CAT 2018 Question Paper Slot 1

We have been given details about the quarterly sales figures. Also, we have been given details about the sales figures every month. Some of the data are missing and some additional conditions have been given in the question. Let us try to complete the pie chart as much as possible with the data available to us.

It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression.
We know that the sales in April is 40.
Let the sales in May be 40+x and the sales in June be 40+2x.
We know that the total sales in Q2 is 150.
=> 40 + 40 + x + 40 + 2x = 150
3x = 30
x = 10

Therefore, sales in May 2016 = 40 + 10 = 50
Sales in June 2016 = 40 + 20 = 60

Similarly, it has been given that the sales in October, November, and December 2016 form an arithmetic progression.
Sales in October = 100
Sales in Q4 = 360

Let the sales in November be 100+y and the sales in December be 100+2y.
100 + 100 + y + 100 + 2y = 360
300 + 3y = 360
=> y = 20

Sales in November 2016 = 120 and Sales in December 2016 = 140
Sales in Q1 of 2016 = Sum of the sales in the months of January, February, and March 2016 
                             = 80 + 60 + 100 
                              = 240 

 Sales in Q3 of 2016 = Sum of the sales in the months of July, August, and September 2016                                                   
                              = 75 + 120 + 55       
                              = 250 

 Sales in Q1 of 2017 = 120 + 100 + 160 = 380
Sales in Q2 of 2017 = 65 + 75 + 60 = 200
We know that sales in Q3 of 2017 = 220 

Let the sales in August of 2017 be ‘a’.
60 + 70 + a = 220
=> a = 90
Sales in August 2017 = 90
We know that sales in Q4 of 2017 = 500
Let the sales in December of 2017 be ‘d’.
150 + 170 + d = 500
=> d = 180 

 Sales in December 2017 = 180

Sales in December 2016 = 140
Sales in December 2017 = 180

Percentage change = (180-140)/140 = 40/140 = 28.57%

Therefore, option C is the right answer.

We have been given details about the quarterly sales figures. Also, we have been given details about the sales figures every month. Some of the data are missing and some additional conditions have been given in the question. Let us try to complete the pie chart as much as possible with the data available to us.

It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression.
We know that the sales in April is 40.
Let the sales in May be 40+x and the sales in June be 40+2x.
We know that the total sales in Q2 is 150.
=> 40 + 40 + x + 40 + 2x = 150
3x = 30
x = 10

Therefore, sales in May 2016 = 40 + 10 = 50
Sales in June 2016 = 40 + 20 = 60

Similarly, it has been given that the sales in October, November, and December 2016 form an arithmetic progression.
Sales in October = 100
Sales in Q4 = 360

Let the sales in November be 100+y and the sales in December be 100+2y.
100 + 100 + y + 100 + 2y = 360
300 + 3y = 360
=> y = 20

Sales in November 2016 = 120 and Sales in December 2016 = 140
Sales in Q1 of 2016 = Sum of the sales in the months of January, February, and March 2016 
                             = 80 + 60 + 100 
                              = 240 

 Sales in Q3 of 2016 = Sum of the sales in the months of July, August, and September 2016                                                   
                              = 75 + 120 + 55       
                              = 250 

 Sales in Q1 of 2017 = 120 + 100 + 160 = 380
Sales in Q2 of 2017 = 65 + 75 + 60 = 200
We know that sales in Q3 of 2017 = 220 

Let the sales in August of 2017 be ‘a’.
60 + 70 + a = 220
=> a = 90
Sales in August 2017 = 90
We know that sales in Q4 of 2017 = 500
Let the sales in December of 2017 be ‘d’.
150 + 170 + d = 500
=> d = 180 

 Sales in December 2017 = 180

Among the given 4 options, we have to find the quarter in which the increase in sale from the previous quarter was the highest. 
Q2:
Sales in 2017 = 200
Sales in 2016 = 150

Q1:
Sales in 2017 = 380
Sales in 2016 = 240

Q3:
Sales in 2017 = 220
Sales in 2016 = 250

Q4:
Sales in 2017 = 500
Sales in 2016 = 360

We can eliminate Q3 since the sales has decreased. 
Growth in Q2 sales = 50/150 = 1/3 = 33.33%

Growth in Q1 sales = (380-240)/240 = 140/240 = 58.33%

Growth in Q4 sales = (500-360)/360 = 140/360 

140/240 > 140/360

Therefore, Q1 has recorded the highest growth in sales and hence, option B is the right answer.

