Instructions
Adriana, Bandita, Chitra, and Daisy are four female students, and Amit, Barun, Chetan, and Deb are four male students. Each of them studies in one of three institutes - X, Y, and Z. Each student majors in one subject among Marketing, Operations, and Finance, and minors in a different one among these three subjects. The following facts are known about the eight students:
1. Three students are from X, three are from Y, and the remaining two students, both female, are from Z.
2. Both the male students from Y minor in Finance, while the female student from Y majors in Operations.
3. Only one male student majors in Operations, while three female students minor in Marketing.
4. One female and two male students major in Finance.
5. Adriana and Deb are from the same institute. Daisy and Amit are from the same institute.
6. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
7. Daisy minors in Operations.
46904.Who are the students from the institute Z?
Chitra and Daisy
Adriana and Bandita
Bandita and Chitra
Adriana and Daisy
Explanation:
There are 8 students in total - 4 male and 4 female. There are 3 institutes X, Y, and Z.
3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject. It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.
Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance. It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance.
Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations.
Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing.
Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z. Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance. We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.Bandita and Chitra are from institute Z. Therefore, option C is the right answer.
There are 8 students in total - 4 male and 4 female. There are 3 institutes X, Y, and Z.
3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject. It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.
Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance. It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance.
Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations.
Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing.
Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z. Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance. We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.Bandita and Chitra are from institute Z. Therefore, option C is the right answer.
46905.Which subject does Deb minor in?
Operations
Finance
Marketing
Cannot be determined uniquely from the given information
Explanation:
There are 8 students in total - 4 male and 4 female. There are 3 institutes X, Y, and Z. 3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject.
It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.
Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance. Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations.
Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing. Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z.
Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance.
We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.Deb minors in Finance. Therefore, option B is the right answer.
There are 8 students in total - 4 male and 4 female. There are 3 institutes X, Y, and Z. 3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject.
It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.
Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance. Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations.
Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing. Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z.
Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance.
We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.Deb minors in Finance. Therefore, option B is the right answer.
46906.Which subject does Amit major in?
Marketing
Operations
Cannot be determined uniquely from the given information
Finance
Explanation:
There are 8 students in total - 4 male and 4 female. There are 3 institutes X, Y, and Z.
3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject.
It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.
Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance.
Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations. Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing.
Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z. Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance. We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.Amit majors in finance. Therefore, option D is the right answer.
There are 8 students in total - 4 male and 4 female. There are 3 institutes X, Y, and Z.
3 students are from institute X, 3 students are from institute Y, and 2 students are from institute Z. No student majors and minors in the same subject.
It has been given that both the students from institute Z are female. Also, it has been given that both the male students from institute Y minor in Finance. Therefore, the third student from institute Y should be female. Institute X should also have 2 male and 1 female student.
Both the male students from Y minor in Finance, while the female student from Y majors in Operations. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
It has been given that one female student and 2 male students major in finance. We know that the male student from Y minors in finance. Therefore, he cannot major in finance. Therefore, both the male students from X should major in finance.
Daisy and Amit are from the same institute. Therefore, Daisy cannot be from institute Z (since Amit is a male student and both the students from Z are female). Daisy minors in operations. The girl from institute Y majors in Operations. Therefore, Daisy cannot be from institute Y as well. Daisy and Amit should be from institute X. 3 female students minor in marketing. Therefore, all girls except Daisy should minor in marketing.
Adriana and Deb are from the same institute. Therefore, both of them should be from institute Y. Bandita and Chitra should be from institute Z. Only one male student majors in Operations. We know that Barun is the student. Two male students major in Finance. We know that Amit and Chetan major in finance. Therefore, Deb should major in Marketing.Amit majors in finance. Therefore, option D is the right answer.
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46907.What is the percentage increase in sales in December 2017 as compared to the sales in December 2016?
