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

1

This post received KUDOS

Expert's post

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This post was BOOKMARKED

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).

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

Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]
04 Feb 2013, 07:23

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.

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

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

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

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