AnkitK wrote:

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

Say H=xy [x is the ten's place digit after being halved and y is the unit's place digit after being halved]

H=10x+y

G=(2x)(2y)

G=10*(2x)+2y

Possible values of 2x=2,4,6,8; 2y=0,2,4,6,8

Possible values of x=1,2,3,4; y=0,1,2,3,4

G+H=20x+2y+10x+y=30x+3y=3(10x+y)

Thus, the sum must be a multiple of 3; options C and E are out.

Let's try other options:

A. 3(10x+y)=153 i.e. 10x+y=51 i.e. 10*5+1=51; x=5(Not a possible value of x) and y=1

B. 3(10x+y)=150 i.e. 10x+y=50 i.e. 10*5+0=50; x=5(Not a possible value of x) and y=0

D. 3(10x+y)=129 i.e. 10x+y=43 i.e. 10*4+3=43; x=4(Possible) and y=3(Possible)

Ans: D

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~fluke

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