so, the digits of G must be even.
suppose G = 10A+B where A- tens digit, and B units digit.
now we are told that 5A+0.5B=H.
we are asked for the sum of G and H, or 15A+1.5B
knowing that A and B must be even, there aren't that many ways it can be arranged in that way. 4*4 = 16 possibilities.
nevertheless, since the results are over 100, we can definitely rule out 2, 4, and 6 as A. Thus, A must be 8.
Now, we have 15*8 = 120.
We're getting closer to the answer choice.
B can be 2 = so the min digit of H can be 1.
B can be 8 = so the max digit of H can be 4.
suppose B = 8 => 1.5B = 12, and the sum of G+H=132.
suppose B = 2 => 1.5B = 3, and the sum of G+H=123.
now we know for sure that G+H must be a number between 123 and 132. only one answer choice satisfies this condition.
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