Is Quant Based DI an important topic in XAT?

Yes, Quant Based DI is an important topic in the XAT DILR section. It combines data interpretation with quantitative concepts, requiring candidates to apply mathematical reasoning alongside standard data analysis skills.

What types of Quant Based DI questions appear in XAT?

XAT Quant Based DI questions involve data sets where answering questions requires the application of mathematical concepts such as algebraic equations, number theory, probability, or logical quantitative constraints embedded within a DI framework.

How can I prepare for Quant Based DI in XAT?

Candidates should strengthen their quantitative foundations alongside DI skills. Practise sets that require applying mathematical formulas or concepts to interpret and analyse data. Developing comfort with integrating quantitative calculations into a DI context is essential for this topic.

Are Quant Based DI questions difficult in XAT?

Quant Based DI questions are moderate to difficult in XAT. Their dual-skill requirement makes them more demanding than standard DI questions, and they can be particularly time-intensive for candidates who are not equally strong in both quantitative ability and data interpretation.

How do Cracku's XAT Quant Based DI questions help in preparation?

Cracku's XAT Quant Based DI questions are curated to match the integrated nature of such questions in the actual XAT exam. The step-by-step solutions clearly demonstrate how quantitative concepts are applied within a DI context, helping candidates develop proficiency in handling these hybrid question types.

XAT Quant Based DI Questions

Instructions

Read the following scenario and answer the THREE questions that follow.
An investment company, WinLose, recruits employees to trade in the share market. For newcomers, they have a one-year probation period. During this period, the employees are given Rs. 1 lakh per month to invest the way they see fit. They are evaluated at the end of every month, using the following criteria:
1. If the total loss in any span of three consecutive months exceeds Rs. 20,000, their services are terminated at the end of that 3-month period,
2. If the total loss in any span of six consecutive months exceeds Rs. 10,000, their services are terminated at the end of that 6-month period.
Further, at the end of the 12-month probation period, if there are losses on their overall investment, their services are terminated.
Ratan, Shri, Tamal and Upanshu started working for WinLose in January. Ratan was terminated after 4 months, Shri was terminated after 7 months, Tamal was terminated after 10 months, while Upanshu was not terminated even after 12 months. The table below, partially, lists their monthly profits (In Rs. ‘000) over the 12-month period, where x, y and z are masked information.
Note:
1. A negative profit value indicates a loss.
2. The value in any cell is an integer.
Illustration: As Upanshu is continuing after March, that means his total profit during January-March (2z+2z+0) ≥ ─ Rs.20,000. Similarly, as he is continuing after June, his total profit during January-June ≥ ─ Rs.10,000, as well as his total profit during April-June ≥ ─ Rs.20,000.

Question 1

What BEST can be said about the value of x?

x is a non-negative even number less than or equal to 10

x is an even number between 10 and 14, both inclusive

x is a positive even number

x is a non-negative even number less than or equal to 14

x = 10

Solution

As Ratan was terminated after 4 months, he did not register a loss of more than 20000 at the end of 3 months but registered a loss of more than 20000 in 2nd, 3rd and 4th month. This implies: x - 30 >= -20 and x - 35 < -20 ==> 15 > x >= 10. As x is an integer and x/2 must also be an integer x could be 10, 12 or 14.

For Shri, as he was terminated after 7 months, his loss for the first 6 months did not exceed 10000. It implies that x + 1.5y - 65 >= -10 ==> x+1.5y >= 55.
Also, for the first 6 months, his loss did not exceed 20000 for any group of 3 consecutive months. It implies that: x+y-25 >= -20 ==> x+y >= 5
y - 45 >= -20 ==> y >= 25 and 0.5y - 45 >= -20 ==> y >= 50.
Now, he could be terminated for 2 reasons i.e. 1. his loss for May, June, July exceeded 20000 OR 2. his loss for last 6 months exceeded 10000.
Case 1: 0.5y - 50 < -20 ==> y < 60.
Case 2: 1.5y - 95 < -10 ==> y < 56.667.
As y is also even, possible values for y are 50, 52, 54, 56 and 58.

As Tamal was terminated after 10 months, in ay consecutive period of 3 months for the first 9 months, his loss did not exceed 20000 and for any consecutive period of 6 months, his loss did not exceed 10000.
This implies: 0.5x + y + x >= -20 ==> 1.5x + y >= -20.
There are 2 cases for Tamal also.
Case 1: 5 + 10 - z < -20 ==> z > 35
Case 2: x - 20 + 20 -z < -10 ==> z -10 > x.

