Arithmetic

Mean, Median & Mode

Medium

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

Entered answer:

✅ Correct Answer: 60

Let us solve this step-by-step using the given conditions about how the average changes when certain tests are excluded.

Let us define our variables clearly:

N = Total number of tests taken

A = Average score across all N tests

Total score of all tests = N × A

We know:

Average of first 10 tests = 20 → Total score of first 10 tests = 10 × 20 = 200

Average of last 10 tests = 30 → Total score of last 10 tests = 10 × 30 = 300

When the first 10 tests are removed, average increases by 1:

Remaining tests = N - 10

Total score of remaining tests = NA - 200

New average = $\tfrac{NA - 200}{N - 10} = A + 1$

This gives us our first equation:

$\tfrac{NA - 200}{N - 10} = A + 1$

When the last 10 tests are removed, average decreases by 1:

Remaining tests = N - 10

Total score of remaining tests = NA - 300

New average = $\tfrac{NA - 300}{N - 10} = A - 1$

This gives us our second equation:

$\tfrac{NA - 300}{N - 10} = A - 1$

From equation 1:

$NA - 200 = (A + 1)(N - 10)$

$NA - 200 = AN - 10A + N - 10$

$NA - 200 = AN + N - 10A - 10$

$-200 = N - 10A - 10$

$N - 10A = -190$ ... (3)

From equation 2:

$NA - 300 = (A - 1)(N - 10)$

$NA - 300 = AN - 10A - N + 10$

$NA - 300 = AN - N - 10A + 10$

$-300 = -N - 10A + 10$

$N + 10A = 310$ ... (4)

Now we will solve equations (3) and (4) simultaneously:

From equation (3): N - 10A = -190

From equation (4): N + 10A = 310

Adding both equations:

$2N = -190 + 310$

$2N = 120$

$N = 60$

Substituting back to find A:

$60 + 10A = 310$

$10A = 250$

$A = 25$

Therefore, the total number of tests taken = 60

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