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

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Chembeti
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The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post  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
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Bunuel
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post  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 \(2a+b=2*35+14=84\).

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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pharm
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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, 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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Bunuel
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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, 08:23
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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, 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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Bunuel
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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, 08:42
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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?


I don't understand what you mean. Please, elaborate your question.
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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, 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] New post  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] New post  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

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] New post  02 May 2017, 16:48
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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


We are given that:

a/b = 5/2 or a : b = 5 : 2

and

a/c = 7/5 or a : c = 7 : 5

We need to determine the smallest value of 2a + b.

Since a is a multiple of 5 and of 7, we see it’s a multiple of the LCM of 5 and 7, which is 35.

Thus we now have:

a : b = 5 x 7 : 2 x 7 = 35 : 14

a : c = 7 x 5 : 5 x 5 = 35 : 25

So the minimum value of a is 35 and the minimum value of b is 14 (since a, b, and c must be integers), and thus the minimum value of 2a + b is 2(35) + 14 = 70 = 14 = 84.

Answer: C
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Abhishek009
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post  26 Jul 2018, 11:23
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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


\(\frac{a}{b} = \frac{5}{2}\), & \(\frac{a}{c} = \frac{7}{5}\)

Make numerators same base

So, \(\frac{a}{b} = \frac{35}{14}\), & \(\frac{a}{c} = \frac{35}{25}\)

So We have \(a = 35\), \(b = 14\) & \(c = 25\)

Thus, The value of \(2a + b = 2*35 + 14 = 84\), Answer must be (C)
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Re: The integers a, b, and c are positive a/b = 5/2, and a/c [#permalink] New post  21 Nov 2018, 19:57
JUst in case its not obvious that a is a multiple of 35:

b/a + c/a = 2/5 + 5/7

(b+c)/a = 39/35

this clearly shows that a is a multiple of 35
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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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