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odds and evens!

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Manager
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odds and evens! [#permalink] New post 04 May 2011, 22:37
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

78% (01:38) correct 22% (01:19) wrong based on 9 sessions
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
[Reveal] Spoiler: OA

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Re: odds and evens! [#permalink] New post 04 May 2011, 22:56
the unit digit combinations are 6,3; 8,4; 4,2 giving 9, 2 and 6 as the units digit of the sum.

also if the digits are x and y then G = 10x + y and H = 5x + 0.5y G + H = 1.5 ( 10x + y).

thus with these two criteria. D fits the bill.
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Re: odds and evens! [#permalink] New post 05 May 2011, 00:16
Amit:pls explain lil briefly .Not exactly clear
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Re: odds and evens! [#permalink] New post 05 May 2011, 06:34
02468 - Even digits

12349 - Odd digits

So the set of halves of even digits gives (odds can't be halved):

01234


86
43
---
129

With some trial and error :

Answer - D
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Re: odds and evens! [#permalink] New post 05 May 2011, 07:35
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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Re: odds and evens! [#permalink] New post 05 May 2011, 13:45
Expert's post
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


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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Re: odds and evens! [#permalink] New post 05 May 2011, 15:23
G = 10x + y
H = 5x + y/2

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.

ABC out and E out because not a multiple of 3

Answer is D

Hope this is helpful
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Re: odds and evens! [#permalink] New post 06 May 2011, 14:49
G is even

H can be even or odd

G's unit digit possibilities are 0 2 4 6 8

H's unit digit possibilities are 0 1 2 3 4

No checking the answers we can see only D fits the criteria.

Answer is D.
Re: odds and evens!   [#permalink] 06 May 2011, 14:49
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