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Last year Isabella took 7 math tests and received 7 different scores,
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03 Apr 2019, 04:46
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25% (03:12) correct 75% (03:03) wrong based on 139 sessions
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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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Re: Last year Isabella took 7 math tests and received 7 different scores,
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03 Apr 2019, 11:09
Noshad wrote: 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 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*955r)=5(78r). 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
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Last year Isabella took 7 math tests and received 7 different scores,
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19 May 2019, 16:20
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.



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Re: Last year Isabella took 7 math tests and received 7 different scores,
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19 May 2019, 20:41
Noshad wrote: 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 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)
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Re: Last year Isabella took 7 math tests and received 7 different scores,
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22 Jul 2019, 06:58
chetan2u wrote: Noshad wrote: 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 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*955r)=5(78r). 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 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



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Re: Last year Isabella took 7 math tests and received 7 different scores,
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23 Jul 2019, 08:12
Luca1111111111111 wrote: chetan2u wrote: Noshad wrote: 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 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*955r)=5(78r). 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 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 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.




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