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w, x and y are positive integers such that w ≤ x ≤ y. If the average

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w, x and y are positive integers such that w ≤ x ≤ y. If the average  [#permalink]

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Updated on: 18 Oct 2017, 21:01
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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

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Originally posted by adkikani on 18 Oct 2017, 19:05.
Last edited by Bunuel on 18 Oct 2017, 21:01, edited 2 times in total.
Renamed the topic.
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w, x and y are positive integers  [#permalink]

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Updated on: 18 Oct 2017, 21:02
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

it 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 insufficient

Statement 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 insufficient

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

Option C

Originally posted by niks18 on 18 Oct 2017, 20:27.
Last edited by niks18 on 18 Oct 2017, 21:02, edited 1 time in total.
w, x and y are positive integers &nbs [#permalink] 18 Oct 2017, 20:27
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w, x and y are positive integers such that w ≤ x ≤ y. If the average

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