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Sum of two digit number G+H

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Sum of two digit number G+H

by championspunch » Sun Apr 14, 2013 8:23 pm
Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H?

a)153
b)150
c) 137
d) 129
e) 89

Soln:

I can get to a point where I derived

G+H = 15x + 1.5y


How do I go about eliminating answer choices? I tried getting 15 as common

= 15(10x+y/10) but this does not get me any far as well....

Any help on reasoning such questions will be appreciated.
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by Anju@Gurome » Sun Apr 14, 2013 9:03 pm
championspunch wrote:Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H?
Although it is not explicitly mentioned, but let us assume that the order of the digits remain same. i.e. units digit of G is halved to get the units digit of H and tens digit of G is halved to get the tens digit of H.

As digits of G are halved to form the number H, both digits of G must be even.
Let us assume, G = 10*2a + 2b, where 2a is the tens digit and 2b is the units digit of G.
Hence, H = 10*a + b

Now, G = 2H

So, (G + H) = 20a + 2b + 10a + b = 30a + 3b = 3(10a + b) = 3H = 3G/2
As (G + H) must be integer and 3 is not divisible by 2, (G + H) must be divisible by 3.

Also G is a two digit number with both digits even.
So, maximum value of G is 88.
Hence, maximum value of 3G/2 is 3*88/2 = 3*44 = 132

Hence, the correct answer must be a multiple of 3 and less than or equal to 132.
Only options D and E are less than 132 and among them option E is not divisible by 3.

The correct answer is D.
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by GMATGuruNY » Mon Apr 15, 2013 1:12 am
championspunch wrote:Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H?

a)153
b)150
c) 137
d) 129
e) 89
Since each of G's digits is HALVED to form H, both the tens digit and the units digit of G must be EVEN.
Four of the five answer choices are greater than 120.
Thus, it is almost certain that G≥80, so that H≥40 and G+H≥120.

Options for G+H:
80+40 = 120.
82+41 = 123.
84+42 = 126.
86+43 = 129 -- answer choice D.

The correct answer is D.
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by vipulgoyal » Mon Apr 15, 2013 10:07 pm
33 </ ans </ 132 (88+44) and multiple of 3 hence D
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by sumitshrestha » Tue Jan 23, 2018 6:22 am
Essentially, I see this as x + 1/2 x = answer., so 3/2 x = answer. This means answer is divisible by 3.

Cross out C and E.

Divide 153 by 3 = 51. Multiply by 2= 102 so incorrect as we need it to be 2 digit.
Same applies to option B.

Left with D.

Check answer: 129/3 = 43.
2 X 43 = 86. Fits within our rules of being even and only 2 digit.
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by GMATGuruNY » Tue Jan 23, 2018 10:01 am
I posted the following solution back in 2013, but it has inexplicably disappeared:
Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H?

a)153
b)150
c) 137
d) 129
e) 89
Since each of G's digits is HALVED to form H, both the tens digit and the units digit of G must be EVEN.
Four of the five answer choices are greater than 120.
Thus, it is almost certain that G≥80, so that H≥40 and G+H≥120.

Options for G+H:
80+40 = 120.
82+41 = 123.
84+42 = 126.
86+43 = 129 -- answer choice D.

The correct answer is D.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
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by Jeff@TargetTestPrep » Wed Jan 24, 2018 9:45 am
championspunch wrote:Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H?

a)153
b)150
c) 137
d) 129
e) 89
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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answer

by masoom j negi » Fri Dec 21, 2018 8:34 pm
Each digit of G is halved to make H. So, H is half of G.
So, G + H = 2H + H = 3H
So, the sum should be divisible by 3. Thus, C and E can't be the answer because they are not divisible by 3.
Now consider 153 = 102 + 51
It can't be the value because G should be a 2-digit no.
150 = 100 + 50
It also could not be the answer as G is a 2-digit number.
Hence, the only option that satisfies the given conditions is 129.
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