Similarly, for Upanshu, the equations will be that the total profits should exceed 0 : 5z - x + y - 81 >= 0
z -48 >= -20 ==> z >= 28 and y - x - 58 >= -10 ==> y -x >= 48. This is only possible when x = 10 and y = 58.

So, value for x is 10 i.e. Option E.

Instructions

Question 2

What BEST can be said about the value of y?

y = 58

y = 56 or 58

y is an even number less than 60

y = 56

y is an even number greater than 50

Solution

So, value of y is 58 i.e. Option A.

Instructions

Question 3

What BEST can be said about the profit generated by Upanshu at the end of his probation period?

At least Rs. 107,000

At least Rs. 147,000

Rs. 109,000 exactly

Rs. 107,000 exactly

At most Rs. 110,000

Solution

Profits generated by Upanshu at the end of his probation period are 5z + y-x - 81. Now, z >= 28 and y-x = 48 which gives 5z +y-x >= 188 and 5z +y-x -81 >= 107.
Thus, the correct answer is Option A i.e. atleast 107000.

Instructions

Read the following scenario and answer the THREE questions that follow.

Brijbhushan, a microfinancier, lends money at the rate of Rs.10 per square meter to small farmers at a village. He charges an annual interest rate of 10%. All the farming plots in that village are rectangular, with areas varying between a minimum of 1000 square meters and a maximum of 10,000 square meters.

This year, Brijbhushan has lent money only to five farmers: Aditya, Binod, Chhuttan, Dabloo and Govind. The perimeter of Chhuttan’s plot is 250 meters, with the length and width being at a ratio of 4:1. Aditya’s plot has an area three times the area of Govind’s plot. The area of Aditya’s plot is also the average of the areas of Govind’s plot and Dabloo’s plot. The plots belonging to Aditya, Binod and Dabloo are of the same width, but of different lengths. Moreover, the length of Binod’s plot is the sum of the lengths of Aditya’s plot and Dabloo’s plot.

Question 4

What is the interest owed by Chhuttan to Brijbhushan at the end of the year?

Rs. 2000

Rs. 3500

Rs. 5000

Rs. 4000

Rs. 2500

Video Solution

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 5

What is the MAXIMUM possible value of the total loan Brijbhushan has given to these five farmers?

None of the other options is correct

Rs. 250,000

Rs. 237,500

Rs. 232,500

Rs. 242,500

Video Solution

Solution

Let the name of farmers be A,B,C,D,G for simplicity.

Let the area of G's plot be x $$m^2$$.

Area of A's plot = 3x $$m^2$$

Let area of D's plot be y $$m^2$$.

The area of A's plot is the average of the area of D's and G's plot.

3x = $$\ \frac{\ x+y}{2}$$

y = 5x $$m^2$$

The width of the plot of A,B and D is the same and the lengths are different. Let the length of the plot of A,B and D be l1,l2 and l3 respectively and the width be y.

We got the following table.

From A's plot, $$l1\times\ y=3x$$

$$l1=\ \frac{\ 3x}{y}$$

From D's plot, $$l3\times\ y=5x$$

$$l3=\ \frac{\ 5x}{y}$$

The length of B's plot is the sum of the length of A's plot and D's plot.

l2 = l1 + l3 = $$\frac{\ 8x}{y}$$

Hence, the area of B's plot = $$l2\times\ y=8x$$

The required final table is :

Total area of all plots = 2500 + 17x

Total loan given by Brijbhushan = $$Rs\ \left(2500+17x\right)\times\ 10$$

Since, we have to maximize the loan amount, we have to maximize the value of x.

The largest area of a land = $$10000\ m^2$$ (Given)

Among the farmers, B will have the land with the largest area.

8x = 10000 $$m^2$$

x = 1250 $$m^2$$

Hence, maximum loan given by Brijbhushan = $$Rs\ \left(\left(1250\times\ 17\right)+2500\right)\times\ 10$$ = Rs 237500

$$\therefore\ $$ The required answer is C.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 6

If the width of the Aditya’s plot is 25 meters, what is the MINIMUM possible length of Binod’s plot?

360 meters

320 meters

250 meters

400 meters

280 meters

Solution

Let the name of farmers be A,B,C,D,G for simplicity.

Let the area of G's plot be x $$m^2$$.

Area of A's plot = 3x $$m^2$$

Let area of D's plot be y $$m^2$$.

The area of A's plot is the average of the area of D's and G's plot.

3x = $$\ \frac{\ x+y}{2}$$

y = 5x $$m^2$$

The width of the plot of A,B and D is the same and the lengths are different. Let the length of the plot of A,B and D be l1,l2 and l3 respectively and the width be y.

We got the following table.

