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A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is
Correct answer is '60'. Can you explain this answer?
Verified Answer
A CAT aspirant appears for a certain number of tests. His average scor...
Let the total number of tests be 'n' and the average by 'A'
Total score = n*A
When 1st 10 tests are excluded, decrease in total value of scores = (nA - 20 * 10) = (nA - 200)
Also, (n - 10)(A + 1) = (nA - 200)
On solving, we get 10A - n = 190..........(i)
When last 10 tests are excluded, decrease in total value of scores = (nA - 30 * 10) = (nA - 300)
Also, (n - 10)(A - 1) = (nA - 300)
On solving, we get 10A + n = 310..........(ii) From (i) and (ii), we get n = 60
Hence, 60 is the correct answer.
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Most Upvoted Answer
A CAT aspirant appears for a certain number of tests. His average scor...
Solution:

Let the total number of tests taken by the CAT aspirant be 'n'.

Given, his average score increases by 1 if the first 10 tests are not considered.
So, the average score of remaining (n-10) tests = previous average + 1

Similarly, the average score decreases by 1 if the last 10 tests are not considered.
So, the average score of remaining (n-10) tests = previous average - 1

Now, we have to find the value of 'n'.

Given, the average score for the first 10 tests = 20
So, the total score of first 10 tests = 20*10 = 200

Given, the average score for the last 10 tests = 30
So, the total score of last 10 tests = 30*10 = 300

Let the total score of remaining (n-20) tests be 'x'.

So, the total score of all n tests = 200 + x + 300 = x + 500

Also, we know that the average score of remaining (n-10) tests = previous average + 1
=> x/(n-10) = (20+1) = 21

=> x = 21(n-10)

Similarly, the average score of remaining (n-10) tests = previous average - 1
=> x/(n-10) = (30-1) = 29

=> x = 29(n-10)

Equating both the expressions of x, we get:

21(n-10) = 29(n-10)
=> 8n = 80
=> n = 10

Therefore, the total number of tests taken by the CAT aspirant is 60.
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A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him isCorrect answer is '60'. Can you explain this answer?
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