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Q is the set of all integers between A and B, inclusive, where A is an

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Bunuel
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Q is the set of all integers between A and B, inclusive, where A is an [#permalink] New post   02 May 2021, 00:45
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Q is the set of all integers between A and B, inclusive, where A is an integer. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150
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Re: Q is the set of all integers between A and B, inclusive, where A is an [#permalink] New post   09 May 2021, 02:30
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Bunuel wrote:
Q is the set of all integers between A and B, inclusive, where A is an integer. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150



So, we are told that the average of integers between integer A and 190 is m. Notice that Q is a set of consecutive integers, thus its average, m, equals to the median.

(1) Q contains 40 terms that are greater than m.

If A is odd, then the number of integers between odd A and 190 would be even, so in this case m would be the average of two middle terms, so not an integer. Thus, in this case we would have that there are 80 integers in the set, 40 less than m and 40 greater than m. This gives the value of A as 111.

If A is eve, then the number of integers between even A and 190 would be odd, so in this case m would be the middle terms, so an integer. Thus, in this case we would have that there are 81 integers in the set, 40 less than m, m itself and 40 greater than m. This gives the value of A as 110.

(2) m=150 --> m = (A + B)/2 --> 150 = (A + 190)/2 --> A = 110. Sufficient.

Answer: B.

Hope it's clear.
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Q is the set of all integers between A and B, inclusive, where A is an [#permalink] New post   03 May 2021, 18:37
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Bunuel wrote:
Q is the set of all integers between A and B, inclusive, where A is an integer. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150


We know both A and B are integers, thus the integers from A to B is an arithmetic progression. The average is m, which must be the median as well so \(\frac{A + B}{2} = m\).

Statement 1:

We may have m = 190 - 40 = 150, so that there are 40 integers greater than the median/average. We may also have Q = 150.5, which is a possible case when the set has an even amount of integers (exactly 80 integers). Insufficient.

Statement 2:

We know B and m, plug those in \(\frac{A + B}{2} = m\) and we can find A. Sufficient.

Ans: B
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Re: Q is the set of all integers between A and B, inclusive, where A is an [#permalink] New post   05 May 2021, 22:41
TestPrepUnlimited wrote:
Bunuel wrote:
Q is the set of all integers between A and B, inclusive, where A is an integer. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150


We know both A and B are integers, thus the integers from A to B is an arithmetic progression. The average is m, which must be the median as well so \(\frac{A + B}{2} = m\).

Statement 1:

We may have m = 190 - 40 = 150, so that there are 40 integers greater than the median/average. We may also have Q = 150.5, which is a possible case when the set has an even amount of integers (exactly 80 integers). Insufficient.

Statement 2:

We know B and m, plug those in \(\frac{A + B}{2} = m\) and we can find A. Sufficient.

Ans: B


I don't quite understand your solution to Statement 1: if we know that there are 40 integers greater than the average and that numbers within the set must be consecutive, shouldn't there also be 40 integers below the average as well. Wouldn't we be able to derive back to A knowing this (e.g., the range is from A = 110 to 190)? Thanks in advanced!
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Re: Q is the set of all integers between A and B, inclusive, where A is an [#permalink]
05 May 2021, 22:41
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