. Dheeraj should have visited another city apart from Arun’s city on Thursday. Rajan visited Pune on Monday whereas Rajan and Charan met on Friday. Hence, we can say that Charan is not from Pune. Charan also met the person from Mumbai and Chennai. Hence, we can say that Charan is not from Mumbai, Pune, Hyderabad, or Chennai. Therefore, Charan should be from Bangalore. We know that Dheeraj and Arun are from Mumbai and Pune in any order (Since Rajan meets the persons from Mumbai and Pune on Monday). Also, we know that Charan met the person from Mumbai a day after he met the person from Chennai. Charan should have visited the 2 cities he visited on Friday (Since we know that Rajan visited the cities on Monday). Arun should have visited the cities on some day which is not Monday, Tuesday, Thursday, and Friday. Therefore, Arun should have visited the 2 cities he visited on Wednesday and Tilak should have visited the 2 cities he visited on Tuesday. Since Dheeraj did not meet anyone on Tuesday, Dheeraj and Tilak should have met on Thursday. Dheeraj and Charan should have met on Friday. It has been given that the person from Mumbai met Charan a day after Charan met the person from Chennai. Therefore, Tilak should be the person from Chennai and Arun should be the person from Mumbai. Dheeraj should be from Pune. Charan is from Bangalore. He went to Hyderabad and Pune on Friday. Therefore, Charan did not go to Mumbai and Chennai and hence, option D is the right answer. Question 67: Which city is Tilak from? a) Hyderabad b) Chennai c) Bangalore d) Cannot be determined 67) Answer (B) Solution: There are 5 persons. Therefore, there will be 5C2 = 10 ways in which 2 persons can meet. Let us denote the persons by the first letter of their names. The 10 possible combinations are A-R, A-D, A-T, A-C, R-D, R-T, R-C, D-T, D-C, and T-C. Rajan visited Pune and Mumbai on Monday. Therefore, Rajan should be the visitor on Monday. Tilak and Rajan met on Hyderabad on Tuesday. Since we knew that Rajan visited Pune and Mumbai on Monday, he should have been at his hometown on Tuesday. Hyderabad should be the hometown of Rajan. The person from Hyderabad met Charan on Friday. Therefore, the Rajan and Charan should have met each other on Friday. Arun and Dheeraj met on Thursday. Arun did not go to Dheeraj’s city. Therefore, Dheeraj should have visited Arun’s city. Dheeraj should have visited another city apart from Arun’s city on Thursday. Rajan visited Pune on Monday whereas Rajan and Charan met on Friday. Hence, we can say that Charan is not from Pune. Charan also met the person from Mumbai and Chennai. Hence, we can say that Charan is not from Mumbai, Pune, Hyderabad, or Chennai. Therefore, Charan should be from Bangalore. We know that Dheeraj and Arun are from Mumbai and Pune in any order (Since Rajan meets the persons from Mumbai and Pune on Monday). Also, we know that Charan met the person from Mumbai a day after he met the person from Chennai. Charan should have visited the 2 cities he visited on Friday (Since we know that Rajan visited the cities on Monday). Arun should have visited the cities on some day which is not Monday, Tuesday, Thursday, and Friday. Therefore, Arun should have visited the 2 cities he visited on Wednesday and Tilak should have visited the 2 cities he visited on Tuesday. Since Dheeraj did not meet anyone on Tuesday, Dheeraj and Tilak should have met on Thursday. Dheeraj and Charan should have met on Friday. It has been given that the person from Mumbai met Charan a day after Charan met the person from Chennai. Therefore, Tilak should be the person from Chennai and Arun should be the person from Mumbai. Dheeraj should be from Pune. Tilak is from Chennai. Therefore, option B is the right answer. Question 68: The person from Bangalore met the person from Chennai on a) Tuesday b) Wednesday c) Thursday d) Friday 68) Answer (A) Solution: There are 5 persons. Therefore, there will be 5C2 = 10 ways in which 2 persons can meet. Let us denote the persons by the first letter of their names. The 10 possible combinations are A-R, A-D, A-T, A-C, R-D, R-T, R-C, D-T, D-C, and T-C. Rajan visited Pune and Mumbai on Monday. Therefore, Rajan should be the visitor on Monday. Tilak and Rajan met on Hyderabad on Tuesday. Since we knew that Rajan visited Pune and Mumbai on Monday, he should have been at his hometown on Tuesday. Hyderabad should be the hometown of Rajan. The person from Hyderabad met Charan on Friday. Therefore, the Rajan and Charan should have met each other on Friday. Arun and Dheeraj met on Thursday. Arun did not go to Dheeraj’s city. Therefore, Dheeraj should have visited Arun’s city. Dheeraj should have visited another city apart from Arun’s city on Thursday. Rajan visited Pune on Monday whereas Rajan and Charan met on Friday. Hence, we can say that Charan is not from Pune. Charan also met the person from Mumbai and Chennai. Hence, we can say that Charan is not from Mumbai, Pune, Hyderabad, or Chennai. Therefore, Charan should be from Bangalore. We know that Dheeraj and Arun are from Mumbai and Pune in any order (Since Rajan meets the persons from Mumbai and Pune on Monday). Also, we know that Charan met the person from Mumbai a day after he met the person from Chennai. Charan should have visited the 2 cities he visited on Friday (Since we know that Rajan visited the cities on Monday). Arun should have visited the cities on some day which is not Monday, Tuesday, Thursday, and Friday. Therefore, Arun should have visited the 2 cities he visited on Wednesday and Tilak should have visited the 2 cities he visited on Tuesday. Since Dheeraj did not meet anyone on Tuesday, Dheeraj and Tilak should have met on Thursday. Dheeraj and Charan should have met on Friday. It has been given that the person from Mumbai met Charan a day after Charan met the person from Chennai. Therefore, Tilak should be the person from Chennai and Arun should be the person from Mumbai. Dheeraj should be from Pune. The person from Bangalore (Charan) met the person from Chennai (Tilak) on Tuesday. Therefore, option A is the right answer. Instructions A team of at least three students has to be selected from eight students available-A, B, C, D, E, F, G and H. The selection should be based on the following rules. 