jabhatta@umail.iu.edu wrote:
EgmatQuantExpert wrote:
Bunuel wrote:
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
Hey,
Though there are a number of ways to solve this. I would like to use a slightly different method which involved options elimination, using the
concept of even-odd.
We are given that a+b = 2c, now
2c is EVEN, therefore the sum of
a + b should also be
EVEN.
Look at the answer choices and focus on
only the first two numbers and their sum -
A. \(1+2 = 3\) - ODD
B. \(1 + 3 = 4\)- EVEN
C. \(3+ 2 = 5\) - ODD
D. \(1+ 4 =5\) - ODD
E.\(1+2 =3\) - ODD
Hence
Option B is the correct answer.
It is always a good idea to look for patterns in the options if one is not sure how to approach using the conventional method.
Thanks,
Saquib
Quant Expert
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Hello
EgmatQuantExpert Bunuel - follow up on this method of doing the problem
VeritasKarishmaI 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.
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