From A's plot, $$l1\times\ y=3x$$

$$l1=\ \frac{\ 3x}{y}$$

From D's plot, $$l3\times\ y=5x$$

$$l3=\ \frac{\ 5x}{y}$$

The length of B's plot is the sum of the length of A's plot and D's plot.

l2 = l1 + l3 = $$\frac{\ 8x}{y}$$

Hence, the area of B's plot = $$l2\times\ y=8x$$

The required final table is :

The width of A's plot = 25m

y = 25m

For B's plot, $$l2\times\ 25=8x$$

To minimize the value of l2, we have to minimize the value of x.

Minimum possible area of land among the farmers = 1000 $$m^2$$

Smallest area of land = G's land = x $$m^2$$

x = 1000

Hence, minimum value of l2 = $$\ \frac{\ 8\times\ 1000}{25}=320\ m$$

$$\therefore\ $$ The required answer is B.

Instructions

Read the following scenario and answer the THREE questions that follow.

Comprehension:
GadRev is a firm that reviews different latest gadgets through a team of four reviewers (R1, R2, R3, and R4). Recently the reviewers reviewed four different tech gadgets (A, B, C, and D) on a scale of 1 to 5 (all integer values) where 1 denotes poor and 5 denotes excellent. These review ratings were then tabulated. However, due to a technical glitch, some of these ratings got deleted. The average rating given by each reviewer, and the average rating given to each gadget were earlier communicated to the team management in a separate email and hence can be useful to retrieve the deleted ratings. The available ratings along with the average ratings are represented in the following table:

Question 1

What rating provided by Reviewer R1 to Gadget A can help determining the remaining ratings uniquely?

Solution

The first thing we should do while solving this set is we should replace the average with the sum of the ratings as we have done in the following table:

We know that the ratings can be 1/2/3/4/5 only.

R1 has given 4 and 3 to B and D respectively and the sum of the ratings given by R1 in total is 16.

So, the sum of the ratings given by R1 to A and C is (16-4-3 = 9)

Now, this sum is only possible when he has given 4 to one of them and 5 to another.

Let's make both the cases:

Now, with this, we can derive the score given by R3 to C.

In case 1, it will be (16-4-4-5) = 3

In case 2, it will be (16-5-4-5) = 2

Now, R2 gave 3 and 4 score to A and C respectively. Hence, he would have given a sum of (16 - 3 - 4 = 9) to B and D. It could be possible when he gave 4 to one and 5 to the other one.

Hence, we can make the possible cases and sub cases:

With this only, we can fill the complete table. Hence, we get:

However, Case - 2 is not possible as there no reviewer can give 6 points to any gadget.

Now, according to the question, we have to find what rating R1 give to A so that we can find remaining ratings uniquely.

Now, R1 give 5 ratings according to two cases and 4 rating to A according to the one another.

If R1 gives 5 rating, we still can't figure out the remaining ratings uniquely. Hence, R1 has to give 4 to A in order to find the remaining ratings uniquely.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 2

In how many different ways could Reviewer R2 have rated Gadget B so that the ratings lead to the same averages for the gadgets and the reviewers as shown in the table?

Solution

The first thing we should do while solving this set is we should replace the average with the sum of the ratings as we have done in the following table:

We know that the ratings can be 1/2/3/4/5 only.

R1 has given 4 and 3 to B and D respectively and the sum of the ratings given by R1 in total is 16.

So, the sum of the ratings given by R1 to A and C is (16-4-3 = 9)

Now, this sum is only possible when he has given 4 to one of them and 5 to another.

Let's make both the cases:

Now, with this, we can derive the score given by R3 to C.

In case 1, it will be (16-4-4-5) = 3

In case 2, it will be (16-5-4-5) = 2

Now, R2 gave 3 and 4 score to A and C respectively. Hence, he would have given a sum of (16 - 3 - 4 = 9) to B and D. It could be possible when he gave 4 to one and 5 to the other one.

Hence, we can make the possible cases and sub cases:

With this only, we can fill the complete table. Hence, we get:

However, Case - 2 is not possible as there no reviewer can give 6 points to any gadget.

Now, according to the three possible cases, R2 give either 4 or 5 to B.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 3

How many different valid combinations of the missing ratings are possible?

Solution

The first thing we should do while solving this set is we should replace the average with the sum of the ratings as we have done in the following table:

We know that the ratings can be 1/2/3/4/5 only.

R1 has given 4 and 3 to B and D respectively and the sum of the ratings given by R1 in total is 16.

So, the sum of the ratings given by R1 to A and C is (16-4-3 = 9)

Now, this sum is only possible when he has given 4 to one of them and 5 to another.

