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# If a, b, and c are three different numbers and ax:by:cz = 1:2:(–3), th

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Math Expert
Joined: 02 Sep 2009
Posts: 58396
If a, b, and c are three different numbers and ax:by:cz = 1:2:(–3), th  [#permalink]

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10 Apr 2019, 00:34
00:00

Difficulty:

5% (low)

Question Stats:

94% (00:47) correct 6% (01:02) wrong based on 35 sessions

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If a, b, and c are three different numbers and ax:by:cz = 1:2:(–3), then ax + by + cz =

(A) 0
(B) 1/2
(C) 3
(D) 6
(E) 9

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GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5027
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: If a, b, and c are three different numbers and ax:by:cz = 1:2:(–3), th  [#permalink]

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10 Apr 2019, 02:49
Bunuel wrote:
If a, b, and c are three different numbers and ax:by:cz = 1:2:(–3), then ax + by + cz =

(A) 0
(B) 1/2
(C) 3
(D) 6
(E) 9

ratio sum is 0 so sum of ax+by+cz = 0
IMO A
Intern
Joined: 22 Apr 2019
Posts: 31
Re: If a, b, and c are three different numbers and ax:by:cz = 1:2:(–3), th  [#permalink]

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28 Apr 2019, 14:19
Bunuel wrote:
If a, b, and c are three different numbers and ax:by:cz = 1:2:(–3), then ax + by + cz =

(A) 0
(B) 1/2
(C) 3
(D) 6
(E) 9

Hi Bunuel,

Can you please explain this question? The user above states that the ratio sum is always 0. Correct me if I'm wrong, but I don't think that's always the case. I realize you can just add 1 + 2 + (-3) = 0, but I'm having trouble seeing that as what you're supposed to do. Are there any similar questions you could link? Thanks!
Re: If a, b, and c are three different numbers and ax:by:cz = 1:2:(–3), th   [#permalink] 28 Apr 2019, 14:19
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