1. If D is selected, then G must also be selected. 2. If G is selected, then H must also be selected. 3. If A is selected, then D cannot be selected. 4. If E is selected, then H cannot be selected. 5. F cannot be in a team of more than three members including him. 6. C cannot be in a team of more than four members including him. 7. At least one person among B and C has to be selected. Question 69: What is the number of teams possible if F is definitely selected? 69) Answer: 7 Solution: If D is selected, then G must also be selected. If G is selected, then H must also be selected. If A is selected, then D cannot be selected. If E is selected, then H cannot be selected. Thus the case groups possible are 1. DGH— ( A and E both cannot be included) 2. GH— ( A can be included but E cannot ) i) AGH is possible 3. H— ( A can be included but E cannot ) i) AH is possible F cannot be in a team of more than three members including him. Aside from B,C and F all these cases are possible: A, E, AE, DGH, GH, H, AH, AGH At least one person among B and C has to be selected. We can create the following table to categorize all the cases: C cannot be in a team of more than four members including him. We can see from the table that 7 teams are possible with F as a member. Question 70: What is the number of teams possible if C is definitely selected? 70) Answer: 15 Solution: If D is selected, then G must also be selected. If G is selected, then H must also be selected. If A is selected, then D cannot be selected. If E is selected, then H cannot be selected. Thus the case groups possible are 1. DGH— ( A and E both cannot be included) 2. GH— ( A can be included but E cannot ) i) AGH is possible 3. H— ( A can be included but E cannot ) i) AH is possible F cannot be in a team of more than three members including him. Aside from B,C and F all these cases are possible: A, E, AE, DGH, GH, H, AH, AGH At least one person among B and C has to be selected. We can create the following table to categorize all the cases: C cannot be in a team of more than four members including him. Teams with C as a member =8+7=15 Answer 15 Question 71: How many 4 member teams are possible? 71) Answer: 7 Solution: If D is selected, then G must also be selected. If G is selected, then H must also be selected. If A is selected, then D cannot be selected. If E is selected, then H cannot be selected. Thus the case groups possible are 1. DGH— ( A and E both cannot be included) 2. GH— ( A can be included but E cannot ) i) AGH is possible 3. H— ( A can be included but E cannot ) i) AH is possible F cannot be in a team of more than three members including him. Aside from B,C and F all these cases are possible: A, E, AE, DGH, GH, H, AH, AGH At least one person among B and C has to be selected. We can create the following table to categorize all the cases: C cannot be in a team of more than four members including him. Answer 7 Question 72: What is the number of teams possible if A is definitely selected? 72) Answer: 11 Solution: If D is selected, then G must also be selected. If G is selected, then H must also be selected. If A is selected, then D cannot be selected. If E is selected, then H cannot be selected. Thus the case groups possible are 1. DGH— ( A and E both cannot be included) 2. GH— ( A can be included but E cannot ) i) AGH is possible 3. H— ( A can be included but E cannot ) i) AH is possible F cannot be in a team of more than three members including him. Aside from B,C and F all these cases are possible: A, E, AE, DGH, GH, H, AH, AGH At least one person among B and C has to be selected. We can create the following table to categorize all the cases: C cannot be in a team of more than four members including him. Answer 11 Instructions The following area graph represents the production of certain vegetables (in kgs) between 2015 and 2019 in a particular district. The X- axis represents the time period. The area of a particular vegetable, between a chosen time period gives the quantity of the particular vegetable produced in that time period. In X axis, the line segment from 2015 to 2016 measures 100 units. In Y axis, the line segment from 0 to 20 measures 20 units. 1 sq unit area for a particular vegetable denotes 1kg production of that particular vegetable. The values from the graph on the y axis are in multiples of 5. Question 73: How many kgs of potatoes are produced in the time period between 2015 and 2017? a) 12750 b) 12500 c) 12250 d) 12000 73) Answer (C) Solution: We find that the area of the figure between any two given years would always be a trapezium. For example consider the time period between 2015 and 2016. The lines x= 2015 and x= 2016 are parallel. Hence, opposite sides are parallel and the figure is a quadrilateral for any vegetable in between any two years. Hence, the area would be found by the formula (height * average of parallel sides). Onion: 2015-2016 :$\frac{1}{2}\left(40+45\right)\cdot100\ =\ 4250$2016-2017 :$\frac{1}{2}\left(45+50\right)\cdot100\ =\ 4750$2017-2018 :$\frac{1}{2}\left(50+55\right)\cdot100\ =\ 5250$2018-2019 :$\frac{1}{2}\left(55+95\right)\cdot100\ =\ 7500$Tomato : 2015-2016 :$\frac{1}{2}\left(35+25\right)\cdot100\ =\ 3000$2016-2017 :$\frac{1}{2}\left(25+25\right)\cdot100\ =\ 2500$2017-2018 :$\frac{1}{2}\left(25+35\right)\cdot100\ =\ 3000$2018-2019 :$\frac{1}{2}\left(35+15\right)\cdot100\ =2500$Beetroot: 2015-2016 :$\frac{1}{2}\left(50+60\right)\cdot100\ =5500$2016-2017 :$\frac{1}{2}\left(60+75\right)\cdot100\ =\ 6750$2017-2018 :$\frac{1}{2}\left(75+90\right)\cdot100\ =8250$2018-2019 :$\frac{1}{2}\left(90+105\right)\cdot100\ =9750$Cabbage: 2015-2016 :$\frac{1}{2}\left(20+35\right)\cdot100\ =2750$2016-2017 :$\frac{1}{2}\left(35+25\right)\cdot100\ =\ 3000$2017-2018 :$\frac{1}{2}\left(25+40\right)\cdot100\ =3250$2018-2019 :$\frac{1}{2}\left(40+60\right)\cdot100\ =5000$Using this, the below table is computed. Where, each value represents kgs of the vegetable produced. 