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

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The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post 25 Feb 2012, 23:21
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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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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post 25 Feb 2012, 23:30
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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.

Answer: C.

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).

Answer: C.
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post 04 Feb 2013, 07:15
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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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post 04 Feb 2013, 07:23
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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] New post 04 Feb 2013, 07:39
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] New post 04 Feb 2013, 07:42
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post 04 Feb 2013, 07:58
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] New post 16 Feb 2013, 14:57
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] New post 09 Apr 2014, 09:34
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post 22 Apr 2014, 02:38
\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

Answer = C

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] 22 Apr 2014, 02:38
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