38.46
22.22
28.57
50.00
Explanation:
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
$\Rightarrow$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
$\Rightarrow$ 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
$\Rightarrow$d = 180
Sales in December 2017 = 180
Sales in December 2016 = 140
Sales in December 2017 = 180
Percentage change = (180-140)/180 = 40/140 = 28.57%
Therefore, option C is the right answer.Question 52
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
$\Rightarrow$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
$\Rightarrow$ 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
$\Rightarrow$d = 180
Sales in December 2017 = 180
Sales in December 2016 = 140
Sales in December 2017 = 180
Percentage change = (180-140)/180 = 40/140 = 28.57%
Therefore, option C is the right answer.Question 52
46908.In which quarter of 2017 was the percentage increase in sales from the same quarter of 2016 the highest?
Q2
Q1
Q4
Q3
Explanation:
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.$\Rightarrow$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
$\Rightarrow$ 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
$\Rightarrow$ 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
$\Rightarrow$ 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 = 500Sales 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.$\Rightarrow$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
$\Rightarrow$ 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
$\Rightarrow$ 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
$\Rightarrow$ 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 = 500Sales 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.
uring which quarter was the percentage decrease in sales from the previous quarter’s sales the highest?
Q2 of 2017
Q4 of 2017
Q2 of 2016
Q1 of 2017
Explanation:
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.
$\Rightarrow$ 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
$\Rightarrow$ 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
= 250Sales 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
$\Rightarrow$ 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
$\Rightarrow$ 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.
$\Rightarrow$ 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
$\Rightarrow$ 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
= 250Sales 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
$\Rightarrow$ 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
$\Rightarrow$ 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.
uring which month was the percentage increase in sales from the previous month’s sales the highest?
March of 2017
October of 2017
March of 2016
October of 2016
Explanation:
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.
$\Rightarrow$ 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
$\Rightarrow$ 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
$\Rightarrow$ 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
$\Rightarrow$ 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.
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.
$\Rightarrow$ 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
$\Rightarrow$ 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
$\Rightarrow$ 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
$\Rightarrow$ 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.
#pagebreak
Instructions
Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:
- 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.
- 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.
- 60% of the politicians are in the administration committee, and 20% are in the teaching committee.
46911.Based on the given information, which of the following statements MUST be FALSE?
In the teaching committee the number of educationalists is equal to the number of politicians
In the administration committee the number of bureaucrats is equal to the number of educationalists
The size of the research committee is less than the size of the teaching committee
The size of the research committee is less than the size of the administration committee
Explanation:
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 cant take any natural number.
Hence, we can say that x = 1.
At x = 1, 3y+5z = 24. 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 9 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 cant take any natural number.
Hence, we can say that x = 1.
At x = 1, 3y+5z = 24. 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 9 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.
46912.What is the number of bureaucrats in the administration committee?
3
4
5
2
Explanation:
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 cant take any natural number.
Hence, we can say that x = 1.
At x = 1, 3y+5z = 24. 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 cant take any natural number.
Hence, we can say that x = 1.
At x = 1, 3y+5z = 24. 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.
46913.What is the number of educationalists in the research committee?
3
4
5
2
Explanation:
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
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 cant take any natural number.
Hence, we can say that x = 1.
At x = 1, 3y+5z = 24. 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
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 cant take any natural number.
Hence, we can say that x = 1.
At x = 1, 3y+5z = 24. 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.
46914.Which of the following CANNOT be determined uniquely based on the given information?
The size of the teaching committee
The size of the research committee
The total number of bureaucrats in the three committees
The total number of educationalists in the three committees
Explanation:
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,
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 cant take any natural number.
Hence, we can say that x = 1.
At x = 1, 3y+5z = 24. 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.
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,
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 cant take any natural number.
Hence, we can say that x = 1.
At x = 1, 3y+5z = 24. 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.
#pagebreak
Instructions
1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites:
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.
4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.
46915.What best can be said about the number of satellites serving C?
Must be at least 100
Cannot be more than 800
Must be between 450 and 725
Must be between 400 and 800
Explanation:
It is given that a satellite serving either B, or C, or S does not serve O. So we can say that its 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.
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 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 = 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.
10x + 2z + 100 = 2$\Rightarrow$ z + 100 + 3x + y = 1
$\Rightarrow$ 10x + 2z + 100 = 2(z + 100 + 3x + y)
$\Rightarrow$ 4x = 100 + 2y
$\Rightarrow$ 2x = 50 + y? 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$\in$ [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 its 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.
- The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
- The number of satellites serving all three of B, C, and S is 100.