12250 kgs of potatoes are produced during 2015-2017 Question 74: By what percent has the production of beetroot between the time period 2018-2019 increased from the time period 2017-2018? a) 12.5% b) 11.43% c) 15.15% d) 18.18% 74) Answer (D) Solution: We find that the area of the figure between any two given years would always be a trapezium. For example consider the time period between 2015 and 2016. The lines x= 2015 and x= 2016 are parallel. Hence, opposite sides are parallel and the figure is a quadrilateral for any vegetable in between any two years. Hence, the area would be found by the formula (height * average of parallel sides). Onion: 2015-2016 :$\frac{1}{2}\left(40+45\right)\cdot100\ =\ 4250$2016-2017 :$\frac{1}{2}\left(45+50\right)\cdot100\ =\ 4750$2017-2018 :$\frac{1}{2}\left(50+55\right)\cdot100\ =\ 5250$2018-2019 :$\frac{1}{2}\left(55+95\right)\cdot100\ =\ 7500$Tomato : 2015-2016 :$\frac{1}{2}\left(35+25\right)\cdot100\ =\ 3000$2016-2017 :$\frac{1}{2}\left(25+25\right)\cdot100\ =\ 2500$2017-2018 :$\frac{1}{2}\left(25+35\right)\cdot100\ =\ 3000$2018-2019 :$\frac{1}{2}\left(35+15\right)\cdot100\ =2500$Beetroot: 2015-2016 :$\frac{1}{2}\left(50+60\right)\cdot100\ =5500$2016-2017 :$\frac{1}{2}\left(60+75\right)\cdot100\ =\ 6750$2017-2018 :$\frac{1}{2}\left(75+90\right)\cdot100\ =8250$2018-2019 :$\frac{1}{2}\left(90+105\right)\cdot100\ =9750$Cabbage: 2015-2016 :$\frac{1}{2}\left(20+35\right)\cdot100\ =2750$2016-2017 :$\frac{1}{2}\left(35+25\right)\cdot100\ =\ 3000$2017-2018 :$\frac{1}{2}\left(25+40\right)\cdot100\ =3250$2018-2019 :$\frac{1}{2}\left(40+60\right)\cdot100\ =5000$Using this, the below table is computed. Where, each cell represents kgs of the vegetable produced in that particular year. In 2017-18 the production is 8250 kgs . In 2018-19 the production is 9750 kgs. Increase in production is 18.18% Question 75: What is the difference in the amount of onion produced in 2018-2019 to the amount of cabbage produced in 2016-2017? a) 2500 b) 4500 c) 3250 d) 3750 75) Answer (B) Solution: We find that the area of the figure between any two given years would always be a trapezium. For example consider the time period between 2015 and 2016. The lines x= 2015 and x= 2016 are parallel. Hence, opposite sides are parallel and the figure is a quadrilateral for any vegetable in between any two years. Hence, the area would be found by the formula (height * average of parallel sides). Onion: 2015-2016 :$\frac{1}{2}\left(40+45\right)\cdot100\ =\ 4250$2016-2017 :$\frac{1}{2}\left(45+50\right)\cdot100\ =\ 4750$2017-2018 :$\frac{1}{2}\left(50+55\right)\cdot100\ =\ 5250$2018-2019 :$\frac{1}{2}\left(55+95\right)\cdot100\ =\ 7500$Tomato : 2015-2016 :$\frac{1}{2}\left(35+25\right)\cdot100\ =\ 3000$2016-2017 :$\frac{1}{2}\left(25+25\right)\cdot100\ =\ 2500$2017-2018 :$\frac{1}{2}\left(25+35\right)\cdot100\ =\ 3000$2018-2019 :$\frac{1}{2}\left(35+15\right)\cdot100\ =2500$Beetroot: 2015-2016 :$\frac{1}{2}\left(50+60\right)\cdot100\ =5500$2016-2017 :$\frac{1}{2}\left(60+75\right)\cdot100\ =\ 6750$2017-2018 :$\frac{1}{2}\left(75+90\right)\cdot100\ =8250$2018-2019 :$\frac{1}{2}\left(90+105\right)\cdot100\ =9750$Cabbage: 2015-2016 :$\frac{1}{2}\left(20+35\right)\cdot100\ =2750$2016-2017 :$\frac{1}{2}\left(35+25\right)\cdot100\ =\ 3000$2017-2018 :$\frac{1}{2}\left(25+40\right)\cdot100\ =3250$2018-2019 :$\frac{1}{2}\left(40+60\right)\cdot100\ =5000$Using this, the below table is computed. Where, each cell represents kgs of the vegetable produced in that particular year. The difference between the amount of Onion produced in 2018-19 and the amount of cabbage produced in 2016-17 is 4500. Question 76: What is the total production of tomatoes during 2015-2019? a) 11000 kgs b) 10000kgs c) 10500kgs d) 10750kgs 76) Answer (A) Solution: We find that the area of the figure between any two given years would always be a trapezium. For example consider the time period between 2015 and 2016. The lines x= 2015 and x= 2016 are parallel. Hence, opposite sides are parallel and the figure is a quadrilateral for any vegetable in between any two years. Hence, the area would be found by the formula (height * average of parallel sides). Onion: 2015-2016 :$\frac{1}{2}\left(40+45\right)\cdot100\ =\ 4250$2016-2017 :$\frac{1}{2}\left(45+50\right)\cdot100\ =\ 4750$2017-2018 :$\frac{1}{2}\left(50+55\right)\cdot100\ =\ 5250$2018-2019 :$\frac{1}{2}\left(55+95\right)\cdot100\ =\ 7500$Tomato : 2015-2016 :$\frac{1}{2}\left(35+25\right)\cdot100\ =\ 3000$2016-2017 :$\frac{1}{2}\left(25+25\right)\cdot100\ =\ 2500$2017-2018 :$\frac{1}{2}\left(25+35\right)\cdot100\ =\ 3000$2018-2019 :$\frac{1}{2}\left(35+15\right)\cdot100\ =2500$Beetroot: 2015-2016 :$\frac{1}{2}\left(50+60\right)\cdot100\ =5500$2016-2017 :$\frac{1}{2}\left(60+75\right)\cdot100\ =\ 6750$2017-2018 :$\frac{1}{2}\left(75+90\right)\cdot100\ =8250$2018-2019 :$\frac{1}{2}\left(90+105\right)\cdot100\ =9750$Cabbage: 2015-2016 :$\frac{1}{2}\left(20+35\right)\cdot100\ =2750$2016-2017 :$\frac{1}{2}\left(35+25\right)\cdot100\ =\ 3000$2017-2018 :$\frac{1}{2}\left(25+40\right)\cdot100\ =3250$2018-2019 :$\frac{1}{2}\left(40+60\right)\cdot100\ =5000$Using this, the below table is computed. Where, each cell represents kgs of the vegetable produced in that particular year. The total production of tomatoes is 3000+2500+3000+2500=11000 kgs Instructions A group of 8 developing countries namely P, Q, R, S, T, U and V decided to calculate their development index(DI) based on 3 parameters: Income Index(I), Health Index(H), Education Index(E). The formula for development index is given below: DI = xI+yH+zE, where the value each of x, y and z can be either 2 or 3. The data regarding the Income Index, Health Index and Education Index is given in the table. Some data might be missing from the table. The following information is also known: 1. DI for no two countries is the same. 2. DI for the country Q is 38. 3. The values of I, H and E are integral and vary from 1 to 10. 4. The value of DI is neither less than 31 nor greater than 40. 5. The score for education index is the same for P and Q. 6. Exactly one country has a score of 10 in one parameter. 