adkikani wrote:

w, x and y are positive integers such that w ≤ x ≤ y. If the average (arithmetic mean) of w, x and y is 20, is w > 15 ?

(1) y = 28

(2) One of the three numbers is 17

Hi

adkikaniit seems OA, Option B, is not correct here. Can you confirm. IMO it should be C

Given \(w+x+y=60\), or \(w=60-x-y\)

Statement 1: implies \(w=60-x-28=32-x\), so if \(x=16\), then \(w=16\) but if \(x=20\) \(w=12\). Hence

insufficientStatement 2: if \(w=17\), then we have a "\(Yes\)" for our question stem

but if \(x=17\), then \(w=60-17-y=43-y\), now if y=30\(\), \(w=13\) and if \(y=27\), then \(w=16\). Hence

insufficientCombining 1 & 2, we get the value of two variable, hence third variable can be calculated.

SufficientOption

C