Let's make both the cases:

Now, with this, we can derive the score given by R3 to C.

In case 1, it will be (16-4-4-5) = 3

In case 2, it will be (16-5-4-5) = 2

Now, R2 gave 3 and 4 score to A and C respectively. Hence, he would have given a sum of (16 - 3 - 4 = 9) to B and D. It could be possible when he gave 4 to one and 5 to the other one.

Hence, we can make the possible cases and sub cases:

With this only, we can fill the complete table. Hence, we get:

However, Case - 2 is not possible as there no reviewer can give 6 points to any gadget. Hence, there are only three possible combinations.

Instructions

Read the following scenario and answer the THREE questions that follow.

Comprehension:
In an 8-week course, a professor administered a test at the end of each week. Each of the eight tests was scored out of 4 marks, and a student could only receive a non-negative integer score.
Two students, Ravi and Sumana, took the eight tests.
In the first test, Ravi and Sumana scored the same marks. From the second to eighth tests, Ravi scored the exact same non-zero marks. Sumana scored the same marks as Ravi from the fifth test onwards. Ravi’s total marks in the first three tests was the same as Sumana’s total marks in the first two tests. Also, Sumana’s marks in the first test, total marks of the first two tests, and total marks of the eight tests are in a geometric progression.

Question 4

Which of the following CAN be true?

Ravi scored 4 marks in the third test

Sumana scored 2 marks in the first test

Ravi scored 0 marks in the fifth test

Sumana scored 3 marks in the second test

Sumana scored 4 marks in the eighth test

Solution

According to the set, there are 8 tests and each test can have a maximum of 4 marks so any student can get a maximum of 32 marks.

Now, the number of marks received by any student in a test can be 0/1/2/3/4.

We are given information about two students, Ravi and Sumana.

In the first test, they both scored the same marks. Let the marks scored by both is $$x$$.

Further, from the second to eighth tests, Ravi scored the exact same non-zero marks. Let the marks scored by Ravi from second to eighth test is $$y$$ where y can be 1/2/3/4. Also, Sumana scored the same marks as Ravi from the fifth test onwards.

With this, we get the following table:

Ravi’s total marks in the first three tests was the same as Sumana’s total marks in the first two tests.

Let the marks of Sumana in the second test be $$z$$.

We are given that, $$x\ +\ 2y=x+z$$

Therefore, we can say, z = 2y.

We don't have any other information about their marks in specific tests so let the marks of Sumana in test 3 and test 4 be $$a$$ and $$b$$ respectively.

With this, we have:

At last, we are given that Sumana’s marks in the first test, total marks of the first two tests, and total marks of the eight tests are in a geometric progression.

i.e. x, (x + 2y) and (x + a + b + 6y) are in a geometric progression.

With this, we can infer that x can't be 0 as if x is 0, any number multiplied with it will be 0 and (x+2y) and (x + a + b + 6y) can't be in a geometric progression.

Hence, x can be 1 / 2 / 3 / 4.

Let's try to make all possible geometric progression with this information:

1. (1, 3, 9) when x = 1, y = 1 and a + b = 2

2. (2, 4, 8) when x = 2, y = 1 and a + b = 0

3. (2, 6, 8) when x = 2, y = 2 and a + b = 4

4. (4, 8, 16) when x = 4, y = 2 and a + b = 0

In the question, we need to answer which of the mentioned statements can be true:

A: Ravi scored 4 marks in the third test

y can't be 4 in any of the possible case. Hence, the statement can't be true.

B: Sumana scored 2 marks in the first test

Sumana scored x marks in the first test and x can be 2. Hence, this can be true.

C: Ravi scored 0 marks in the fifth test

y can't be 0 as we discussed above. Hence, this statement can't be true.

D: Sumana scored 3 marks in the second test

Sumana scored 2y in the second test and if 2y = 3, y will be a non integer number which is not possible.

E: Sumana scored 4 marks in the eighth test

y can't be 4 in any of the possible case. Hence, this can't be true.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 5

If Ravi scored 4 marks in the first test, how many marks did Sumana score in the third test?

Solution

According to the set, there are 8 tests and each test can have a maximum of 4 marks so any student can get a maximum of 32 marks.

Now, the number of marks received by any student in a test can be 0/1/2/3/4.

We are given information about two students, Ravi and Sumana.

In the first test, they both scored the same marks. Let the marks scored by both is $$x$$.

With this, we get the following table:

Ravi’s total marks in the first three tests was the same as Sumana’s total marks in the first two tests.

Let the marks of Sumana in the second test be $$z$$.

We are given that, $$x\ +\ 2y=x+z$$

Therefore, we can say, z = 2y.