7. DI for W is not the highest. Question 77: What is the value of DI for S? a) 31 b) 32 c) 33 d) 34 77) Answer (A) Solution: For country V, the values of I, H and E are known. Since the value of each of x, y and z can be 2 or 3. Case: x=y=z=3, then the values of DI for each country will be a multiple of 3. (Rejected) Case: x=y=3, z=2, the value of DI for V will become 38. (Rejected, DI of Q is already 38) Case: y=z=3, x=2, the value of DI for Q cannot be 38 for any integral value of E. Case: x=z=3, y=2, the value of DI for V will become 30. Only case possible will be when x=z=2 and y=3. For the rest of the combinations, the value of DI for V will become less than 30. Hence, x=2, y=3, z=2 Then, the value of DI for V = 2*3+3*9+2*1 = 35 The value of E for the country Q, 2*3+3*6+7E = 38 => E = 7 From 5, the value of E for P = 7, => DI for P = 2*5+3*4+2*7 = 36 Now, the value of DI for U = 1*2+3*H+2*3 = 3H+8. The only possible value for H to satisfy both conditions 1 and 4 will be 8 and DI for U will be 32. The table can be filled as follows: From 6, the value of I for the country R can be 10. (Only possibility) So, DI for R = 10*2+3*3+5*2 = 39 Then, DI for T = 3*2+3H+2*6 = 3H+18 (Multiple of 3, 33 is only possibility for DI, H=5) DI for country W = 2*8+3*H+2*9 = 3H+34. Hence, DI for W can be 37, when H=1. (Using clue 7, H cannot be 2) Now, DI for S = 7*2+3*3+2E = 2E+23 (The value of E can only be odd, hence 31 is the only possibility when E =4) The final table can be formed as follows: The value of DI for S is 31. Question 78: Which country scored 10 in a parameter? a) P b) R c) T d) U 78) Answer (B) Solution: For country V, the values of I, H and E are known. Since the value of each of x, y and z can be 2 or 3. Case: x=y=z=3, then the values of DI for each country will be a multiple of 3. (Rejected) Case: x=y=3, z=2, the value of DI for V will become 38. (Rejected, DI of Q is already 38) Case: y=z=3, x=2, the value of DI for Q cannot be 38 for any integral value of E. Case: x=z=3, y=2, the value of DI for V will become 30. Only case possible will be when x=z=2 and y=3. For the rest of the combinations, the value of DI for V will become less than 30. Hence, x=2, y=3, z=2 Then, the value of DI for V = 2*3+3*9+2*1 = 35 The value of E for the country Q, 2*3+3*6+7E = 38 => E = 7 From 5, the value of E for P = 7, => DI for P = 2*5+3*4+2*7 = 36 Now, the value of DI for U = 1*2+3*H+2*3 = 3H+8. The only possible value for H to satisfy both conditions 1 and 4 will be 8 and DI for U will be 32. The table can be filled as follows: From 6, the value of I for the country R can be 10. (Only possibility) So, DI for R = 10*2+3*3+5*2 = 39 Then, DI for T = 3*2+3H+2*6 = 3H+18 (Multiple of 3, 33 is only possibility for DI, H=5) DI for country W = 2*8+3*H+2*9 = 3H+34. Hence, DI for W can be 37, when H=1. (Using clue 7, H cannot be 2) Now, DI for S = 7*2+3*3+2E = 2E+23 (The value of E can only be odd, hence 31 is the only possibility when E =4) The final table can be formed as follows: R scored 10 in Income Index. Question 79: What is the total sum of score of all countries in the health index? a) 37 b) 40 c) 38 d) 39 79) Answer (D) Solution: For country V, the values of I, H and E are known. Since the value of each of x, y and z can be 2 or 3. Case: x=y=z=3, then the values of DI for each country will be a multiple of 3. (Rejected) Case: x=y=3, z=2, the value of DI for V will become 38. (Rejected, DI of Q is already 38) Case: y=z=3, x=2, the value of DI for Q cannot be 38 for any integral value of E. Case: x=z=3, y=2, the value of DI for V will become 30. Only case possible will be when x=z=2 and y=3. For the rest of the combinations, the value of DI for V will become less than 30. Hence, x=2, y=3, z=2 Then, the value of DI for V = 2*3+3*9+2*1 = 35 The value of E for the country Q, 2*3+3*6+7E = 38 => E = 7 From 5, the value of E for P = 7, => DI for P = 2*5+3*4+2*7 = 36 Now, the value of DI for U = 1*2+3*H+2*3 = 3H+8. The only possible value for H to satisfy both conditions 1 and 4 will be 8 and DI for U will be 32. The table can be filled as follows: From 6, the value of I for the country R can be 10. (Only possibility) So, DI for R = 10*2+3*3+5*2 = 39 Then, DI for T = 3*2+3H+2*6 = 3H+18 (Multiple of 3, 33 is only possibility for DI, H=5) DI for country W = 2*8+3*H+2*9 = 3H+34. Hence, DI for W can be 37, when H=1. (Using clue 7, H cannot be 2) Now, DI for S = 7*2+3*3+2E = 2E+23 (The value of E can only be odd, hence 31 is the only possibility when E =4) The final table can be formed as follows: The sum of total score for the health index = 4+6+3+3+5+8+9+1 = 39 Question 80: What is the median value of DI for all the countries? a) 35 b) 36.5 c) 36 d) 35.5 80) Answer (D) Solution: For country V, the values of I, H and E are known. Since the value of each of x, y and z can be 2 or 3. Case: x=y=z=3, then the values of DI for each country will be a multiple of 3. (Rejected) Case: x=y=3, z=2, the value of DI for V will become 38. (Rejected, DI of Q is already 38) Case: y=z=3, x=2, the value of DI for Q cannot be 38 for any integral value of E. Case: x=z=3, y=2, the value of DI for V will become 30. Only case possible will be when x=z=2 and y=3. For the rest of the combinations, the value of DI for V will become less than 30. Hence, x=2, y=3, z=2 Then, the value of DI for V = 2*3+3*9+2*1 = 35 The value of E for the country Q, 2*3+3*6+7E = 38 => E = 7 From 5, the value of E for P = 7, => DI for P = 2*5+3*4+2*7 = 36 Now, the value of DI for U = 1*2+3*H+2*3 = 3H+8. The only possible value for H to satisfy both conditions 1 and 4 will be 8 and DI for U will be 32. The table can be filled as follows: From 6, the value of I for the country R can be 10. (Only possibility) So, DI for R = 10*2+3*3+5*2 = 39 Then, DI for T = 3*2+3H+2*6 = 3H+18 (Multiple of 3, 33 is only possibility for DI, H=5) DI for country W = 2*8+3*H+2*9 = 3H+34. Hence, DI for W can be 37, when H=1. Now, DI for S = 7*2+3*3+2E = 2E+23 (The value of E can only be odd, hence 31 is only possibility when E =4) The final table can be formed as follows: Arranging the DI scores in increasing order, 31,32,33,35,36,37,38,39 The median value =$\ \frac{\ 35+36}{2}\ =35.5\$