We have been given details about the quarterly sales figures. Also, we have been given details about the sales figures every month. Some of the data are missing and some additional conditions have been given in the question. Let us try to complete the pie chart as much as possible with the data available to us.

It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression.
We know that the sales in April is 40.
Let the sales in May be 40+x and the sales in June be 40+2x.
We know that the total sales in Q2 is 150.
=> 40 + 40 + x + 40 + 2x = 150
3x = 30
x = 10

Therefore, sales in May 2016 = 40 + 10 = 50
Sales in June 2016 = 40 + 20 = 60

Similarly, it has been given that the sales in October, November, and December 2016 form an arithmetic progression.
Sales in October = 100
Sales in Q4 = 360

Let the sales in November be 100+y and the sales in December be 100+2y.
100 + 100 + y + 100 + 2y = 360
300 + 3y = 360
=> y = 20

Sales in November 2016 = 120 and Sales in December 2016 = 140
Sales in Q1 of 2016 = Sum of the sales in the months of January, February, and March 2016 
                             = 80 + 60 + 100 
                              = 240 

 Sales in Q3 of 2016 = Sum of the sales in the months of July, August, and September 2016                                                   
                              = 75 + 120 + 55       
                              = 250 

 Sales in Q1 of 2017 = 120 + 100 + 160 = 380
Sales in Q2 of 2017 = 65 + 75 + 60 = 200
We know that sales in Q3 of 2017 = 220 

Let the sales in August of 2017 be ‘a’.
60 + 70 + a = 220
=> a = 90
Sales in August 2017 = 90
We know that sales in Q4 of 2017 = 500
Let the sales in December of 2017 be ‘d’.
150 + 170 + d = 500
=> d = 180 

 Sales in December 2017 = 180

Q2 of 2017:

Sales in Q2 of 2017 = 200
Sales in Q1 of 2017 = 380
% decrease = 180/380

Q4 of 2017:
We can eliminate this option since the sales has increased in Q4 of 2017 as compared to the previous quarter.

Q2 of 2016:
Sales in Q2 of 2016 = 150
Sales in Q1 of 2016 = 240
% decrease = 90/240

Q1 of 2017:
Sales in Q1 of 2017 has increased as compared to sales in the previous quarter. We can eliminate this option as well. 

180/380 is very close to 50%. 90/240 is closer to 33.33%. Therefore, option A is the right answer. 

We have been given details about the quarterly sales figures. Also, we have been given details about the sales figures every month. Some of the data are missing and some additional conditions have been given in the question. Let us try to complete the pie chart as much as possible with the data available to us.

It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression.
We know that the sales in April is 40.
Let the sales in May be 40+x and the sales in June be 40+2x.
We know that the total sales in Q2 is 150.
=> 40 + 40 + x + 40 + 2x = 150
3x = 30
x = 10

Therefore, sales in May 2016 = 40 + 10 = 50
Sales in June 2016 = 40 + 20 = 60

Similarly, it has been given that the sales in October, November, and December 2016 form an arithmetic progression.
Sales in October = 100
Sales in Q4 = 360

Let the sales in November be 100+y and the sales in December be 100+2y.
100 + 100 + y + 100 + 2y = 360
300 + 3y = 360
=> y = 20

Sales in November 2016 = 120 and Sales in December 2016 = 140
Sales in Q1 of 2016 = Sum of the sales in the months of January, February, and March 2016 
                             = 80 + 60 + 100 
                              = 240 

 Sales in Q3 of 2016 = Sum of the sales in the months of July, August, and September 2016                                                   
                              = 75 + 120 + 55       
                              = 250 

 Sales in Q1 of 2017 = 120 + 100 + 160 = 380
Sales in Q2 of 2017 = 65 + 75 + 60 = 200
We know that sales in Q3 of 2017 = 220 

Let the sales in August of 2017 be ‘a’.
60 + 70 + a = 220
=> a = 90
Sales in August 2017 = 90
We know that sales in Q4 of 2017 = 500
Let the sales in December of 2017 be ‘d’.
150 + 170 + d = 500
=> d = 180 

 Sales in December 2017 = 180

March of 2017:
Sales in March of 2017 = 160
Sales in February of 2017 = 100
% increase = 60/100 = 60%

October of 2017:
Sales in October of 2017 = 150
Sales in September of 2017 = 70
As we can see, the sales has increased by more than 100%.

