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

Nice way of solving it.. I couldn't. I had to back test the options one by one

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...

in such kind of problems its easy to back solving it if we plug answers then the only values which satisfy is 1:3:2 which option B

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.

We are given that:

a + b = 2c

a + c = b

b + c = 5a

Subtracting the second equation from the first equation, we have:

(a + b = 2c) - (a + c = b)

b - c = 2c - b

2b = 3c

Subtracting the third equation from the second equation, we have:

(a + c = b) - (b + c = 5a)

a - b = b - 5a

6a = 2b

Since 2b = 3c and 2b = 6a, we can say that 6a = 2b = 3c

Furthermore, we see that 6a = 2b = 3c holds when a = 1, b = 3, and c = 2.

Thus, the ratio of a to b to c = 1 to 3 to 2.

Answer: B

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).

You don't necessarily even have to substitute- just plug in a value for a b and c that works and then you have a ratio; for example, consider a = 1, b = 3 , c = 2

1 + 3 = 2(2)

Albeit, it is still important to know how to substitute values into equations which I think is the concept being tested. Learn the rules only to break them when convenient.

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

That's kind what I thought too

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 ?

Here, I used only two of the three equations, avoiding b + c = 5a entirely.

b is the only variable with coefficient 1 in all three equations.

1. With b as initial focus, rewrite a + c = b ----> b = a + c

2. Two equations to choose from now: a + b = 2c and b + c = 5a.

Coefficient 2 looked easier than coefficient 5. So I substituted value of b into a + b = 2c ---->

a + (a + c) = 2c, subtract c from both sides to get

2a = c or c = 2a

3. c now has a coefficient of 1, and b has a coefficient of 1 in equation from Step 1: b = a + c, substitute c = 2a ------> b = a + 2a, b = 3a

4. b and c are now both written in terms of a, thus

a = 1a
b = 3a
c = 2a.

Ratio of a:b:c is therefore 1:3:2

Answer B

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.!

Hello EgmatQuantExpert Bunuel - follow up on this method of doing the problem VeritasKarishma

I understand a + b has to be an even multiple of 2

Question : in all the answer choices, its quite possible that the unknown multiplier in all the options are even

Option 1) is 1: 2 :3

Assuming an even multiplier, lets say k = 6, thus the actual values are 6 : 12 : 18

Hence a + b in this ratio is also even (i.e. 6 + 12 = 18)

Why is this not a scenario / option we need to look at.

Thank you !

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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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.

a + b = 2c
a + c = b
b + c = 5a

Equate the equations :
b = 3a - (i)

From b + c = 5a and above eq:
c= 2a - (ii)

Hence, a:b:c =a:3a:2a => a:b:c =1:3:2

Option B

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