Instructions

Harish went to a circus where four popular jokers from the countries A, B, C and D performed.
Each joker wore 4 different garments namely the Bow, Coat, Trousers and Shoes.
While returning he tried to recall the colour of each garment: Bow, Coat, Trousers and Shoes worn by the jokers.
He notes down the colours of each garment in a notebook as given below:

Colours noted for each garment for different jokers is represented in the form of colour code as B1, B2…C1, C2…..etc.

Later it was known that only one colour of each garment was correct and rest three of them were incorrectly positioned by Harish.

The following facts are true about the correct order:
1. The colour B3 is in a column which is to the right of the column of colour C1.
2. C4 is in the column which is 2 places to the right of the column of T1 which in turn is to the right of the column of S3.
3. T3 is in the column which is 2 places to the right of the column of B1 which in turn is to the left of the column of C3.
4. B2 is in the column which is to the right of the column of colour S2.

Question 81: What is the colour code of Bow which belongs to the joker from A?

a) B1

b) B2

c) B3

d) B4

Solution:

The colours of each garment are interchanged except one, we can make the following table to represent the actual colour.

C4 is in the column second right of the column of T1 which in turn is in right to the column of S3.

Only one case is possible:

(The grey column is the correct order)

Only one colour of each garment was correct and rest three of them were incorrectly positioned. Thus, C4 is the only colour in the coat which is positioned correctly.

T3 is in the column second right of the column of B1 which in turn is in left to the column of C3.

T3 can take only two positions, either with C or D.

Two cases are possible:

Case 1:

But C3 and C4 will be correctly positioned, which cannot be possible.

Case 2:

C1 cannot be at the correct position thus it will take under C1.

Similarly, T4 will be under column A.

The colour B3 is in the column right to the column of colour C1. B2 is in the column right to the column of S2.

Option A

Question 82: Joker from which country wore S4 colour-coded shoes?

a) A

b) B

c) C

d) D

Solution:

The colours of each garment are interchanged except one, we can make the following table to represent the actual colour.

C4 is in the column second right of the column of T1 which in turn is in right to the column of S3.

Only one case is possible:

(The grey column is the correct order)

Only one colour of each garment was correct and rest three of them were incorrectly positioned. Thus, C4 is the only colour in the coat which is positioned correctly.

T3 is in the column second right of the column of B1 which in turn is in left to the column of C3.

T3 can take only two positions, either with C or D.

Two cases are possible:

Case 1:

But C3 and C4 will be correctly positioned, which cannot be possible.

Case 2:

C1 cannot be at the correct position thus it will take under C1.

Similarly, T4 will be under column A.

The colour B3 is in the column right to the column of colour C1. B2 is in the column right to the column of S2.

Option C

Question 83: What is the country of the joker whose shoes were correctly noted by Harish?

a) A

b) B

c) C

d) D

Solution:

The colours of each garment are interchanged except one, we can make the following table to represent the actual colour.

C4 is in the column second right of the column of T1 which in turn is in right to the column of S3.

Only one case is possible:

(The grey column is the correct order)

Only one colour of each garment was correct and rest three of them were incorrectly positioned. Thus, C4 is the only colour in the coat which is positioned correctly.

T3 is in the column second right of the column of B1 which in turn is in left to the column of C3.

T3 can take only two positions, either with C or D.

Two cases are possible:

Case 1:

But C3 and C4 will be correctly positioned, which cannot be possible.

Case 2:

C1 cannot be at the correct position thus it will take under C1.

Similarly, T4 will be under column A.

The colour B3 is in the column right to the column of colour C1. B2 is in the column right to the column of S2.

Option B

Question 84: Joker of which country wore T4 colour coded trouser?

a) A

b) B

c) C

d) D

Solution:

The colours of each garment are interchanged except one, we can make the following table to represent the actual colour.

C4 is in the column second right of the column of T1 which in turn is in right to the column of S3.

Only one case is possible:

(The grey column is the correct order)

Only one colour of each garment was correct and rest three of them were incorrectly positioned. Thus, C4 is the only colour in the coat which is positioned correctly.

T3 is in the column second right of the column of B1 which in turn is in left to the column of C3.

T3 can take only two positions, either with C or D.

Two cases are possible:

Case 1:

But C3 and C4 will be correctly positioned, which cannot be possible.

Case 2:

C1 cannot be at the correct position thus it will take under C1.

Similarly, T4 will be under column A.

The colour B3 is in the column right to the column of colour C1. B2 is in the column right to the column of S2.

Option A

Instructions

P, Q, R, and S are four brothers who went to an ice cream parlour. Each of them ate a different number of ice creams among 1, 2, 3, and 4. Since their mother do not allow more than one ice cream per day, they decided to report distorted information to her. So, each one of them decided to be a truth-teller, or a liar, or an alternator. Truth-tellers always tell the truth and liars always make false statements. Alternators, however, alternate between making true and false statements, starting with either truth or a lie. Among the four, there is at least one truth teller, one liar, and one alternator. Each of the four brothers make the following statements –
P:
(i) The difference between the number of ice creams eaten by me and that by R is two.
(ii) S is not a liar.
Q:
(i) Exactly two of us tell the truth.
(ii) R ate one ice cream while P ate 3 ice creams.
R:
(i) Only one of us is an alternator.
(ii) S ate four ice creams.
S:
(i) I am an alternator
(ii) Only one of us is a liar.

Question 85: Who ate the most number of ice creams?

a) P

b) Q

c) S

d) Either A or C

Solution:

If the first statement made by S is true, then his second statement must definitely be false. But, if his first statement is false, then he definitely must be a liar. So, his second statement will again be false. In any case, the second statement made by S is definitely false. So, there are 2 liars which leaves us with one truth-teller and alternator.
So, this means that Q’s first statement is definitely false and R’s first statement is definitely true. This means that Q definitely made 1 false statement and R definitely made 1 true statement.
If S is an alternator, then two among P, Q, and R must be a liar. Now, P’s second statement says that S is not a liar which will be true. So, S will also have 1 true statement. So, there will be at least 2 alternators or 2 truth-tellers in the group which is not possible.
So, S must be one of the liars. So, P’s second statement is false. This means one among P and Q must be the second liar. This means R is a truth teller. This means S ate 4 ice creams.
Let’s assume that P’s 1st statement is true. So, P and R must have eaten 1 or 3 ice creams leaving Q with 2 ice creams. So, both of Q’s statement should be false. This means that P must have eaten 1 and R must have eaten 3 ice creams.
Let’s assume the other case when P’s 2nd statement is false. This will also leave Q’s 2nd statement as incorrect. So, there will be no alternator. Thus, this case is not possible.
So, Q and S are liars, P is an alternator and R is a truth teller. The number of icecreams eaten by each of them is-
P-1, Q-2, R-3, and S-4

Question 86: Among the eight statements made, how many were lies?

