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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
Joined: 04 Sep 2016
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Location: India
WE: Engineering (Other)
w, x and y are positive integers such that w ≤ x ≤ y. If the average [#permalink]

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18 Oct 2017, 19:05
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45% (medium)

Question Stats:

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.

Kudos [?]: 83 [0], given: 235

Director
Joined: 25 Feb 2013
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Location: India
Schools: Mannheim"19 (S)
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w, x and y are positive integers [#permalink]

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18 Oct 2017, 20:27
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

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

Kudos [?]: 257 [0], given: 35

w, x and y are positive integers   [#permalink] 18 Oct 2017, 20:27
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