- 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.
- 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 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 = 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.
10x + 2z + 100 = 2$\Rightarrow$ z + 100 + 3x + y = 1
$\Rightarrow$ 10x + 2z + 100 = 2(z + 100 + 3x + y)
$\Rightarrow$ 4x = 100 + 2y
$\Rightarrow$ 2x = 50 + y? 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$\in$ [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.
46916.What is the minimum possible number of satellites serving B exclusively?
250
100
500
200
Explanation:
It is given that a satellite serving either B, or C, or S does not serve O. So we can say that its 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.
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 = 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.
10x + 2z + 100
... (2)
$\Rightarrow$ z + 100 + 3x + y =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
$\Rightarrow x \geq$ 80
Therefore, we can say that x $\epsilon$[25, 80].
The number of satellites serving B exclusively = 10x. This will be minimum when x is minimum.
At $x_{min}$ = 25, The number of satellites serving B exclusively = 10*25 =250. 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 its 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.
- The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
- The number of satellites serving all three of B, C, and S is 100.
- 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.
- 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 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 = 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.
10x + 2z + 100
... (2)
$\Rightarrow$ z + 100 + 3x + y =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
$\Rightarrow x \geq$ 80
Therefore, we can say that x $\epsilon$[25, 80].
The number of satellites serving B exclusively = 10x. This will be minimum when x is minimum.
At $x_{min}$ = 25, The number of satellites serving B exclusively = 10*25 =250. Hence, option A is the correct answer.
46917.If at least 100 of the 1600 satellites were serving O, what can be said about the number of satellites serving S?
At most 475
Exactly 475
No conclusion is possible based on the given information
At least 475
Explanation:
It is given that a satellite serving either B, or C, or S does not serve O. So we can say that its 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.
It is given that a satellite serving either B, or C, or S does not serve O. So we can say that its 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.
46918.If the number of satellites serving at least two among B, C, and S is 1200, which of the following MUST be FALSE?
The number of satellites serving B exclusively is exactly 250
The number of satellites serving B is more than 1000
The number of satellites serving C cannot be uniquely determined
All 1600 satellites serve B or C or S
Explanation:
It is given that a satellite serving either B, or C, or S does not serve O. So we can say that its 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.
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 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 = 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.
10x + 2z + 100 2
$\Rightarrow$z + 100 + 3x + y = 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>=0
$\Rightarrow$x >= 25
Also, 800 - 10x >=0
$\Rightarrow$x<= 80
Therefore, we can say that x $\in$[25, 80].
It is given that the number of satellites serving at least two among B, C, and S is 1200.
$\Rightarrow$800 - 10x + 800 - 10x + 2x -50 + 100 = 1200
$\Rightarrow$18x = 450
$\Rightarrow$x = 25.
We can determine number of satellites in each of the following category. Hence, option C is definitely false. Therefore, we can say that option C is incorrect.
It is given that a satellite serving either B, or C, or S does not serve O. So we can say that its 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.
- The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
- The number of satellites serving all three of B, C, and S is 100.
- 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.
- 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 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 = 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.
10x + 2z + 100 2
$\Rightarrow$z + 100 + 3x + y = 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>=0
$\Rightarrow$x >= 25
Also, 800 - 10x >=0
$\Rightarrow$x<= 80
Therefore, we can say that x $\in$[25, 80].
It is given that the number of satellites serving at least two among B, C, and S is 1200.
$\Rightarrow$800 - 10x + 800 - 10x + 2x -50 + 100 = 1200
$\Rightarrow$18x = 450
$\Rightarrow$x = 25.
We can determine number of satellites in each of the following category. Hence, option C is definitely false. Therefore, we can say that option C is incorrect.
#pagebreak
Instructions
A company administers a written test comprising of three sections of 20 marks each - Data Interpretation (DI), Written English (WE) and General Awareness (GA), for recruitment. A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections. Candidates who score less than 70% marks in two or more sections are disqualified. From among the rest, the four with the highest composite scores are recruited. If four or less candidates qualify, all who qualify are recruited. Ten candid ates appeared for the written test. Their marks in the test are given in the table below. Some marks in the table are missing, but the following facts are known:
- No two candidates had the same composite score.