Solution:

If the first statement made by S is true, then his second statement must definitely be false. But, if his first statement is false, then he definitely must be a liar. So, his second statement will again be false. In any case, the second statement made by S is definitely false. So, there are 2 liars which leaves us with one truth-teller and alternator.
So, this means that Q’s first statement is definitely false and R’s first statement is definitely true. This means that Q definitely made 1 false statement and R definitely made 1 true statement.
If S is an alternator, then two among P, Q, and R must be a liar. Now, P’s second statement says that S is not a liar which will be true. So, S will also have 1 true statement. So, there will be at least 2 alternators or 2 truth-tellers in the group which is not possible.
So, S must be one of the liars. So, P’s second statement is false. This means one among P and Q must be the second liar. This means R is a truth teller. This means S ate 4 ice creams.
Let’s assume that P’s 1st statement is true. So, P and R must have eaten 1 or 3 ice creams leaving Q with 2 ice creams. So, both of Q’s statement should be false. This means that P must have eaten 1 and R must have eaten 3 ice creams.
Let’s assume the other case when P’s 2nd statement is false. This will also leave Q’s 2nd statement as incorrect. So, there will be no alternator. Thus, this case is not possible.
So, Q and S are liars, P is an alternator and R is a truth teller. The number of icecreams eaten by each of them is-
P-1, Q-2, R-3, and S-4
Since there are 2 liars and 1 alternator, 5 statements must be lies.

Question 87: Which of the following is definitely true?

a) S made the most number of false statements.

b) P made the most number of true statements.

c) Q ate more ice creams than R.

d) The number of ice creams eaten by P and S together is more than that eaten by the other two.

Solution:

If the first statement made by S is true, then his second statement must definitely be false. But, if his first statement is false, then he definitely must be a liar. So, his second statement will again be false. In any case, the second statement made by S is definitely false. So, there are 2 liars which leaves us with one truth-teller and alternator.
So, this means that Q’s first statement is definitely false and R’s first statement is definitely true. This means that Q definitely made 1 false statement and R definitely made 1 true statement.
If S is an alternator, then two among P, Q, and R must be a liar. Now, P’s second statement says that S is not a liar which will be true. So, S will also have 1 true statement. So, there will be at least 2 alternators or 2 truth-tellers in the group which is not possible.
So, S must be one of the liars. So, P’s second statement is false. This means one among P and Q must be the second liar. This means R is a truth teller. This means S ate 4 ice creams.
Let’s assume that P’s 1st statement is true. So, P and R must have eaten 1 or 3 ice creams leaving Q with 2 ice creams. So, both of Q’s statement should be false. This means that P must have eaten 1 and R must have eaten 3 ice creams.
Let’s assume the other case when P’s 2nd statement is false. This will also leave Q’s 2nd statement as incorrect. So, there will be no alternator. Thus, this case is not possible.
So, Q and S are liars, P is an alternator and R is a truth teller. The number of icecreams eaten by each of them is-
P-1, Q-2, R-3, and S-4
We can see that among the given options, only A is true.

Question 88: What is the total number of ice creams eaten by Q and S together?

Solution:

If the first statement made by S is true, then his second statement must definitely be false. But, if his first statement is false, then he definitely must be a liar. So, his second statement will again be false. In any case, the second statement made by S is definitely false. So, there are 2 liars which leaves us with one truth-teller and alternator.
So, this means that Q’s first statement is definitely false and R’s first statement is definitely true. This means that Q definitely made 1 false statement and R definitely made 1 true statement.
If S is an alternator, then two among P, Q, and R must be a liar. Now, P’s second statement says that S is not a liar which will be true. So, S will also have 1 true statement. So, there will be at least 2 alternators or 2 truth-tellers in the group which is not possible.
So, S must be one of the liars. So, P’s second statement is false. This means one among P and Q must be the second liar. This means R is a truth teller. This means S ate 4 ice creams.
Let’s assume that P’s 1st statement is true. So, P and R must have eaten 1 or 3 ice creams leaving Q with 2 ice creams. So, both of Q’s statement should be false. This means that P must have eaten 1 and R must have eaten 3 ice creams.
Let’s assume the other case when P’s 2nd statement is false. This will also leave Q’s 2nd statement as incorrect. So, there will be no alternator. Thus, this case is not possible.
So, Q and S are liars, P is an alternator and R is a truth teller. The number of icecreams eaten by each of them is-
P-1, Q-2, R-3, and S-4
We can see that they ate 6 ice creams together.

Instructions

In a university, there are a total of 500 students who are pursuing either of the two streams medical or sciences but not both.
In a university-wide survey conducted by their canteen committee, they asked the students their inclination for tea and coffee.
1. Among those who pursue medical, 25% like only coffee.
2. An equal number of students in both streams do not like both tea and coffee.
3. A total of 250 students like tea and 220 students like coffee.
4. Science has 50 more tea lovers than Medical.
5. The number of students who like both tea and coffee in medical is 30 less than that in the science stream.
6. In the science stream, the number of students who like coffee is equal to the number of students who like Tea.

Question 89: What is the number of students in science stream?

Solution:

Among those who pursue medical, 25% like only coffee.

Let the number of students in the medical stream be 4a, then the number of students who like only coffee = a

A total of 250 students like tea.
Science has 50 more tea lovers than Medical.

The number of tea lovers in Science = x

Then the number of tea lovers in Medical= x-50

x+x-50= 250

x= 150

In the science stream, the number of students who like coffee is equal to the number of students who like Tea.

The number of coffee lovers in the Science stream = x=150

An equal number of students do not like both tea or coffee in both the streams.

Number of students who does not like both tea and coffee= 4a – 100 – a= 3a -100

The number of students who like both tea and coffee in medical is 30 less than that in the science stream.

Let the number of students in Medical who like both tea and coffee = b

The number of students in Science who like both tea and coffee = b+30

Total number of students= 500

Medical + Science =500

4a     +   150+150- (b+30) + 3a -100=500

7a-b = 330………(1)

Number of students who like coffee= 220

150+a+b= 220

a+b = 70……….(2)

a= 50 & b= 20

Number of students in science stream= 100+100+50+50 =300

Question 90: What is the number of students who neither like tea nor coffee?

Solution:

Among those who pursue medical, 25% like only coffee.

Let the number of students in the medical stream be 4a, then the number of students who like only coffee = a

A total of 250 students like tea.
Science has 50 more tea lovers than Medical.

The number of tea lovers in Science = x

Then the number of tea lovers in Medical= x-50

x+x-50= 250

x= 150

In the science stream, the number of students who like coffee is equal to the number of students who like Tea.

The number of coffee lovers in the Science stream = x=150

An equal number of students do not like both tea or coffee in both the streams.

Number of students who does not like both tea and coffee= 4a – 100 – a= 3a -100

The number of students who like both tea and coffee in medical is 30 less than that in the science stream.

Let the number of students in Medical who like both tea and coffee = b

The number of students in Science who like both tea and coffee = b+30

Total number of students= 500

Medical + Science =500

4a + 150+150- (b+30) + 3a -100=500

7a-b = 330………(1)

Number of students who like coffee= 220

150+a+b= 220

a+b = 70……….(2)

a= 50 & b= 20

The number of students who like both tea and coffee= 50+50=100

Question 91: What is the number of students who like both tea and coffee?

