G=10a+b
H=10c+d, here c=a/2 & d=b/2 or a=2c and b=2d substitute these values
G=20c+2d;
G+H =30c+3d = 3(10c+d)
now in options check for which options are divisible by 3;
options A, B and D are divisible by 3;
A. 153 => 153/3=51 => c cannot be 5 as d will be a single digit integer.
B. 150 => 150/3=50 => c cannot be 5 as d will be a single digit integer.
D. 129 => 129/3=43 ---- CORRECT ANS. option D

There is a way of saving 60% time by eliminating the first three answers (A, B & C):
Since both digits of G must be even (so as to yield an integer when halved), the highest possible value of G is 88. So the highest possible value of G+H is 88+44=132 which eliminates the first three answers. So, we now have to test only the last two answers to find out which satisfies the equation 3G/2 for an integer value of G.

Another angle, not sure if some one has already covered this.
1) G can end with 0,2,4,6,8
2) Half of each is 5,1,2,3,4
3) Sum of units could be 5 or 3 or 6 or 9 or 2
So A or D or E
4) Sum of Tens 5,3,6,9,12
Only D remains

The easiest way to do is thinking of the biggest 2 digit number that can be halved .. obviously it has to be even as u can’t divide odd to get whole number .. so it can be 88 or 86 .. 132 is not an option so we are left with 129 (86 + 43)


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We can let a = the tens digit of H and b = units digit of H; thus, H = 10a + b and G = 20a + 2b and the sum of H and G is:

H + G = (10a + b) + (20a + 2b) = 30a + 3b = 3(10a + b) = 3H

Since the sum G + H is a multiple of 3, we can eliminate choices C and E. Now let’s analyze the remaining three choices:

A) 153

3H = 153

H = 51 and G = 102

However, G is a two-digit number, so A couldn’t be the answer.

B) 150

3H = 150

H = 50 and G = 100

However, G is a two-digit number, so B couldn’t be the answer.

Therefore, the answer must be D. Let’s verify it anyway.

D) 129

3H = 129

H = 43 and G = 86

Answer: D
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