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

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

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New post 18 Oct 2017, 19:05
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56% (01:27) correct 44% (00:56) wrong based on 25 sessions

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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
[Reveal] Spoiler: OA

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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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New post 18 Oct 2017, 20:27
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 adkikani

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

Last edited by niks18 on 18 Oct 2017, 21:02, edited 1 time in total.

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w, x and y are positive integers   [#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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