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# Each digit in the two-digit number G is halved to form a new

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Manager
Joined: 17 Jun 2015
Posts: 206
GMAT 1: 540 Q39 V26
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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28 Dec 2015, 11:56
G = 10x + y
H = 10 (x/2) + (y/2)

Adding the two, the number is in the form 1.5 (10x + y)

Look for options that can be represented in this format and is also a two digit number. Only D satisfies the condition. Hence, D
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Each digit in the two-digit number G is halved to form a new  [#permalink]

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12 Mar 2016, 15:21
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
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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12 Jan 2017, 12:48
Let G = 10a + b hence [halved G] = H = 10a/2 + b/2 = 5a + 3/2

Then G + H = 15a + 3/2b . Multiplying both sides by 2 we get 2(G+H) = 30a + 3b = 3(10a+b)

This means that (G+H) HAS TO BE A MULTIPLE OF 3 and this is fulfilled by answer D.
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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22 Feb 2017, 22:55
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.
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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12 Mar 2017, 12:26
let's find firstly the max number: 88+44=132
A, B, C are out
D and E
when it is easy to get 9 (6+3)
is it possible to get 8? x+x/2=8 then x is no integer
is it possible to get 12? x+x/2=12 x=8
Answer is D
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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19 Apr 2017, 23:53
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
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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28 Dec 2017, 07:47
Hi Guys,
This is the first time I am giving a solution

given G=xy and H=(x/2).(y/2)

so plugging in x=4 and y=6 as these have to be even numbers( as we can divide them by 2).

we get 46 and its half is 23 and total is 69 .We see that the total is 3 times H.
applying same rule to all the numbers :

153 ---> 51 and 102 ( rejected as we are looking for 2 digit numbers)
129 ---> 43 and 86 ( accepted)
137 --> not divisible by 3
150 -->50 and 100 (same as 1)
89 --> not divisible by 3

hence only 129 or option D in the question.
Intern
Joined: 28 Oct 2015
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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28 Dec 2017, 13:31
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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Manager
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Posts: 64
Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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28 Dec 2017, 18:04
Let G be the 2 digit number represented by xy, thus H would be x/2 y/2
Now possible pairs of (y,y/2) = (0,0) (2,1) (4,2) (6,3) (8,4)
thus possible sum of units place (0,3,6,9,2 with one carried)

For (x,x/2) = (2,1) (4,2) (6,3) (8,4)
Now possible values (3,4 - for one carried, 6,7 ; 9,10 ; 12,13)

Now browse through the options for this combination
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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24 Jan 2018, 09:39
u2lover 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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Manager
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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21 Dec 2018, 20:42
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.
Re: Each digit in the two-digit number G is halved to form a new   [#permalink] 21 Dec 2018, 20:42

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# Each digit in the two-digit number G is halved to form a new

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