We don't have any other information about their marks in specific tests so let the marks of Sumana in test 3 and test 4 be $$a$$ and $$b$$ respectively.

With this, we have:

At last, we are given that Sumana’s marks in the first test, total marks of the first two tests, and total marks of the eight tests are in a geometric progression.

i.e. x, (x + 2y) and (x + a + b + 6y) are in a geometric progression.

With this, we can infer that x can't be 0 as if x is 0, any number multiplied with it will be 0 and (x+2y) and (x + a + b + 6y) can't be in a geometric progression.

Hence, x can be 1 / 2 / 3 / 4.

Let's try to make all possible geometric progression with this information:

1. (1, 3, 9) when x = 1, y = 1 and a + b = 2

2. (2, 4, 8) when x = 2, y = 1 and a + b = 0

3. (2, 6, 8) when x = 2, y = 2 and a + b = 4

4. (4, 8, 16) when x = 4, y = 2 and a + b = 0

According to the question, if Ravi scored 4 marks in the first test i.e. if x = 4, the possible case is case 4,

Hence, sumana scored $$a$$ marks in the third test and we know that in case 4, a + b = 0. Hence, both $$a$$ and $$b$$ are 0.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 6

If Ravi scored 1 mark in the second test, what is the maximum possible value of Sumana’s total marks in all the eight tests together?

Cannot be uniquely determined from the given information

Solution

According to the set, there are 8 tests and each test can have a maximum of 4 marks so any student can get a maximum of 32 marks.

Now, the number of marks received by any student in a test can be 0/1/2/3/4.

We are given information about two students, Ravi and Sumana.

In the first test, they both scored the same marks. Let the marks scored by both is $$x$$.

With this, we get the following table:

Ravi’s total marks in the first three tests was the same as Sumana’s total marks in the first two tests.

Let the marks of Sumana in the second test be $$z$$.

We are given that, $$x\ +\ 2y=x+z$$

Therefore, we can say, z = 2y.

We don't have any other information about their marks in specific tests so let the marks of Sumana in test 3 and test 4 be $$a$$ and $$b$$ respectively.

With this, we have:

At last, we are given that Sumana’s marks in the first test, total marks of the first two tests, and total marks of the eight tests are in a geometric progression.

i.e. x, (x + 2y) and (x + a + b + 6y) are in a geometric progression.

With this, we can infer that x can't be 0 as if x is 0, any number multiplied with it will be 0 and (x+2y) and (x + a + b + 6y) can't be in a geometric progression.

Hence, x can be 1 / 2 / 3 / 4.

Let's try to make all possible geometric progression with this information:

1. (1, 3, 9) when x = 1, y = 1 and a + b = 2

2. (2, 4, 8) when x = 2, y = 1 and a + b = 0

3. (2, 6, 18) when x = 2, y = 2 and a + b = 4

4. (4, 8, 16) when x = 4, y = 2 and a + b = 0

If Ravi scored 1 mark in the second test i.e. y = 1, it can be either case 1 and case 2.

The total marks scored by Sumana across the tests according to case 1:

x + a + b + 6y = 1 + 2 + 6 = 9

The total marks scored by Sumana across the tests according to case 2:

x + a + b + 6y = 2 + 0 + 6 = 8

Hence, the maximum possible marks that can be scored by Sumana is 9.

Instructions

Read the following scenario and answer the THREE questions that follow.

A T20 cricket match consists of two teams playing twenty overs each, numbered 1 to 20. The runs scored in any over is a non-negative integer. The run rate at the end of any over is the average runs scored up to and including that over, i.e., the run rate at the end of the k-th over is the average number of runs scored in overs numbered 1, 2, …, k, where 1 ≤ k ≤ 20, k a positive integer.

The following table indicates the run rate of a team at the end of some of the overs during a T20 cricket match (correct up to 2 decimal places), where 1 ≤ N - 2 < N + 6 ≤ 20, N a positive integer. It is also known that the team did not score less than 6 runs and more than 15 runs in any over.

Question 1

What is the value of N?

Solution

It is given that the runs in any over is an integer. So, the run rate multiplied by overs should give us an integer.

1 ≤ N - 2 < N + 6 ≤ 20

The range of N is [3, 14].

After N overs, the run rate is given as 7.43. So, 7.43*N has to be an integer. Let's try putting all the possible values N and see where we get an integer.

3*7.43=22.29

4*7.43=29.72

5*7.43=36.15

6*7.43=44.58

7*7.43=52.01

Here, we got an integer; since the run rate is reduced to 2 decimal points, We can not get an exact integer.

So, N can be 7. Let's also check the remaining possible values of N.

