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The ratio of a to b to c is 2 to 3 to 4, and a, b, c are pos

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by Max@Math Revolution » Wed Jun 29, 2016 7:23 pm
The ratio of a to b to c is 2 to 3 to 4, and a, b, c are positive integers. If the average (arithmetic mean) of them is 18, what is the value of a?
A. 3
B. 6
C. 9
D. 12
E. 15

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by 800_or_bust » Thu Jun 30, 2016 6:20 am
Max@Math Revolution wrote:The ratio of a to b to c is 2 to 3 to 4, and a, b, c are positive integers. If the average (arithmetic mean) of them is 18, what is the value of a?
A. 3
B. 6
C. 9
D. 12
E. 15

*An answer will be posted in 2 days.
I'm sure it can be solved algebraically, but I just recognized that since a is an even number in the given ratio, it would have to be an even number in order to make b an integer. That left just B & D, and B seemed way too low to give an arithmetic mean of 18. So I tested 12 and it worked perfectly. 12:18:24. Mean = 18. Answer: D
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by Max@Math Revolution » Sun Jul 03, 2016 4:43 am
a=2k, b=3k, c=4k, a+b+c=2k+3k+4k=(18)(3), 9k=18(3), k=6. Therefore, a=2k=2(6)=12, and the correct answer is D.
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by Danny@GMATAcademy » Sun Jul 03, 2016 7:08 am
If the numbers in a list are 'evenly spaced out,' then the average equals the median. (e.g. in the list {7, 14, 21, 28, 25}, 21 is immediately apparent as the average because each term is separated from the preceding term by the same number)

In this question, a, b and c are evenly spaced out, so we know that b is the average. Therefore b's real value is 18. this means our multiplier is 6. So a is 2*6=12.
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