Solution:

Among those who pursue medical, 25% like only coffee.

Let the number of students in the medical stream be 4a, then the number of students who like only coffee = a

A total of 250 students like tea.
Science has 50 more tea lovers than Medical.

The number of tea lovers in Science = x

Then the number of tea lovers in Medical= x-50

x+x-50= 250

x= 150

In the science stream, the number of students who like coffee is equal to the number of students who like Tea.

The number of coffee lovers in the Science stream = x=150

An equal number of students do not like both tea or coffee in both the streams.

Number of students who does not like both tea and coffee= 4a – 100 – a= 3a -100

The number of students who like both tea and coffee in medical is 30 less than that in the science stream.

Let the number of students in Medical who like both tea and coffee = b

The number of students in Science who like both tea and coffee = b+30

Total number of students= 500

Medical + Science =500

4a + 150+150- (b+30) + 3a -100=500

7a-b = 330………(1)

Number of students who like coffee= 220

150+a+b= 220

a+b = 70……….(2)

a= 50 & b= 20

The number of students who like both tea and coffee= 50+20=70

Question 92: How many students in science stream like only tea?

Solution:

Among those who pursue medical, 25% like only coffee.

Let the number of students in the medical stream be 4a, then the number of students who like only coffee = a

A total of 250 students like tea.
Science has 50 more tea lovers than Medical.

The number of tea lovers in Science = x

Then the number of tea lovers in Medical= x-50

x+x-50= 250

x= 150

In the science stream, the number of students who like coffee is equal to the number of students who like Tea.

The number of coffee lovers in the Science stream = x=150

An equal number of students do not like both tea or coffee in both the streams.

Number of students who does not like both tea and coffee= 4a – 100 – a= 3a -100

The number of students who like both tea and coffee in medical is 30 less than that in the science stream.

Let the number of students in Medical who like both tea and coffee = b

The number of students in Science who like both tea and coffee = b+30

Total number of students= 500

Medical + Science =500

4a + 150+150- (b+30) + 3a -100=500

7a-b = 330………(1)

Number of students who like coffee= 220

150+a+b= 220

a+b = 70……….(2)

a= 50 & b= 20

Instructions

2 friends Arun and Bipin are playing a game. A certain number of coins are lying on a table. Each player must pick up at least 1 coin and at most 15 coins in each turn. No player can skip his turn. The player who picks up the last coin on the table loses the game. Assume that both the players play logically.

Question 93: Arun and Bipin decide that Arun will always pick up the first coin. Further, Bipin offers Arun 4 choices for the total number of coins. Which of the following number of coins should Arun not agree to ?

a) 1347

b) 1495

c) 1249

d) 1177

Solution:

Arun will always pick up the first coin.
Let us approach the problem from the last step.
If 1 coin is left on the table, Arun will lose (Since he has to pick up the last coin)
If 2 coins are left on the table, Arun will win.
If 3 coins are left on the table, Arun will win.
Extending the logic, if 15 coins are present on the table, Arun will win.
If 16 coins are present on the table, Arun will pick up 15 coins and let Bipin pick up the last coin.
If 17 coins are present on the table, Arun cannot win.
Irrespective of the number of coins that Arun picks, Bipin will pick up some coins such that the sum of the coins picked up by Arun and Bipin add up to 16, leaving Arun to pick up the last coin.

Therefore, if the total number of coins are of the form 16k + 1, Arun will end up losing the game.
Let us evaluate the options one by one.

1347 = 1346 + 1.
1346 is not divisible by 16. Therefore, Arun will win a game that starts with 1347 coins.

1495 = 1494 + 1.
1494 is not divisible by 16.

1177 = 1176 + 1
1176 is not divisible by 16.

1249 = 1248 + 1
1248 is divisible by 16. Therefore, if the game starts with 1249 coins, it will be impossible for Arun to win. Therefore, Arun should not agree to have 1249 coins and hence, option C is the right answer.

Question 94: If it is known that Bipin was the first one to pick up the coin and he picked up 9 coins, which of the following cannot be the total number of coins?

a) 250

b) 330

c) 282

d) 500

Solution:

We know that both the players play logically.
So, if Bipin picked up 9 coins, it should have been done to bring the number of coins to the format 16k + 1.
Therefore, we know that the total number of coins must be of the form 16k + 1 + 9 = 16k + 10.

250 = 240 + 10.
240 is divisible by 16. Therefore, the total number of coins could have been 250.

330 = 320 + 10.
320 is divisible by 16. Therefore, the total number of coins could have been 320.

282 = 272 + 10
272 is divisible 16. Therefore, the total number of coins could have been 282.

500 = 490 + 10
490 is not divisible by 16. Therefore, the game could not have started with 500 coins and hence, option D is the right answer.

Question 95: The players agree to rework the rules of the game. The last player to pick up the coin from the table wins the game. Bipin will be the first person to pick up the coin. If all other rules remain unchanged and if it is known that Arun won the game, then the number of coins on the table initially could be all but

a) 3792

b) 3888

c) 3824

d) 3144

Solution:

We know that the last person to pick up the coin wins the game.
If 1 coin is left, the person will pick up the coin and win the game.
If 2 coins are left, the person will pick up the coin and win the game.
Extending the logic, if 15 coins are left, the person will pick up the coin and win the game.
If 16 coins are left, the person can pick up a maximum of 15 coins and a minimum of 1 coin. Irrespective of the number of coins picked up by the person, he will end up losing the game.
However, if 17 coins are left, he can pick up 1 coin, leaving 16 coins on the table. Now, the other person will end up losing the game.

As we can see, the player who faces a total number of coins of the form 16k ends up losing the game. Bipin is the first person to pick up the coin but ends up losing the game. Therefore, the total number of coins initially should have been of the form 16k.

3792, 3888 and 3824 are divisible by 16. Only 3144 is not divisible by 16. Therefore, 3144 could not have been the total number of coins at the beginning of the game and hence, option D is the right answer.

Question 96: The players agree to rework the rules of the game. Each player can pick up a minimum of 3 coins and a maximum of 10 coins. The last player to pick up the coin loses the game. However, if the last player does not have enough coins to pick up (say 1 or 2 coins are remaining on the table), the other player loses the game. Further, it is known that Bipin picked up the first coin. If it is known that Arun wins the game, which of the following could not be the number of coins on the table initially?

a) 1276

b) 1277

c) 1278

d) 1279

Solution:

Let us start solving from the last step. Let us consider 2 person A and B. Let us assume that it is the turn of A to pick up the coin.
If 1 coin was left on the table, A would have won.
If 2 coins were left on the table, A would have won.
If 3 coins were left on the table, A would have lost.
If 4 coins were left on the table, A would have lost.
If 5 coins were left on the table, A would have lost.
If 6 coins were present on the table, A would have picked up 3 coins, forcing B to pick up the remaining 3. A would have won in this case.
If 7 coins were present on the table, A would have picked up 4 coins and won the game.

