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When you use both statements here, you find j+p+m+n+q = 280, and since the sum of all six numbers is 336, the remaining number k must be equal to 336 - 280 = 56. So using both statements, at least one value in the set is exactly 56. But it's certainly possible other values are also 56. Looking at Statement 1, we might have j = 56 and p = 72, for example, in which case we'd have at least one other value equal to 56, but there's no need for one of j or p to be 56 -- we could have j = 0 and p = 128, say. Similarly, some of m, n and q might equal 56, but there's no need for that to be true. So the answer is E.
It would perhaps be a more interesting question if it asked whether at least one of the numbers was equal to 56. In that case, the two statements would be sufficient together.
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Re: If the average (arithmetic mean) of six numbers j, k, m, n, p, and q [#permalink]
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04 Aug 2017, 04:53
sananoor wrote:
If the average (arithmetic mean) of six numbers j, k, m, n, p, and q is 56, how many of the numbers are equal to 56?
(1) The sum of j and p is 128. (2) The sum of m, n, and q is 152.
I would go for E
Since 1) J and P can be any value ( No idea about M, N and P) Insufficient 2) same as above M, N, Q can be any value ( No idea about J and P) Insufficient
both 1 and 2 ( J, P, M, N, Q) can again be any value to make the sum 290 Insufficient
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61% (01:11) correct
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