- Ajay was the unique highest scorer in WE.
- Among the four recruited, Geeta had the lowest composite score.
- Indu was recruited.
- Danish, Harini, and Indu had scored the same marks the in GA.
- Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
46919.Which of the following statements MUST be true?
1. Jatins composite score was more than that of Danish.
2. Indu scored less than Chetna in DI.
3. Jatin scored more than Indu in GA.
1. Jatins composite score was more than that of Danish.
2. Indu scored less than Chetna in DI.
3. Jatin scored more than Indu in GA.
Both 2 and 3
Only 1
Only 2
Both 1 and 2
Explanation:
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI.
Therefore, Jatins composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indus composite score = 70- 10 = 60.
Indu also scored 100% in exactly one section.
Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indus score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu cant be recruited . Hence, we can reject this case.
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indus score in DI =$\dfrac{60-8-20}{2}=16$
It is also given that Danish, Harini, and Indu had scored the same marks the in GA.
We are given that, among the four recruited, Geeta had the lowest composite score.
Maximum composite score that Geeta can get = 2*14 + 6 + 20 = 54 {Assuming 100% marks in WE}. Since, Geeta was recruited at a composite score of 54 or less we can say that Ester was definitely recruited.
It is given that no two candidates had the same composite score.
We can see that Chetnas composite score is 54.
Hence, Geeta cant have a composite score of 54. Therefore, we can say that Geetas composite score is 53 or less.
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we can say that Danish was rejected at a composite score of 51. Hence, we can say that Geetas composite score in 52 or more.
Consequently, we can say that Geetas composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE.
Ajay was the unique highest scorer in WE.
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajays composite score =2*8 + 20 + 16 = 52. Which is a possible case.
Case 2: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI.
If Ajay scored 20 marks in DI then in that case Ajays composite score = 2*8 + 20 + 16 = 52 which will be same as Geetas composite score. Hence, we can say that in this case Ajay cant score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajays composite score = 2*8 + 19 + 16 = 51 which will be same as Danishs composite score. Hence, we can say that in this case Ajay cant score 19 marks.
Therefore, we can say that case 2 is not possible at all.
Let us check all the statement one by one.
Statement 1: Jatins composite score was more than that of Danish. We can see that this statement is correct.
Statement 2: Indu scored less than Chetna in DI. We can see that Indu scored 16 marks in DI whereas Chetna scored 19 marks in DI. Hence, we can say that this statement is also correct.
Statement 3: Jatin scored more than Indu in GA. We can see that Jatin scored 14 marks in GA whereas Indu scored 20 marks in GA. Hence, we can say that this statement is incorrect.
Hence, we can say that option D is the correct answer.
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI.
Therefore, Jatins composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indus composite score = 70- 10 = 60.
Indu also scored 100% in exactly one section.
Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indus score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu cant be recruited . Hence, we can reject this case.
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indus score in DI =$\dfrac{60-8-20}{2}=16$
It is also given that Danish, Harini, and Indu had scored the same marks the in GA.
We are given that, among the four recruited, Geeta had the lowest composite score.
Maximum composite score that Geeta can get = 2*14 + 6 + 20 = 54 {Assuming 100% marks in WE}. Since, Geeta was recruited at a composite score of 54 or less we can say that Ester was definitely recruited.
It is given that no two candidates had the same composite score.
We can see that Chetnas composite score is 54.
Hence, Geeta cant have a composite score of 54. Therefore, we can say that Geetas composite score is 53 or less.
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we can say that Danish was rejected at a composite score of 51. Hence, we can say that Geetas composite score in 52 or more.
Consequently, we can say that Geetas composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE.
Ajay was the unique highest scorer in WE.
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajays composite score =2*8 + 20 + 16 = 52. Which is a possible case.
Case 2: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI.
If Ajay scored 20 marks in DI then in that case Ajays composite score = 2*8 + 20 + 16 = 52 which will be same as Geetas composite score. Hence, we can say that in this case Ajay cant score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajays composite score = 2*8 + 19 + 16 = 51 which will be same as Danishs composite score. Hence, we can say that in this case Ajay cant score 19 marks.
