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11 Feb 2021, 01:15
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Q is the set of all integers between A and B, inclusive. 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
Updated on: 12 Feb 2021, 22:18
Originally posted by TarunKumar1234 on 11 Feb 2021, 09:26.
Last edited by TarunKumar1234 on 12 Feb 2021, 22:18, edited 1 time in total.
Last edited by TarunKumar1234 on 12 Feb 2021, 22:18, edited 1 time in total.
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A, (A+1),----------------------188, 189, 190= B
Stat1: Q contains 40 terms that are greater than m.
then, case 1: m is a part of set Q , Q= 81 consecutive numbers, m = 189-40 +1 = 150 and A = (150-40+1) -1 = 110.
case 2: m isn't a part of set Q , Q= 80 consecutive numbers and m = (149 +150)/2 = 149.5, then A = (189-80+1)+1 = 111 [/u] Not Sufficient
Stat2: m=150
then, A = (150 -40+1) -1 = 110. Sufficient
So, It is B.
Stat1: Q contains 40 terms that are greater than m.
then, case 1: m is a part of set Q , Q= 81 consecutive numbers, m = 189-40 +1 = 150 and A = (150-40+1) -1 = 110.
case 2: m isn't a part of set Q , Q= 80 consecutive numbers and m = (149 +150)/2 = 149.5, then A = (189-80+1)+1 = 111 [/u] Not Sufficient
Stat2: m=150
then, A = (150 -40+1) -1 = 110. Sufficient
So, It is B.
12 Feb 2021, 01:30
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Bunuel
Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?
First of all, in order the answer to be B, it should be mentioned that A is an integer itself. Suppose we are told that.
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.
