Last year Isabella took 7 math tests and received 7 different scores,

Question

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

(A) 92

(B) 94

(C) 96

(D) 98

(E) 100

gacuto · Accepted Answer

Since the 7th test score of 95 kept the average as an integer, the first 6 tests must have an average of 95.

So the total score for the first 6 tests must be 6 * 95 = 570.

If the 6th test score is subtracted from 570, the remaining total of the first 5 test scores must still average to an integer. Algebraically, we get:

570 - x = 5 * a, where x is the 6th test score which is an integer and a is the integer average of the first 5 tests

From this, we can observe that 570 - x must result in an integer that is a multiple of 5. In order to satisfy this condition, x must also be a multiple of 5. The only answer choice that is a multiple of 5 is 100.

The answer is E, 100.

chetan2u · Answer

A quick answer would by applying number properties..

If after adding 95, the average is an integer, it means the average was 95 in first 6 tests..
You can also test by plugging values of x from 91 till 100 in 6x+95=7y. The value of x that makes left side 6x+95 multiple of 7 is 95.

Thus, total in 6 test is 6*95. We can again form an equation by taking the average of 5 tests to be r and the score in 6th to be t, so 5r+t=6*95 or t=(6*95-5r)=5(78-r).
So, t has to be a multiple of 5. But 95 is the 7th number, so only 100 is the other multiple of 5 from 91 to 100.

E

KarishmaB · Answer

A little bit of logical thinking will give you the answer.

The 7th number is 95. The average was an integer before and stays an integer now. Since all the are between 91 - 100, let's assume the average we obtained of the 6 numbers before was 95 too so adding 95 just keeps the average at 95.

Note that the numbers cannot be repeated and that the average is closer to 91 than to 100. So let's start stacking the smaller numbers first.

Let the first number be 91. Next, to get an integer average, add 93.
Add the third number at the average so that it stays an integer i.e. add 92 in.
Now we have 3 numbers with average of 92 so we add 96 to keep the average integer. The new average of 4 numbers has become 93.
Now we need to add 98 to it so that the avg of the 5 numbers in integer. The new average will be 94. (Just focus on the units digits to calculate since 90 will be divisible by 5)
Now we need the average to be 95 after adding the 6th number. We will get 95 average if we add 100 now.

So 6th number should be 100.

Answer (E)

Luca1111111111111 · Answer

Hello,
Can you explain why this is the case: "If after adding 95, the average is an integer, it means the average was 95 in first 6 tests.."
Thanks

gacuto · Answer

Good question. It is not obvious that the average of the first 6 tests is 95 until you consider that each of the first 6 test scores must be between 91 and 100. The reasoning is as follows.

The key is understanding why adding a new test score would keep the average as an integer in the first place - the new score has to distribute the surplus over the average or deficit below the average evenly among the first 6 scores. Specifically in this case, that means the 7th test score has to be a multiple of 7 more or a multiple of 7 less than the average of the first 6 scores.

Take the simplified example of the set of numbers {1,1}, which has an average of 1. If a 3rd number is added and the average needs to stay as an integer, some options are {1,1,1} and {1,1,4}. Why do 1 and 4 work? 1 has 0 surplus so it distributes nothing across the whole set. 4 has a surplus 3 so it can increase each number in the set by 1 and hence the average of the whole set by 1.

Now look at the test scores in question. Of course if the first 6 scores averaged to 95, adding another 95 would keep the average at 95. However, if the first 6 scores didn't average to 95, what sets of numbers would work? For simplicity sake, let's pretend the 6 test scores are the same.

{88, 88, 88, 88, 88, 88, 95} and {102, 102, 102, 102, 102, 102, 95}

The first started with an average of 88, and since 95 has a surplus of 7 over that average, it is able to increase the average by 1 to 89. The opposite is true for the set with an average of 102 - that set's average decreased by 1 to 101.

Since there cannot be an average of 88 without a number below 91 and there cannot be an average of 102 without a number above 100, these averages cannot be valid even when not all of the test scores are the same. The only average that works is, therefore, 95 for the first 6 tests.

effatara · Answer

Luca1111111111111

Let AS6 and AS7 be the average scores after the 6th and 7th tests respectively. Thus:
AS7 = (6*AS6 + 95)/7. For this to be an integer, AS6 must be equal to 95 or increments or decrements of 7 from 95. In other words, AS6 must be equal to 95 or (95+7) or (95-7). But it cannot be the last two (102 or 88) because that would mean some test scores would be 102 or more or 88 or less. So AS6 must be equal to 95.
Hope its clear.

Jbbaille · Answer

Another way to look at it is as follows:

We could consider the problem with 1, 2, 3, 4 till 10 instead of 90, 91 ... 100. At the end of the day, the only thing that we need to determine is the unit digit.

You know that the average of the 7 scores is an integer, so the sum of the 7 scores is a multiple of 7. You also know that the last score was 95 (equivalent: 5). So, by checking all the multiples below 10! = 55, you can see that the sum of the first 6 scores is 30 by checking which one is not a multiple of 6:

7 - 5 = 2 Not a multiple of 6
14 - 5 = 9 Not a multiple of 6
21 - 5 = 16 Not a multiple of 6
28 - 5 = 23 Not a multiple of 6
35 - 5 = 30 Multiple of 6 => that's our sum
42 - 5 = 37 Not a multiple of 6
49 - 5 = 44 Not a multiple of 6.

Now you need to go from 30 to a multiple of 5 below 30, so either 25, 20 or 15. We can't get either 25 because 5 is already used or 15 because it is too far away. 20 is the only possibility, which means that the score is 10, or 100. Answer E.

kumarsunit · Answer

91/98,92,93,94,95,96,97
91/98,94,95,96,97,99,100

These 4 cases are possible.
So as per answer 96 is correct

Posted from my mobile device

stoned · Answer

though it took me 5 minutes almost. 
here is my thinking.

as the average has to be from 91 to 100 in all the cases. in the case of 5 tests, it can be either 95 or 100.
when it is 95 for 5 tests- for 6 tests average to be integer there is no number in the answer choices that can be added to 95*5 and the sum is divisible by 6.

but when we take the average of 5 tests to be 100 then the only option e i.e. 100 can give a number (100*5+100=600) that is divisible by 6.

so whatever be the 7th test score, for average to be an integer for 6 tests only option e is valid.

am I right?

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Last year Isabella took 7 math tests and received 7 different scores,

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Last year Isabella took 7 math tests and received 7 different scores,

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards