## Last year Isabella took 7 math tests and received 7 differen

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### Last year Isabella took 7 math tests and received 7 differen

by arnob » Wed May 29, 2019 2:39 am
Last year Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?

(A) 92

(B) 94

(C) 96

(D) 98

(E) 100

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by Ian Stewart » Mon Jun 03, 2019 6:56 am
Notice a few things here:

- If her average after n tests is always an integer, then after n tests, the sum of her scores will always be divisible by n.
- So after six tests, the sum of her score is a multiple of 6. Let's call that sum, after six tests, "S"
- If her scores are between 91 and 100 every time, and her scores are always different, S is definitely between 6*93 and 6*98. Since S is also a multiple of 6, it can only be 6*94, 6*95, 6*96 or 6*97
- Finally, when we add her seventh test score, 95 to S, we must get a multiple of 7. The remainder is 4 when we divide 95 by 7, so when we divide S by 7, the remainder will need to be 3, if S+95 will give us a multiple of 7.

So we just need to work out which of the few possible values of S, 6*94, 6*95, 6*96 or 6*97, gives us a remainder of 3 when we divide it by 7. Using remainder arithmetic, or inspection, you can find that S = 6*95 = 570.

So after six tests, her scores add up to 570. When we subtract from that her sixth test score, we need to get a multiple of 5. But 570 is already a multiple of 5, so her sixth test score must be a multiple of 5 too. It can't be 95, because that's her sixth score, and she never repeated scores. The only other possibility between 91 and 100 inclusive is 100, so that must be the answer.
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