Therefore, we can say that case 2 is not possible at all.
Let us check all the statement one by one.
Statement 1: Jatins composite score was more than that of Danish. We can see that this statement is correct.
Statement 2: Indu scored less than Chetna in DI. We can see that Indu scored 16 marks in DI whereas Chetna scored 19 marks in DI. Hence, we can say that this statement is also correct.
Statement 3: Jatin scored more than Indu in GA. We can see that Jatin scored 14 marks in GA whereas Indu scored 20 marks in GA. Hence, we can say that this statement is incorrect.
Hence, we can say that option D is the correct answer.
46920.Which of the following statements MUST be FALSE?
Bala scored same as Jatin in DI
Harini’s composite score was less than that of Falak
Bala’s composite score was less than that of Ester
Chetna scored more than Bala in DI
Explanation:
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI.
Therefore, Jatins composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indus composite score = 70- 10 = 60.
Indu also scored 100% in exactly one section.
Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indus score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu cant be recruited . Hence, we can reject this case.
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indus score in DI = $\dfrac{60-8-20}{2}$=16
It is also given that Danish, Harini, and Indu had scored the same marks the in GA.
We are given that, among the four recruited, Geeta had the lowest composite score.
Maximum composite score that Geeta can get = 2*14 + 6 + 20 = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited.
It is given that no two candidates had the same composite score. We can see that Chetnas composite score is 54.
Hence, Geeta cant have a composite score of 54. Therefore, we can say that Geetas composite score is 53 or less.
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geetas composite score in 52 or more.
Consequently, we can say that Geetas composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE.
Ajay was the unique highest scorer in WE.
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajays composite score = 2*8 + 20 + 16 = 52. Which is a possible case.
Case 2: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI.
If Ajay scored 20 marks in DI then in that case Ajays composite score = 2*8 + 20 + 16 = 52 which will be same as Geetas composite score. Hence, we can say that in this case Ajay cant score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajays composite score = 2*8 + 19 + 16 = 51 which will be same as Danishs composite score. Hence, we can say that in this case Ajay cant score 19 marks.
Therefore, we can say that case 2 is not possible at all.
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI.
Therefore, Jatins composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indus composite score = 70- 10 = 60.
Indu also scored 100% in exactly one section.
Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indus score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu cant be recruited . Hence, we can reject this case.
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indus score in DI = $\dfrac{60-8-20}{2}$=16
It is also given that Danish, Harini, and Indu had scored the same marks the in GA.
We are given that, among the four recruited, Geeta had the lowest composite score.
Maximum composite score that Geeta can get = 2*14 + 6 + 20 = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited.
It is given that no two candidates had the same composite score. We can see that Chetnas composite score is 54.
Hence, Geeta cant have a composite score of 54. Therefore, we can say that Geetas composite score is 53 or less.
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geetas composite score in 52 or more.
Consequently, we can say that Geetas composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE.
Ajay was the unique highest scorer in WE.
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajays composite score = 2*8 + 20 + 16 = 52. Which is a possible case.
Case 2: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI.
If Ajay scored 20 marks in DI then in that case Ajays composite score = 2*8 + 20 + 16 = 52 which will be same as Geetas composite score. Hence, we can say that in this case Ajay cant score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajays composite score = 2*8 + 19 + 16 = 51 which will be same as Danishs composite score. Hence, we can say that in this case Ajay cant score 19 marks.
Therefore, we can say that case 2 is not possible at all.
46921.If all the candidates except Ajay and Danish had different marks in DI, and Balas composite score was less than Chetnas composite score, then what is the maximum marks that Bala could have scored in DI?
12
11
13
9
Explanation:
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI.
Therefore, Jatins composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indus composite score = 70 - 10 = 60.
Indu also scored 100% in exactly one section.
Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indus score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu cant be recruited . Hence, we can reject this case.
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indus score in DI =$\dfrac{60-8-20}{2}$ = 16
It is also given that Danish, Harini, and Indu had scored the same marks the in GA.
We are given that, among the four recruited, Geeta had the lowest composite score.
Maximum composite score that Geeta can get = 2*14 + 6 + 20 = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited.
It is given that no two candidates had the same composite score. We can see that Chetnas composite score is 54.
