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The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]
 26 Feb 2012, 00:21
Question Stats:
82% (02:03) correct
17% (01:55) wrong based on 35 sessions
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
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]
26 Feb 2012, 00:30
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]
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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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]
04 Feb 2013, 08:23
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]
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]
04 Feb 2013, 08:42
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink]
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]
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
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