8*7.43=59.44

9*7.43=66.87

10*7.43=74.3

11*7.43=81.73

12*7.43= 89.16

13*7.43=96.59

14*7.43=102.02

N can also be 14.

If N=14, the (N+2)*8.11=16*8.11 should also be an integer. But it is 129.76. So, N is not 14. The only possible value is 7.

N-2 --- 8.0 = 5*8=40

N --- 7.43 = 52.01

N+2 --- 8.11 = 9*8.11=72.99

N+4--- 8.45 = 11*8.45 = 92.95

N+6 --- 8.08 = 13*8.08= 105.04

Every number is almost an integer, so N will be 7.

Now, let's calculate the runs scored in each of these overs.

After five overs, the score is 40

After seven overs, the score is 52

So, the sum of scores in 6 and 7 overs is 12. They have to score a minimum of 6 in an over. So, 6 is the sixth over, and 6 in the seventh over.

After the 9th over, the score was 73. They scored 21 in the 8th and 9th over.

After the 11th over, the score was 93, they scored 20 in the 10th and 11th.

After the 13th over, the score was 105, they scored 12 in the 12th and 13th over.

The lowest number of runs is scored in 6, 7, 12, and 13 overs. So, 7 is the correct answer.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 2

In which of these pairs of over numbers, the team could have scored 22 runs in total?

6 and 7

7 and 8

8 and 9

9 and 10

10 and 11

Solution

It is given that the runs in any over is an integer. So, the run rate multiplied by overs should give us an integer.

1 ≤ N - 2 < N + 6 ≤ 20

The range of N is [3, 14].

After N overs, the run rate is given as 7.43. So, 7.43*N has to be an integer. Let's try putting all the possible values N and see where we get an integer.

3*7.43=22.29

4*7.43=29.72

5*7.43=36.15

6*7.43=44.58

7*7.43=52.01

Here, we got an integer; since the run rate is reduced to 2 decimal points, We can not get an exact integer.

So, N can be 7. Let's also check the remaining possible values of N.

8*7.43=59.44

9*7.43=66.87

10*7.43=74.3

11*7.43=81.73

12*7.43= 89.16

13*7.43=96.59

14*7.43=102.02

N can also be 14.

If N=14, the (N+2)*8.11=16*8.11 should also be an integer. But it is 129.76. So, N is not 14. The only possible value is 7.

N-2 --- 8.0 = 5*8=40

N --- 7.43 = 52.01

N+2 --- 8.11 = 9*8.11=72.99

N+4--- 8.45 = 11*8.45 = 92.95

N+6 --- 8.08 = 13*8.08= 105.04

Every number is almost an integer, so N will be 7.

Now, let's calculate the runs scored in each of these overs.

After five overs, the score is 40

After seven overs, the score is 52

So, the sum of scores in 6 and 7 overs is 12. They have to score a minimum of 6 in an over. So, 6 is the sixth over, and 6 in the seventh over.

After the 9th over, the score was 73. They scored 21 in the 8th and 9th over.

After the 11th over, the score was 93; they scored 20 in the 10th and 11th.

After the 13th over, the score was 105; they scored 12 in the 12th and 13th over.

Option A) The sum of scores in 6 and 7 overs is 12.

Option B) The maximum score possible in the 7th and 8th overs is 21 because we know that 6 runs were scored in the 7th over, and the maximum possible score in the 8th over is 15, which makes the sum 21, and it is not sufficient to score 22 runs in the 7th and 8th overs.

Option C) The sum of scores in 8 and 9 overs is 21.

Option D) The sum of scores in the 9 and 10 overs cannot be determined uniquely from the information given, but we know that in the 8 and 9 overs, they scored 21 runs, and in the 10th and 11th overs, they scored 20 runs. So, them scoring 22 runs in 9th and 10th overs is numerically possible and this is the correct option.

Option E) The sum of scores in 10 and 11 overs is 20.

So, only option D could be the answer.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 3

In which of the following over numbers, the team MUST have scored the least number of runs?

Solution

It is given that the runs in any over is an integer. So, the run rate multiplied by overs should give us an integer.

1 ≤ N - 2 < N + 6 ≤ 20

The range of N is [3, 14].

After N overs, the run rate is given as 7.43. So, 7.43*N has to be an integer. Let's try putting all the possible values N and see where we get an integer.

3*7.43=22.29

4*7.43=29.72

5*7.43=36.15

6*7.43=44.58

7*7.43=52.01

Here, we got an integer; since the run rate is reduced to 2 decimal points, We can not get an exact integer.

So, N can be 7. Let's also check the remaining possible values of N.

