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02 May 2021, 00:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
25% (01:58) correct
75%
(02:07)
wrong
based on 63
sessions
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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
(1) Q contains 40 terms that are greater than m.
(2) m=150
09 May 2021, 02:30
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Bunuel
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.
General Discussion
03 May 2021, 18:37
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Bunuel
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