Hence, Geeta cant have a composite score of 54. Therefore, we can say that Geetas composite score is 53 or less.
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geetas composite score in 52 or more.
Consequently, we can say that Geetas composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE.
Ajay was the unique highest scorer in WE.
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajays composite score = 2*8 + 20 + 16 = 52. Which is a possible case.
Case 2: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI.
If Ajay scored 20 marks in DI then in that case Ajays composite score = 2*8 + 20 + 16 = 52 which will be same as Geetas composite score. Hence, we can say that in this case Ajay cant score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajays composite score = 2*8 + 19 + 16 = 51 which will be same as Danishs composite score. Hence, we can say that in this case Ajay cant score 19 marks.
Therefore, we can say that case 2 is not possible at all.
It is given that all the candidates except Ajay and Danish had different marks in DI and Balas composite score was less than Chetnas composite score. Let us assume that Bala scored x marks in DI.
$\Rightarrow$2*x + 9 + 11 < 54
$\Rightarrow x$ < 17
We can see that Balas score will be less than 17. Balas maximum score in DI will be the largest possibel number less than 17 which is not same as any other candidates score in DI. From, the table we can see that 16, 15 and 14 are already taken by Indu, Falak and Geeta respectively.
Therefore, we can say that Bala can score a maximum of 13 marks in DI.
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI.
Therefore, Jatins composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indus composite score = 70 - 10 = 60.
Indu also scored 100% in exactly one section.
Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indus score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu cant be recruited . Hence, we can reject this case.
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indus score in DI =$\dfrac{60-8-20}{2}$ = 16
It is also given that Danish, Harini, and Indu had scored the same marks the in GA.
We are given that, among the four recruited, Geeta had the lowest composite score.
Maximum composite score that Geeta can get = 2*14 + 6 + 20 = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited.
It is given that no two candidates had the same composite score. We can see that Chetnas composite score is 54.
Hence, Geeta cant have a composite score of 54. Therefore, we can say that Geetas composite score is 53 or less.
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geetas composite score in 52 or more.
Consequently, we can say that Geetas composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE.
Ajay was the unique highest scorer in WE.
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajays composite score = 2*8 + 20 + 16 = 52. Which is a possible case.
Case 2: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI.
If Ajay scored 20 marks in DI then in that case Ajays composite score = 2*8 + 20 + 16 = 52 which will be same as Geetas composite score. Hence, we can say that in this case Ajay cant score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajays composite score = 2*8 + 19 + 16 = 51 which will be same as Danishs composite score. Hence, we can say that in this case Ajay cant score 19 marks.
Therefore, we can say that case 2 is not possible at all.
It is given that all the candidates except Ajay and Danish had different marks in DI and Balas composite score was less than Chetnas composite score. Let us assume that Bala scored x marks in DI.
$\Rightarrow$2*x + 9 + 11 < 54
$\Rightarrow x$ < 17
We can see that Balas score will be less than 17. Balas maximum score in DI will be the largest possibel number less than 17 which is not same as any other candidates score in DI. From, the table we can see that 16, 15 and 14 are already taken by Indu, Falak and Geeta respectively.
Therefore, we can say that Bala can score a maximum of 13 marks in DI.
46922.If all the candidates scored different marks in WE then what is the maximum marks that Harini could have scored in WE?
12
11
14
9
Explanation:
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI.
Therefore, Jatins composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indus composite score = 70 - 10 = 60.
Indu also scored 100% in exactly one section.
Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indus score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu cant be recruited . Hence, we can reject this case.
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indus score in DI =$\dfrac{60-8-20}{2}$ = 16
It is also given that Danish, Harini, and Indu had scored the same marks the in GA.
We are given that, among the four recruited, Geeta had the lowest composite score.
Maximum composite score that Geeta can get = 2*14 + 6 + 20 = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited.
It is given that no two candidates had the same composite score. We can see that Chetnas composite score is 54.
Hence, Geeta cant have a composite score of 54. Therefore, we can say that Geetas composite score is 53 or less.
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geetas composite score in 52 or more.
Consequently, we can say that Geetas composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE.
Ajay was the unique highest scorer in WE.
