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Prime factorising ­\(6a=3b=7c\) to \(2*3a=3b=7c\). From this we can deduce that whatever number is being represented here will have to be a number that is a multiple of 42. So \(6a=3b=7c=42x\) and therefore \(7x = a\), \(14x = b\) and \(6x = c\). Therefore, \(a+b+c = 27x\)

Plugging \(7x = a\), \(14x = b\) and \(6x = c\) into \(5b = 8a + 4c\): 

\(5(14x)=8(7x)+4(6x)\)

\(70x = 56x + 24x\)

\(10x = 0\)

\(x = 0\)

Therefore, \(a+b+c = 0\)

ANSWER A
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Bunuel
­If \(6a=3b=7c\) and \(5b=8a+4c\), what is the value of \(a+b+c\)?

A. 0
B. 16
C. 17
D. 42
E. Cannot be determined from the given information­

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­6a = 3b = 7c and 5b = 8a + 4c
From the equation , a= b/2 and c= 3b/7

5b = 4b + 12b/7
This equation is only possible for b =0.
Implying all a, b, c are 0.
Hence the value of a+b+c = . Option A
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Where i can practice similar questions?

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Check algebra tag:

https://gmatclub.com/forum/search.php?s ... &tag_id=50
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­If \(6a=3b=7c\) and \(5b=8a+4c\), what is the value of \(a+b+c\)?

Let 6a = 3b = 7c = k; a = k/6; b = k/3; c = k/7
5b = 8a + 4c
5k/3 = 8k/6 + 4k/7 = 4k/3 + 4k/7
k/3 = 4k/7; k = 0

a + b + c = k/6 + k/3 + k/7 = 0

IMO A
­
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