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by quantskillsgmat » Tue Feb 28, 2012 4:24 am
Q Each digit in the two digit number G is halved to form a new two digit number H.which of the following could be sum of G and H.
a)153 b)150 c)137 d)129 e)89
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by Anurag@Gurome » Tue Feb 28, 2012 4:38 am
quantskillsgmat wrote:Q Each digit in the two digit number G is halved to form a new two digit number H.which of the following could be sum of G and H.
a)153 b)150 c)137 d)129 e)89
Algebraic method:

Let G = 10x + y
H = (10x/2) + (y/2)
Then G + H = (10x + y) + (10x + y)/2
= (10x + y)[1 + 1/2]
= 3(10x + y)/2
= 3G/2, which implies that G + H must be a multiple of 3. Also G should be an even integer.
Since G is a 2-digit number, so if we take the maximum value of G as 98 and maximum value of H as 49 approx, then G + H = 147. So, answer choices A and B are not the possible answers.

So, from the answer choices, C, D, and E, only 129 is a multiple of 3.

The correct answer is D.
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by GMATGuruNY » Tue Feb 28, 2012 8:47 am
quantskillsgmat wrote:Q Each digit in the two digit number G is halved to form a new two digit number H.which of the following could be sum of G and H.
a)153 b)150 c)137 d)129 e)89
We can reason our way to the correct answer quickly.

Since each digit in G is to be halved, each digit in G must be even.
The answers represent possible values of G+H.
Since A, B, C and D are all greater than 120, G is likely between 80 and 90 (since 80+40=120):
88+44 = 132.
86+43 = 129.
Success!

The correct answer is D.
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