March of 2016:
Sales in March of 2016 = 100
Sales in February of 2016 = 60
% increase in sales is less than 100%.

October of 2016:
Sales in October of 2016 = 100
Sales in September of 2016 = 55
% increase is less than 100%

As we can see, the percentage increase in sale as compared to the previous month was highest in October of 2017 among the given options. Therefore, option B is the right answer. 

Let us draw a table according to the information given. 

It is given that the numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee. Let '4x' be the number of bureaucrats in Administration committee. 

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees. Let us assume that 'y' is the number of educationalists in the research committee and 'd' be the difference in the number of educationalists in Research and teaching committees. 

60% of the politicians are in the administration committee, and 20% are in the teaching committee. Let '5z' be the number of total number of politicians. 

We can say that

$$\Rightarrow$$ 10x+3y+5z = 24

We can see that each of x, y and z has to a natural number integer. If x > 1, then both y and z can't take any natural number.

Hence, we can say that x = 1.

At x = 1, 3y+5z = 14. If y = 1 or 2, Z is not an integer. 

At x = 1 and y = 3, z = 1 which is the only possible solution.

We can see that 'd' can assume two possible values. d = 1 or 2.

Let us check the option one by one.

Option A: In the teaching committee the number of educationalists is equal to the number of politicians. We can see that in the teaching committee the number of educationalists can be equal to the number of politicians when both the numbers are '1'. Hence, this statement can be correct. 

Option B: In the administration committee the number of bureaucrats is equal to the number of educationalists. We can see that in the administration committee the number of bureaucrats can be equal to the number of educationalists when both the numbers are '4'. Hence, this statement can be correct. 

Option C: The size of the research committee is less than the size of the teaching committee. We can see the maximum size of teaching committee can be '6' which is less than the size of the research committee. Hence, the sentence is incorrect.

Option D: The size of the research committee is less than the size of the administration committee. We can see the minimum size of Administration committee can be '11' which is more than the size of the research committee. Hence, this statement is correct. 

Therefore, we can say that option C is the correct answer. 

Let us draw a table according to the information given. 

It is given that the numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee. Let '4x' be the number of bureaucrats in Administration committee. 

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees. Let us assume that 'y' is the number of educationalists in the research committee and 'd' be the difference in the number of educationalists in Research and teaching committees. 

60% of the politicians are in the administration committee, and 20% are in the teaching committee. Let '5z' be the number of total number of politicians. 

We can say that

$$\Rightarrow$$ 10x+3y+5z = 24

We can see that each of x, y and z has to a natural number integer. If x > 1, then both y and z can't take any natural number.

Hence, we can say that x = 1.

At x = 1, 3y+5z = 14. If y = 1 or 2, Z is not an integer. 

At x = 1 and y = 3, z = 1 which is the only possible solution.

We can see that 'd' can assume two possible values. d = 1 or 2.

From the table, we can see that the number of bureaucrats in the administration committee = 4. 

Let us draw a table according to the information given. 

It is given that the numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee. Let '4x' be the number of bureaucrats in Administration committee. 

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees. Let us assume that 'y' is the number of educationalists in the research committee and 'd' be the difference in the number of educationalists in Research and teaching committees. 

60% of the politicians are in the administration committee, and 20% are in the teaching committee. Let '5z' be the number of total number of politicians. 

We can say that

$$\Rightarrow$$ 10x+3y+5z = 24

We can see that each of x, y and z has to a natural number integer. If x > 1, then both y and z can't take any natural number.

Hence, we can say that x = 1.

At x = 1, 3y+5z = 14. If y = 1 or 2, Z is not an integer. 

At x = 1 and y = 3, z = 1 which is the only possible solution.

We can see that 'd' can assume two possible values. d = 1 or 2.

From the table, we can see that the number of educationalists in the research committee = 3. 

Let us draw a table according to the information given. 

It is given that the numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee. Let '4x' be the number of bureaucrats in Administration committee. 

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees. Let us assume that 'y' is the number of educationalists in the research committee and 'd' be the difference in the number of educationalists in Research and teaching committees. 

60% of the politicians are in the administration committee, and 20% are in the teaching committee. Let '5z' be the number of total number of politicians. 

We can say that

$$\Rightarrow$$ 10x+3y+5z = 24

We can see that each of x, y and z has to a natural number integer. If x > 1, then both y and z can't take any natural number.

Hence, we can say that x = 1.

At x = 1, 3y+5z = 14. If y = 1 or 2, Z is not an integer. 