Extending the logic, we can say that A would have won till 13 coins were left on the table.
If 14 coins were left on the table, A would have picked up 10 coins, forcing B to lose. A would have won in this case.
If 15 coins were left on the table, A would have won.
If 16 coins were left on the table, A would have picked up 10 coins leaving 6 coins on the table. B would have picked up 3 coins, making A lose. Had A picked 3 coins, B would have picked 10 coins. Irrespective of the number of coins that A picks, B will pick up a certain number of coins such that the sum of the number of coins picked up by A and B equals 13.
If 17 coins were left on the table, A would have lost.
If 18 coins were left on the table, A would have lost.
If 19 coins were left on the table, A would have picked up 3 coins leaving 16 coins on the table. As we have seen above, the person whose turn it is to pick coins when 16 coins are left on the table will end up losing the game.

The cycle repeats for every 13 values. As we can see, A will end up losing the game when the total number of coins are of the form 13k + 3, 13k + 4 and 13k + 5.

Bipin starts the game and he ends up losing. Therefore, the total number of coins initially must have been of the form 13k + 3, 13k + 4 or 13k + 5.
1277 can be expressed as 13*98 + 3.
1278 can be expressed as 13*98 + 4.
1279 can be expressed as 13*98 + 5.

As we can see, 1276 cannot be expressed as 13k + 3, 13k + 4 or 13k + 5.Therefore, option A is the right answer.

Instructions

Ajay bought six dolls(A, B, C, D, E, F) for his daughter Sara as birthday gifts for different birthdays. The very first gift was given to her when she was 2 years old and the last one on the tenth birthday. By the time she received her last gift, she has lost three of them.
The following information is also known about them
2)Both the dolls given on the third and sixth birthdays were lost.
3)Sara did not lose F but she lost another one in the same year in which she received F.
4)D was gifted three years after B was gifted and exactly one of them was lost.
5)At least one gift was given in every two years and she did not lose her first gift.

Question 97: Which of the following were lost?

a) B

b) C

c) E

d) More than one of the above.

Solution:

From the given data it is known that she have received gifts on the second, third, sixth and the tenth birthdays.
From condition 4, we can conclude that she must have got one gift on her eighth birthday as she needs to get one in every two days. She must have one gift on her fourth or fifth birthday.

From 2 we can conclude that she have lost the gift received on third and sixth birthdays.
From statement 4, she can get D on her fifth or sixth or eighth birthday. Let us assume that she received D on sixth birthday and B on third. Then she must have lost both B and D and that violates condition 4.

Let us assume that she had received D on eighth and B on the sixth birthday. Then C and F cannot be received with a gap of two years. Hence, condition 1 is violated.
Hence, she received D on her fifth birthday and B on her second birthday.
As she did lose the gift given on her second birthday, she must have lost D.

Sara can receive F either on Sixth or the Eighth birthday. But it is given that she did not lose F. He she must have received F on her Eighth and C on her tenth birthday. As there is no other information given about A and E, Sara must have received them on third and sixth but the exact year cannot be determined.

The following is the final tabulated data.

She lost A, D and E.

Question 98: F was gifted on which birthday?

a) 6th

b) 4th

c) 5th

d) 8th

Solution:

From the given data it is known that she have received gifts on the second, third, sixth and the tenth birthdays.
From condition 4, we can conclude that she must have got one gift on her eighth birthday as she needs to get one in every two days. She must have one gift on her fourth or fifth birthday.

From 2 we can conclude that she have lost the gift received on third and sixth birthdays.
From statement 4, she can get D on her fifth or sixth or eighth birthday. Let us assume that she received D on sixth birthday and B on third. Then she must have lost both B and D and that violates condition 4.

Let us assume that she had received D on eighth and B on the sixth birthday. Then C and F cannot be received with a gap of two years. Hence, condition 1 is violated.
Hence, she received D on her fifth birthday and B on her second birthday.
As she did lose the gift given on her second birthday, she must have lost D.

Sara can receive F either on Sixth or the Eighth birthday. But it is given that she did not lose F. He she must have received F on her Eighth and C on her tenth birthday. As there is no other information given about A and E, Sara must have received them on third and sixth but the exact year cannot be determined.

The following is the final tabulated data.

F was gifted on the Eighth birthday.

Question 99: If Sara receives A after she received D then on which birthday did she receive E?

a) Third

b) Fourth

c) Sixth

d) Cannot be determined.

Solution:

From the given data it is known that she have received gifts on the second, third, sixth and the tenth birthdays.
From condition 4, we can conclude that she must have got one gift on her eighth birthday as she needs to get one in every two days. She must have one gift on her fourth or fifth birthday.

From 2 we can conclude that she have lost the gift received on third and sixth birthdays.
From statement 4, she can get D on her fifth or sixth or eighth birthday. Let us assume that she received D on sixth birthday and B on third. Then she must have lost both B and D and that violates condition 4.

Let us assume that she had received D on eighth and B on the sixth birthday. Then C and F cannot be received with a gap of two years. Hence, condition 1 is violated.
Hence, she received D on her fifth birthday and B on her second birthday.
As she did lose the gift given on her second birthday, she must have lost D.

Sara can receive F either on Sixth or the Eighth birthday. But it is given that she did not lose F. He she must have received F on her Eighth and C on her tenth birthday. As there is no other information given about A and E, Sara must have received them on third and sixth but the exact year cannot be determined.

The following is the final tabulated data.

If she received A on sixth, she should have received E on the third birthday.

Question 100: Which among the following was lost by Sara in the year she received F?

a) A

b) B

c) E

d) Cannot be determined.

Solution:

From the given data it is known that she have received gifts on the second, third, sixth and the tenth birthdays.
From condition 4, we can conclude that she must have got one gift on her eighth birthday as she needs to get one in every two days. She must have one gift on her fourth or fifth birthday.

From 2 we can conclude that she have lost the gift received on third and sixth birthdays.
From statement 4, she can get D on her fifth or sixth or eighth birthday. Let us assume that she received D on sixth birthday and B on third. Then she must have lost both B and D and that violates condition 4.

Let us assume that she had received D on eighth and B on the sixth birthday. Then C and F cannot be received with a gap of two years. Hence, condition 1 is violated.
Hence, she received D on her fifth birthday and B on her second birthday.
As she did lose the gift given on her second birthday, she must have lost D.

Sara can receive F either on Sixth or the Eighth birthday. But it is given that she did not lose F. He she must have received F on her Eighth and C on her tenth birthday. As there is no other information given about A and E, Sara must have received them on third and sixth but the exact year cannot be determined.

The following is the final tabulated data.

She must have lost any of the gifts among A, D, E.
Hence cannot be determined is the correct answer.