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If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Then
- a)A = B
- b)A > B
- c)A < B
- d)Data Insufficient
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If the average marks of 17 students in a class is A. The marks of the ...
This question is one standard example of the definition of average in terms of surplus and deficit of the values. (Refer to the definition given) Now, what we were expected to mark here was the fact that whatever is the average (in this case, it is A), surplus generated by the marks of 3rd student will be same as deficit incurred due to 15th student. So, rusticating both of them is not going to create any difference on average marks of the class (remember marks are in AP). And similar will be the impact of rusticating 7th student and 11th student and then finally 9th student.
So, A = B
So, A = B
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If the average marks of 17 students in a class is A. The marks of the ...
Problem
The average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Then
a) A = B
b) A > B
c) A < />
d) Data Insufficient
Solution
The given problem can be solved by using the concept of Arithmetic Progression (AP) and weighted averages.
Let the common difference of the AP be d, and let the first and the last terms of the AP be a and b, respectively. Then, we have
b = a + 16d ......(1)
Also, since the marks are in AP, we have the sum of the marks of all the 17 students as
Sum = 17/2 × (a + b) = 17/2 × (2a + 16d) = 17(a + 8d) ......(2)
Let the average marks of the remaining 12 students be B. Then, we have
Sum of marks of 12 students = 12B
After rustication, the marks of the remaining 12 students will still be in AP, but with a different common difference. Let the new common difference be d'. Then, we have
b - 5d' = a + 6d' ......(3)
b - 4d' = a + 9d' ......(4)
b - 3d' = a + 10d' ......(5)
b - 2d' = a + 14d' ......(6)
b - d' = a + 18d' ......(7)
Adding equations (3) to (7), we get
5b = 5a + 57d'
b = a + 11.4d' ......(8)
Substituting equation (8) in equation (1), we get
a + 16d = a + 11.4d' => d' = 4d/7
Therefore, the new common difference is 4d/7. Let the sum of the marks of the remaining 12 students after rustication be S'. Then, we have
S' = 12/2 × [2B + (11 × 4/7)d] = 12/7 [14B + 44d] ......(9)
Using equations (2) and (9), we get
Sum - S' = 17(a + 8d) - 12/7 [14B + 44d]
=> 119B = 17(a + 8d) - 68d ......(10)
Since the average marks of the class is A, we have
Sum/17 = A => a + b = 34A
Substituting equation (1) in the above equation, we get
2a + 16d = 34A
Therefore, we have
a = 17A - 8d
Substituting this value of a in equation (
The average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Then
a) A = B
b) A > B
c) A < />
d) Data Insufficient
Solution
The given problem can be solved by using the concept of Arithmetic Progression (AP) and weighted averages.
Let the common difference of the AP be d, and let the first and the last terms of the AP be a and b, respectively. Then, we have
b = a + 16d ......(1)
Also, since the marks are in AP, we have the sum of the marks of all the 17 students as
Sum = 17/2 × (a + b) = 17/2 × (2a + 16d) = 17(a + 8d) ......(2)
Let the average marks of the remaining 12 students be B. Then, we have
Sum of marks of 12 students = 12B
After rustication, the marks of the remaining 12 students will still be in AP, but with a different common difference. Let the new common difference be d'. Then, we have
b - 5d' = a + 6d' ......(3)
b - 4d' = a + 9d' ......(4)
b - 3d' = a + 10d' ......(5)
b - 2d' = a + 14d' ......(6)
b - d' = a + 18d' ......(7)
Adding equations (3) to (7), we get
5b = 5a + 57d'
b = a + 11.4d' ......(8)
Substituting equation (8) in equation (1), we get
a + 16d = a + 11.4d' => d' = 4d/7
Therefore, the new common difference is 4d/7. Let the sum of the marks of the remaining 12 students after rustication be S'. Then, we have
S' = 12/2 × [2B + (11 × 4/7)d] = 12/7 [14B + 44d] ......(9)
Using equations (2) and (9), we get
Sum - S' = 17(a + 8d) - 12/7 [14B + 44d]
=> 119B = 17(a + 8d) - 68d ......(10)
Since the average marks of the class is A, we have
Sum/17 = A => a + b = 34A
Substituting equation (1) in the above equation, we get
2a + 16d = 34A
Therefore, we have
a = 17A - 8d
Substituting this value of a in equation (
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If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Thena)A = Bb)A > Bc)A < Bd)Data InsufficientCorrect answer is option 'A'. Can you explain this answer?
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If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Thena)A = Bb)A > Bc)A < Bd)Data InsufficientCorrect answer is option 'A'. Can you explain this answer? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Thena)A = Bb)A > Bc)A < Bd)Data InsufficientCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Thena)A = Bb)A > Bc)A < Bd)Data InsufficientCorrect answer is option 'A'. Can you explain this answer?.
If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Thena)A = Bb)A > Bc)A < Bd)Data InsufficientCorrect answer is option 'A'. Can you explain this answer? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Thena)A = Bb)A > Bc)A < Bd)Data InsufficientCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Thena)A = Bb)A > Bc)A < Bd)Data InsufficientCorrect answer is option 'A'. Can you explain this answer?.
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