At x = 1 and y = 3, z = 1 which is the only possible solution.

We can see that 'd' can assume two possible values. d = 1 or 2.

From the table, we can not uniquely determine the size of the teaching committee. Hence, option A is the correct answer. 

It is given that a satellite serving either B, or C, or S does not serve O. So we can say that it's basically 3 satellites broadcasting (B), communication (C), surveillance (S) which can have intersections. Those satellites which are not part of any category are placed in others. We can draw the Venn diagram as follows. 

1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

Let '10x' be the number of satellites exclusively serving B. Then, the number of satellites exclusively serving C and S = 0.30*10x = 3x

Let 'y' be the number of satellites serving others(O).

Let 'z' be the number of satellites serving B, C but not S. Since the numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. Therefore, we can can say that number of satellites serving B, S but not C = z.

It is given that 

$$\Rightarrow$$ 10x+2z+2y+6x+100 = 1600

$$\Rightarrow$$ 8x+z+y = 750 ... (1)

The numbers of satellites serving B, C, and S (though maybe not exclusively) are in the ratio 2:1:1.

$$\Rightarrow$$ $$\dfrac{10x+2z+100}{z+100+3x+y} = \dfrac{2}{1}$$

$$\Rightarrow$$ $$10x+2z+100=2(z+100+3x+y)$$

$$\Rightarrow$$ $$4x=100+2y$$

$$\Rightarrow$$ $$2x=50+y$$

$$\Rightarrow$$ $$y=2x-50$$   ... (2)

We can substitute this in equation (1)

$$\Rightarrow$$ 8x+z+2x - 50 = 750

$$\Rightarrow$$ z = 800 - 10x    ... (3)

Let us define boundary condition for x, 

$$\Rightarrow$$ 2x - 50 $$\geq$$ 0

$$\Rightarrow$$ x $$\geq$$ 25

Also,  800 - 10x $$\geq$$ 0

$$\Rightarrow$$ x $$\leq$$ 80

Therefore, we can say that x $$\epsilon$$ [25, 80].

The number of satellites serving C = 800 - 10x + 100 + 3x + 2x - 50 = 850 - 5x

At x = 25, The number of satellites serving C = 850 - 5x = 850 - 5*25 = 725 

At x = 80, The number of satellites serving C = 850 - 5x = 850 - 5*80 = 450

Hence, we can say that the number of satellites serving C must be between 450 and 725. Hence, option C is the correct answer. 

It is given that a satellite serving either B, or C, or S does not serve O. So we can say that it's basically 3 satellites broadcasting (B), communication (C), surveillance (S) which can have intersections. Those satellites which are not part of any category are placed in others. We can draw the Venn diagram as follows. 

1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. 2. The number of satellites serving all three of B, C, and S is 100. 3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B. 4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

Let '10x' be the number of satellites exclusively serving B. Then, the number of satellites exclusively serving C and S = 0.30*10x = 3x

Let 'y' be the number of satellites serving others(O).

Let 'z' be the number of satellites serving B, C but not S. Since the numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. Therefore, we can can say that number of satellites serving B, S but not C = z.

It is given that 

$$\Rightarrow$$ 10x+2z+2y+6x+100 = 1600

$$\Rightarrow$$ 8x+z+y = 750 ... (1)

The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.

$$\Rightarrow$$ $$\dfrac{10x+2z+100}{z+100+3x+y} = \dfrac{2}{1}$$

$$\Rightarrow$$ $$10x+2z+100=2(z+100+3x+y)$$

$$\Rightarrow$$ $$4x=100+2y$$

$$\Rightarrow$$ $$2x=50+y$$

$$\Rightarrow$$ $$y=2x-50$$   ... (2)

We can substitute this in equation (1)

$$\Rightarrow$$ 8x+z+2x - 50 = 750

$$\Rightarrow$$ z = 800 - 10x    ... (3)

Let us define boundary condition for x, 

$$\Rightarrow$$ 2x - 50 $$\geq$$ 0

$$\Rightarrow$$ x $$\geq$$ 25

Also,  800 - 10x $$\geq$$ 0

$$\Rightarrow$$ x $$\leq$$ 80

Therefore, we can say that x $$\epsilon$$ [25, 80].

The number of satellites serving B exclusively = 10x. This will be the minimum when 'x' is the minimum. 

At x$$_{min}$$ = 25, The number of satellites serving B exclusively  = 10*25 =250. Hence, option A is the correct answer.