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajays composite score = 2*8 + 20 + 16 = 52. Which is a possible case.
Case 2: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI.
If Ajay scored 20 marks in DI then in that case Ajays composite score = 2*8 + 20 + 16 = 52 which will be same as Geetas composite score. Hence, we can say that in this case Ajay cant score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajays composite score = 2*8 + 19 + 16 = 51 which will be same as Danishs composite score. Hence, we can say that in this case Ajay cant score 19 marks.
Therefore, we can say that case 2 is not possible at all.
It is given that all the candidates scored different marks in WE.
We can see that Ajay, Geeta and Ester has already scored 20, 19 and 18 marks in WE. Therefore, Harini can score a maximum of 17 marks in WE. If Harinis score in WE is 17, then Harinis composite score = 2*5+17+20 = 47 which is same as Falaks composite score. Hence, we can say that Harini cant score 17 marks in WE. Jatin and Danish have already scored 16 and 15 marks respectively. Therefore, we can say that the maximum marks that Harini could have scored in WE = 14.
It is given that Indu and Jatin both scored 100% in exactly one section. We can say that Jatin scored 100% marks in DI.
Therefore, Jatins composite score = 2*20+16+14 = 70
It is given that Jatin’s composite score was 10 more than Indu’s. Therefore, we can say that Indus composite score = 70 - 10 = 60.
Indu also scored 100% in exactly one section.
Case 1: Indu scored 100% marks in DI.
If Indu scored 100% marks in DI, then Indus score in GA = 60 - 2*20 - 8 = 12 which is less than 70% of maximum possible marks. Indu already has less than 70% in WE, therefore we Indu cant be recruited . Hence, we can reject this case.
Consequently, we can say that Indu scored 100% marks in WE. Therefore, Indus score in DI =$\dfrac{60-8-20}{2}$ = 16
It is also given that Danish, Harini, and Indu had scored the same marks the in GA.
We are given that, among the four recruited, Geeta had the lowest composite score.
Maximum composite score that Geeta can get = 2*14 + 6 + 20 = 54 {Assuming 100% marks in WE}. Since, Geeta was recruitedat a composite score of 54 or less we can say that Ester was definitely recruited.
It is given that no two candidates had the same composite score. We can see that Chetnas composite score is 54.
Hence, Geeta cant have a composite score of 54. Therefore, we can say that Geetas composite score is 53 or less.
We already know the four people(Jatin, Indu, Geeta, Ester) which were recruited. Hence, we cab say that Danish was rejected at a composite score of 51. Hence, we can say that Geetas composite score in 52 or more.
Consequently, we can say that Geetas composite score if either 52 or 53. Therefore we can say that Geeta scored either 18 {52-(2*14+6)} or 19 {53-(2*14+6)} marks in WE.
Ajay was the unique highest scorer in WE.
Case 1: Geeta scored 19 marks in WE.
We can say that if Geeta scored 19 marks in WE, then Ajay scored 20 marks in DI. In that case Ajays composite score = 2*8 + 20 + 16 = 52. Which is a possible case.
Case 2: Geeta scored 18 marks in WE.
We can say that if Geeta scored 18 marks in WE, then Ajay can score either 19 or 20 marks in DI.
If Ajay scored 20 marks in DI then in that case Ajays composite score = 2*8 + 20 + 16 = 52 which will be same as Geetas composite score. Hence, we can say that in this case Ajay cant score 20 marks.
If Ajay scored 19 marks in DI then in that case Ajays composite score = 2*8 + 19 + 16 = 51 which will be same as Danishs composite score. Hence, we can say that in this case Ajay cant score 19 marks.
Therefore, we can say that case 2 is not possible at all.
It is given that all the candidates scored different marks in WE.
We can see that Ajay, Geeta and Ester has already scored 20, 19 and 18 marks in WE. Therefore, Harini can score a maximum of 17 marks in WE. If Harinis score in WE is 17, then Harinis composite score = 2*5+17+20 = 47 which is same as Falaks composite score. Hence, we can say that Harini cant score 17 marks in WE. Jatin and Danish have already scored 16 and 15 marks respectively. Therefore, we can say that the maximum marks that Harini could have scored in WE = 14.