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If 2a – b = 3c

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If 2a – b = 3c

by gmatdriller » Tue Nov 15, 2011 12:04 pm
If 2a - b = 3c, where a, b, and c are non-zero integers,
which of the following could be the average (arithmetic mean)
of a and b, if the average must itself be an integer?

(A)-2 (B)-1 (C)1 (D)10 (E)12

I tried PLUG-IN method but wasted time; could anyone suggest
how to arrive at the best set of numbers to substitute?

Algebra method is no issue.

Thanks.E
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by shankar.ashwin » Tue Nov 15, 2011 12:22 pm
You want to find (a+b)/2

Given 2a-b = 3c

2a + 2b -3b = 3c (Adding and Subtracting 2b)

2(a+b) = 3(c+b)

Now, dividing by 4 (because we need (a+b)/2)

(a+b)/2 = 3/4 (c+B)

3/4 * (some number) = integer.

Only E satisfies
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by sam2304 » Tue Nov 15, 2011 12:23 pm
2a - b = 3c
2a + 2b - 2b - b = 3c
2(a+b) - 3b = 3c
2(a+b) = 3b + 3c
(a+b) = 3(b+c)/2
(a+b)/2 = 3(b + c)/4

So multiple of 4 would fit the average. Hence E.
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by neelgandham » Tue Nov 15, 2011 1:11 pm
gmatdriller wrote:If 2a - b = 3c, where a, b, and c are non-zero integers,
which of the following could be the average (arithmetic mean)
of a and b, if the average must itself be an integer?

(A)-2 (B)-1 (C)1 (D)10 (E)12

I tried PLUG-IN method but wasted time; could anyone suggest
how to arrive at the best set of numbers to substitute?

Thanks.E
2a - b = 3c
Implies b = 2a - 3c

(a+b)/2 = (2a-3c+a)/2 = (3a-3c)/2 = 3(a-c)/2 = 3*x . Only multiple of 3 in the options = 12 !
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by GMATGuruNY » Tue Nov 15, 2011 2:25 pm
gmatdriller wrote:If 2a - b = 3c, where a, b, and c are non-zero integers,
which of the following could be the average (arithmetic mean)
of a and b, if the average must itself be an integer?

(A)-2 (B)-1 (C)1 (D)10 (E)12

I tried PLUG-IN method but wasted time; could anyone suggest
how to arrive at the best set of numbers to substitute?

Algebra method is no issue.

Thanks.E
Plug in an easy combination.
Let c=2.
Then 2a-b = 6.
Thus, it's possible that a=4 and b=2.
Average = (4+2)/2 = 3.

Any answer choice that is a multiple of 3 must be a possible average for a and b.
Only answer choice E is a multiple of 3.

The correct answer is E.
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by gmatdriller » Tue Nov 22, 2011 11:28 pm
I could imagine how easy it was after choosing the appropriate plug-in value.
Thanks all.
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