If a+b=2c, a+c=b, and b+c=5a, what is the ratio of a:b:c?

A. 1:2:3
B. 1:3:2
C. 3:2:1
D. 1:4:2
E. 1:2:5
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add c on both side:
a+b+c = 3c
a+5a=3c
6a=3c
2a=c

from substitue:

a+2a=b
3a=b

so ratio will be 1:3:2

answer b

I will consider that as an effective way of solving this question , I do the same as well...

Check using the options..

For option (B) -

a = 1
b = 3
c = 2

a + b = 2c Or, 1 + 3 = 2*2 ( Satisfies )
a + c = b , Or, 1 + 2 = 3 ( Satisfies )
b + c = 5a, Or 3 + 2 = 5*1 ( Satisfies )

Check using other options, none except (B) satisfies, hence thats the correct answer...

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I think the sum can be solved just by scanning the options.

Given a+c =b, two inferences can be made from this option.
1. If all three numbers are negative, b has to be the smallest. None of the ratios in the options have b as the smallest. So the numbers are positive.
2. Since the numbers are positive, b is the largest among a,b, and c. So options A,C, and E can be eliminated as b is not the biggest in them.

Option D says the b is twice that of c. If that's the case then from 1st equation a=2c-b will be 0 so the ratio will fall.

Hence B is the answer.

Option B

Correct ratio should give a + b as Even (Using eq.n I)

"Alternatively"

Correct ratio should satisfy all 3 equations.

E.g., a + b =2c : 1+3 = 2*2 For B).

a+c=b gives the answer away without any math.
This means that b is largest.
The only answers with b being the largest are B & D.
Looking at B you can cleary see that 1+2=3 or a+c=b

Abhishek009 : I tried solving it the traditional way and it took me a considerable amount of time. I liked your method of plugging in nos. from the options given. I wanted to check if it will be fairly OK to just check any one of the three equations as long as only one option gives positive result for that equation and move on without checking the answer choices for the remaining equations ?

My solution is as follows:

try to see the equations and find out which is the largest among a, b, c; clearly by equation 2 we can say b is the largest.
only two options have b as the largest, plug one into the the set of equations, if it passes great! else the other is the answer.!

Yes, you are right. If the multiplier is even, though the sum of ratio terms may not be even, the actual sum may be even.
Instead I would use the knowledge that the addition relation of actual variables holds in case of ratios too.
i.e. if actually a + b = 2c, then even in ratio terms, a + b will be equal to 2c.

Since 1 + 2 is not equal to 2*3, this is not correct.

Only for option (B), 1 + 3 = 2*2

Note why this is so. We routinely use the ratio scale and actual value scale while using ratios as discussed here:
https://www.veritasprep.com/blog/2011/0 ... of-ratios/

Let's say we have the ratio a:b:c = 1:3:2
Then, a = k, b = 3k and c = 2k

Then an addition relation such as a + b = 2c becomes
k + 3k = 2 * 2k
k(1 + 3) = 2 * 2k
(1 + 3) = 2*2

Since we can take the common multiplier out, the same relation will hold in terms of ratios too.
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