8*7.43=59.44

9*7.43=66.87

10*7.43=74.3

11*7.43=81.73

12*7.43= 89.16

13*7.43=96.59

14*7.43=102.02

N can also be 14.

If N=14, the (N+2)*8.11=16*8.11 should also be an integer. But it is 129.76. So, N is not 14. The only possible value is 7.

N-2 --- 8.0 = 5*8=40

N --- 7.43 = 52.01

N+2 --- 8.11 = 9*8.11=72.99

N+4--- 8.45 = 11*8.45 = 92.95

N+6 --- 8.08 = 13*8.08= 105.04

Every number is almost an integer, so N will be 7.

Now, let's calculate the runs scored in each of these overs.

After five overs, the score is 40

After seven overs, the score is 52

So, the sum of scores in 6 and 7 overs is 12. They have to score a minimum of 6 in an over. So, 6 is the sixth over, and 6 in the seventh over.

After the 9th over, the score was 73. They scored 21 in the 8th and 9th over.

After the 11th over, the score was 93, they scored 20 in the 10th and 11th.

After the 13th over, the score was 105, they scored 12 in the 12th and 13th over.

We can only determine the scores uniquely of a few overs from the information given, and from the values that are determined, the lowest number of runs is scored in 6, 7, 12, and 13 overs. So, 7 is the correct answer.

Instructions

Read the following scenario and answer the THREE questions that follow.

The upper hinge of a dataset is the median of all the values to the right of the median of the dataset in an ascending arrangement, while the lower hinge of the dataset is the median of all the values to the left of the median of the dataset in the same arrangement.

For example, consider the dataset 4, 3, 2, 6, 4, 2, 7. When arranged in the ascending order, it becomes 2, 2, 3, 4, 4, 6, 7. The median is 4 (the bold value), and hence the upper hinge is the median of 4, 6, 7, i.e., 6. Similarly, the lower hinge is 2.

A student has surveyed thirteen of her teachers, and recorded their work experience (in integer years). Two of the values recorded by the student got smudged, and she cannot recall those values. All she remembers is that those two values were unequal, so let us write them as A and B, where A<B. The remaining eleven values, as recorded, are: 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29. Moreover, the student also remembers the following summary measures, calculated based on all the thirteen values:

Minimum: 2
Lower Hinge: 6.5
Median: 12
Upper Hinge: 21
Maximum: 29

Question 4

Which of the following is a possible value of B?

Solution

The first clue is the minimum value provided:

That is not one of the eleven values that we already have; this must mean that one of the missing values, A or B, must be 2

Since we are given that A is less than B, it must be A, which is equal to 2.

Now, writing down the twelve known numbers,
2, 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

we are also provided that the median is of the data set was 12, that would be possible if the value 12 was shifter from 6th position to 7th position.
This must mean that the other missing value would also be in the first 6 values of the data set, lying in the range $$2\le B\le12$$

Now, narrowing down our focus to the lower hinge, the value is given to be 6.5
The first six values would be 2, 5, 6, 7, 8 and one unknown value

The lower hinge would be 6.5 only if the missing value appears after 7
Narrowing down the possible values of B to be 7, 8, 9, 10, 11, 12

For the first question, of all the given options, only option C lies in this range and hence would be the answer.

Therefore, option C is the correct answer.

Instructions

Read the following scenario and answer the THREE questions that follow.

Minimum: 2
Lower Hinge: 6.5
Median: 12
Upper Hinge: 21
Maximum: 29

Question 5

Based on the information recorded, which of the following can be the average work experience of the thirteen teachers?

12.5

13.5

Solution

The first clue is the minimum value provided:

That is not one of the eleven values that we already have; this must mean that one of the missing values, A or B, must be 2

Since we are given that A is less than B, it must be A, which is equal to 2.

Now, writing down the twelve known numbers,
2, 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

we are also provided that the median of the data set was 12, which would be possible if the value 12 was shifted from the 6th position to the 7th position.
This must mean that the other missing value would also be in the first 6 values of the data set, lying in the range $$2\le B\le12$$

Now, narrowing down our focus to the lower hinge, the value is given to be 6.5
The first six values would be 2, 5, 6, 7, 8 and one unknown value

The lower hinge would be 6.5 only if the missing value appears after 7
Narrowing down the possible values of B to be 7, 8, 9, 10, 11, 12

In order to find the average work-experience of all the teachers, we must add up all the values
adding up the known values we get
2+5+6+7+8+12+16+19+21+21+27+29 = 173

Looking at the options, the total sum of all 13 values must be
A: 156
B: 162.5
C: 169
D: 175.5
E: 182

The sum of 13 values can only be 182 of all the given options, which would be the case when B is 9

Therefore, option E is the correct answer.

