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# The integers a, b, and c are positive a/b = 5/2, and a/c

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Senior Manager
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The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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26 Feb 2012, 00:21
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The integers a, b, and c are positive, a/b = 5/2, and a/c = 7/5. What is the smallest possible value of 2a + b?

A. 63
B. 70
C. 84
D. 95
E. 105
[Reveal] Spoiler: OA

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-Aravind Chembeti

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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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26 Feb 2012, 00:30
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Chembeti wrote:
The integers a, b, and c are positive, a/b = 5/2, and a/c = 7/5. What is the smallest possible value of 2a + b?

A. 63
B. 70
C. 84
D. 95
E. 105

Since $$\frac {a}{b}=\frac {5}{2}$$, and $$\frac {a}{c}=\frac {7}{5}$$ then $$a$$ must be a multiple of both 5 and 7, so the lowest value of $$a$$ is $$5*7=35$$ (note that $$a$$ is a positive integer). Then the lowest value of $$b$$ would be $$2*7=14$$, as $$\frac {a}{b}=\frac {5}{2}=\frac {5*7}{2*7}=\frac {35}{14}$$, so the lowest value of $$2a+b=2*35+14=84$$.

Or: since $$\frac {a}{b}=\frac {5}{2}$$ then $$a=\frac{5b}{2}$$ and $$2a + b=2*\frac{5b}{2}+b=6b$$, so it's a multiple of 6. The only multiple of 6 among the answer choices is 84 (C).

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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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04 Feb 2013, 08:15
why is the value of '2' for b have '7' multiplying to it , 7 is the the value for the fraction a/c . I understand how you are getting the lowest value for 'a' but not for 'b' .

Thanks
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Kudos [?]: 70328 [0], given: 9843

Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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04 Feb 2013, 08:23
Expert's post
pharm wrote:
why is the value of '2' for b have '7' multiplying to it , 7 is the the value for the fraction a/c . I understand how you are getting the lowest value for 'a' but not for 'b' .

Thanks

The lowest value of a is 35. Now, if a=35, then from a/b = 5/2 we'll have that b=14.

Hope it's clear.
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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04 Feb 2013, 08:39
Ok so since a/b = 5/2 and a/c = 7/5 . You are multiple across the '7' to 5 & 2 in order to get their LCM. Correct?
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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04 Feb 2013, 08:42
Expert's post
pharm wrote:
Ok so since a/b = 5/2 and a/c = 7/5 . You are multiple across the '7' to 5 & 2 in order to get their LCM. Correct?

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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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04 Feb 2013, 08:58
Bunuel wrote:
pharm wrote:
why is the value of '2' for b have '7' multiplying to it , 7 is the the value for the fraction a/c . I understand how you are getting the lowest value for 'a' but not for 'b' .

Thanks

The lowest value of a is 35. Now, if a=35, then from a/b = 5/2 we'll have that b=14.

Hope it's clear.

you said that from a/b = 5/2 , b=14 . Since in a/b , 'b' was = 2 . the common number is '7' that is being multipled to 5 and 2 in order to reach the lowest possible values correct?
so to get " b's " lowest possible value you multiplied = '2 * 7= 14' . Does '7' hold any significance that it was used for the values in 'a' & 'b' to reach there lowest possible values ?
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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16 Feb 2013, 15:57
It is mentioned that a/c=7/5. So from this ratio we know that 7 needs to be a factor of a. From a/b=5/2 we again know that a must contain 5. Hence, the minimum value of a must be 35.

Using the ratio a/b=5/2, we know that b must have 2 and must also contain 7 as a also contained 7 which was cancelled while calculating he simplest form of ratio.

pharm wrote:
Bunuel wrote:
pharm wrote:
why is the value of '2' for b have '7' multiplying to it , 7 is the the value for the fraction a/c . I understand how you are getting the lowest value for 'a' but not for 'b' .

Thanks

The lowest value of a is 35. Now, if a=35, then from a/b = 5/2 we'll have that b=14.

Hope it's clear.

you said that from a/b = 5/2 , b=14 . Since in a/b , 'b' was = 2 . the common number is '7' that is being multipled to 5 and 2 in order to reach the lowest possible values correct?
so to get " b's " lowest possible value you multiplied = '2 * 7= 14' . Does '7' hold any significance that it was used for the values in 'a' & 'b' to reach there lowest possible values ?
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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09 Apr 2014, 10:34
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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22 Apr 2014, 03:38
$$\frac{a}{b} = \frac{5}{2}$$

$$\frac{2a}{b} = \frac{10}{2}$$

Using componendo / dividendo

$$\frac{2a + b}{b} = \frac{10+2}{2}$$

2a +b = 6b

Looking at the options available, only 84 is divisible by 6

Bunuel, can you kindly tell-

Q1: Is this method correct to solve?

Q2: a/c = 7/5 >> Is this term given just to confuse? As I never required that info in solving this problem.
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]

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16 Dec 2015, 02:06
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The integers a, b, and c are positive a/b = 5/2, and a/c   [#permalink] 16 Dec 2015, 02:06
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