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Each digit in the 2 digit number G is halved to form a new 2 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

For the question to make sense, G must be even. So we are adding an even number G to G/2, and the answer will be 3*(G/2), and thus must be a multiple of 3. Further, G < 100, so 3G/2 is less than 150. The only possible answer is thus D.
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so G + H = 15x + 3/2y. Multiply by 2 and you get 2 ( G+H )= 3 (10 x + y) so G+H must be a multiple of 3 and G a multiple of 2 (obviously otherwise we cant divide G) and we know G less than or equal to 88 (highest 2 digits even) and so H less than or equal to 44 (half G), so G+H less than 132.

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

Two-step solution:

G + G/2 = 3G/2 --> the sum is a multiple of 3.

G is a two-digit number --> G < 100 --> 3G/2 < 150.

Among the answer choices the only multiple of 3 which is less than 150 is 129.

Answer: D.

What could be the minimum number ?

Assuming G is a positive number, the least value of G+G/2 will be 20+10=30. G must be even and cannot be less that 20. If it's an even number less than 20, then G/2 will not be a two-digit number.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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06 Sep 2013, 00:08

What could be the minimum number ?[/quote]

Assuming G is a positive number, the least value of G+G/2 will be 20+10=30. G must be even and cannot be less that 20. If it's an even number less than 20, then G/2 will not be a two-digit number.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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21 Nov 2013, 06:08

Let G be XX. Let H be x/2 x/2. G+H= 3 () Its a multiple of 3. Only two numbers fit the bill 153 and 129. 153/3 = 51 ( not possible because G is a two digit number and 51 is half of 102). Hence (D) 129.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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04 Mar 2014, 17:14

Let H be the 2-digit number xy (actually 10x+y). Then G must be 2x2y (actually 10(2x) + 2y or 2(10x+y). In other words, the digits of G must be even single-digit numbers. The maximum value of G can be 88 and thereby H can be 44. Therefore, maximum value of G+H = 132. Therefore, A & B are out. Now G+H = 3(10x+y) implies, G+H must be a multiple of 3. Only D among the remaining answer choices is a multiple of 3. So D is the answer.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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04 Mar 2014, 19:48

prsnt11 wrote:

Let H be the 2-digit number xy (actually 10x+y). Then G must be 2x2y (actually 10(2x) + 2y or 2(10x+y). In other words, the digits of G must be even single-digit numbers. The maximum value of G can be 88 and thereby H can be 44. Therefore, maximum value of G+H = 132. Therefore, A & B are out. Now G+H = 3(10x+y) implies, G+H must be a multiple of 3. Only D among the remaining answer choices is a multiple of 3. So D is the answer.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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28 May 2015, 21:24

Each of digits in a 2 digit number when halved gives another number. This means both the digits must be even. Possible combinations are as follows: Digit Half sum ---------- ---- ----- 8 4 12 6 3 9 4 2 6 2 1 3

Looking at choices only 129 is the possible answer : numbers being 86 & 43.