Instructions

Read the following scenario and answer the THREE questions that follow.

Minimum: 2
Lower Hinge: 6.5
Median: 12
Upper Hinge: 21
Maximum: 29

Question 6

While rechecking her original notes to re-enter the smudged values of A and B in the records, the student found that one of the eleven recorded work experience values that did not get smudged was recorded wrongly as half of its correct value. After re-entering the values of A and B, and correcting the wrongly recorded value, she recalculated all the summary measures. The recalculated average value was 15. What is the value of B?

Cannot be determined from the given information.

Solution

That is not one of the eleven values that we already have; this must mean that one of the missing values, A or B, must be 2

Since we are given that A is less than B, it must be A, which is equal to 2.

Now, writing down the twelve known numbers,
2, 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

Now, narrowing down our focus to the lower hinge, the value is given to be 6.5
The first six values would be 2, 5, 6, 7, 8 and one unknown value

The lower hinge would be 6.5 only if the missing value appears after 7
Narrowing down the possible values of B to be 7, 8, 9, 10, 11, 12

In the context of the third question, we are given that the average of the 13 values was 15, meaning that the sum of the 13 values was 195
As we calculated previously, the sum with the value we had would be 173 +B

we also have to add the mistaken value once more to this number to get 195
Let's take the unknown number to be U, giving us the equation
173 + B+ U = 195
B+U = 22

Where B must be one of 7, 8, 9, 10, 11 or 12
And U must be one of 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

trying to substitute the values of B into the equation, the only value which yields valid values for both B and U would be 10

Therefore, Option C is the correct answer.

Instructions

Read the following scenario and answer the three questions that follow.

A company awards incentives to its employees for successful project performances. It rates successful project performance in categories A*, A, B, and C. Employees, in solo projects rated A*, A, B, and C, are awarded incentives ₹6 lakh, ₹5 lakh, ₹3 lakh, and ₹1 lakh respectively. When a project has multiple team members, the following scheme is used to award the incentives:

For example, for a project rated A, with three members, the team lead gets ₹4 lakh, and the other team members get ₹2.5 lakh each. A project always has a single team lead.

Six employees: Altaf, Bose, Chakrabarthi, Dipa, Ernie, and Fatima receive a total of ₹45 lakh in incentives by participating in a total of eight different projects that does not involve any other person. Not all six employees are involved in all eight projects.

The following are additionally known about these eight projects:
1. One project involves all six employees. Four projects involve three each, and the rest, two each.
2. Exactly three projects are rated C, for which a total of ₹4.8 lakh is paid.
3. Only one project is rated A*.

Question 1

What BEST is known about the team compositions for the projects rated C?

The three teams have two, three and six members respectively.

All are either two-member or three-member teams.

All are three-member teams.

One is the six-member team, the rest are two-member teams.

All are two-member teams.

Solution

Total percentage incentive when number of team members = 1 = 100%

Total percentage incentive when the number of team members =2 =160%

Total percentage incentive when the number of team members =3=180%

Total percentage incentive when the number of team members =4= 190%

Total percentage incentive when the number of team members >4 = 200%

From 1, Number of people in 8 different projects = 6, 3, 3,3,3, 2,2,2 respectively

From 2, Given, exactly three projects are rated C and 4.8 lakh is paid in total

A minimum of 3 lakhs has to be paid for rating C => 3 *1.6 = 4.8lakhs => All 2 member teams have been rated C

From 3, one project has been rated A*. Let that project be handled by the team of 3 members => Incentives = 180% of 6 = 10.8 lakh

Now remaining 6,3,3,3 should be either rated A or B and the total incentives should be equal to 45 - 10.8-4.8 = 29.4 lakhs

Let us assume 6 has been rated B => Incentives = 200% of 3 = 6 lakhs

The remaining 23.4 lakhs should come from 180%. $$\frac{23.4}{1.8}=13\ lakhs$$

Hence the remaining 3,3,3 can be rated as A, A, B

Hence final ratings are and total payouts are

6 - B - 6lakhs

3- A - 9 lakhs

3-A - 9 lakhs

3-B - 5.4 lakhs

3-A* - 10.8lakhs

2-C - 1.6 lakhs

Instructions

Read the following scenario and answer the three questions that follow.

For example, for a project rated A, with three members, the team lead gets ₹4 lakh, and the other team members get ₹2.5 lakh each. A project always has a single team lead.

Question 2

What BEST is known about the team composition for the project rated A*?

A three-member team

Either a three-member team or the six-member team

A two-member team

Either a two-member team or a three-member team

The six-member team