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Does nm =40? (1) 10/n = m/4 (2) 5n = 20 and 8m = 80

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Does nm =40? (1) 10/n = m/4 (2) 5n = 20 and 8m = 80 [#permalink] New post 01 May 2012, 01:29
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Does nm =40?

(1) 10/n = m/4
(2) 5n = 20 and 8m = 80

a) 10/n = m/4
if we consider different signs of n & m
nm=40 or nm=-40
NS

b) 5n=20 => n=4
8m = 80 =>m=10
=> nm=40 ---Sufficient

Guys do we need to consider that a) is providing ratio so we don't need to consider -ve signs.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 01 May 2012, 01:44, edited 2 times in total.
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Re: Why we dont need to consider negative values here [#permalink] New post 01 May 2012, 01:44
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Re: Why we dont need to consider negative values here [#permalink] New post 01 May 2012, 05:25
Bunuel wrote:
Does nm =40?

(1) \frac{10}{n} = \frac{m}{4}--> cross multiply 40=mn. Sufficient. So, as you can see we have only one value for mn

(2) 5n = 20 and 8m = 80 --> n=4 and m=10 --> mn=40. Sufficient.

Answer: D.





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Re: Does nm =40? (1) 10/n = m/4 (2) 5n = 20 and 8m = 80 [#permalink] New post 17 Jun 2012, 16:48
Answer: D

At first glance, I thought 10/n = m/4 was not sufficient. Using cross multiplication, I was able to get mn=40. Rephrasing the question, I thought the question needed values for m and n. Not the case for this DS.
Re: Does nm =40? (1) 10/n = m/4 (2) 5n = 20 and 8m = 80   [#permalink] 17 Jun 2012, 16:48
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Does nm =40? (1) 10/n = m/4 (2) 5n = 20 and 8m = 80

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