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

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

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New post 08 Aug 2006, 13:49
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A
B
C
D
E

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Question Stats:

58% (01:40) correct 42% (01:12) wrong based on 921 sessions

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Aug 2013, 00:40, edited 1 time in total.
Edited the question and added the OA.

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New post 08 Aug 2006, 14:02
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D it is...

The way to do it is to try and break the units digit into numbers that satisfy the criterion that it should be the sum of a number and its double... i.e. x + 2x...

A satisfies, as 3 = 1 + 2, but 15 cannot be broken into such a form.
The next choice that satisfies is D; 9 = 3 + 6. Also, 12 can be written as (8+4)... Hence D
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New post 08 Aug 2006, 14:12
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D 129

Given G =10x+y
H= 10 (x/2) + y/2

G+H = 3/2(10x+y) = 3/2G

G < 100 => G+H < 150

A & B are ruled out

(G+H)x2/3 = G an integer.

Only D is an integer.

Answer: D

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New post 08 Aug 2006, 14:14
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Can you enlighten me why not E? :roll: what am I missing?

46 and 23 is 89

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Re: PS: sum of 2 digit numbers [#permalink]

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New post 08 Aug 2006, 14:16
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divide each answer choice by 3, the result must be 2 digit number and cannot be greater than 44. also the unit digit of the resulting 2 digit number cannot be greater than 4.

A. 153/3 = 51. no good.
B 150/3 = 50. again not.
C 137/3 = fraction doesnot work.
D 129/3 = 43. it works but let wait for E.
E 89/3 = fraction. same as C.

so, D is the answer.

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Re: PS: sum of 2 digit numbers [#permalink]

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New post 08 Aug 2006, 14:21
Professor wrote:
divide each answer choice by 3, the result must be 2 digit number and cannot be greater than 44. also the unit digit of the resulting 2 digit number cannot be greater than 4.


I am sorry... probably a dumb :?: but please bear with me :wink: so, why are we diving these by 3? :dunnow

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New post 08 Aug 2006, 14:21
u2lover wrote:
Can you enlighten me why not E? :roll: what am I missing?

46 and 23 is 89


69, not 89 :lol:
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Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

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New post 08 Aug 2006, 14:22
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D

That means H is half of G.
So G+ H = 3H.
This means sum of G and H must be divisible by 3.
C and E are out.

Lets see other choices

A - 153 then H = 51, G = 102 :no G is a two gidit number
B - 150 then H = 50, G = 100 :no G is a two gidit number
D - 129 then H = 43, G = 86 :yes ANSWER
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New post 08 Aug 2006, 14:24
paddyboy wrote:
u2lover wrote:
Can you enlighten me why not E? :roll: what am I missing?

46 and 23 is 89


69, not 89 :lol:


talking about feeling stupid on some days... can't stop laughing now... I need lunch :food no carbs :lol:

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New post 08 Aug 2006, 14:27
Eat some fruits... :P

u2lover wrote:
I need lunch :food no carbs :lol:

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Re: PS: sum of 2 digit numbers [#permalink]

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New post 08 Aug 2006, 15:09
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u2lover wrote:
Professor wrote:
divide each answer choice by 3, the result must be 2 digit number and cannot be greater than 44. also the unit digit of the resulting 2 digit number cannot be greater than 4.


I am sorry... probably a dumb :?: but please bear with me :wink: so, why are we diving these by 3? :dunnow


probably you got it by the time you read this post.

the question says that H is half of G and G is divisible by 2. if so G = 2H
their sum = G + H = 2H + H = 3H

so it should be divisible by 3.

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New post 08 Aug 2006, 15:17
haas_mba07 wrote:
Eat some fruits... :P

u2lover wrote:
I need lunch :food no carbs :lol:


yup, take some fruit. i just finished 3 bowls cuz i had to divide every answer choice by 3. :wink:

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New post 08 Aug 2006, 15:45
If H has digits x and y, G has digits 2x and 2y.
So G+H =30x+3y and x and y are less than 5

G+H must be a multiple of 3, and units digit must be one of {3,6,9,2}

Unless units digit is 2, number formed by hundreds and tens digits must be a multiple of 3.

Only D fits the bill

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New post 08 Aug 2006, 20:50
I got it in 30 seconds by picking numbers
I started out around 80 as the answer choices are about the 120-130 mark..

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New post 10 Aug 2006, 09:14
I remember a trick from last year, someone from kaplan had told me..whenever u see the "which of the following" type question..start with answer choices in the bottom, i.e d and e..

ok, so lets see

i picked answer choice D ...129..hmm...

sum=2X+x....so lets 129=3x, does it divide by 3 yes...x=43....i will keep the answer choice..i check 89..does it divide by 3..NO..so I am done...129 is my answer...

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New post 10 Aug 2006, 10:23
As a Kaplan instructor, I can tell you that that advice is not helpful. The correct answer is just as likely to be A as E. You should quickly look at all of the answers to see which is the easiest to evaluate/most promising.

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New post 11 Aug 2006, 02:27
D

let the tens digit be x and units digit be y

Hence first number = 10x + y
Second number = 5x + y/2

Sum = 15x + 3y/2

Substitute values for x and y
15*8 + 3*6/2 = 129

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Re: odds and evens! [#permalink]

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New post 04 May 2011, 23:56
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]

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New post 05 May 2011, 07:34
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]

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New post 05 May 2011, 08:35
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] 05 May 2011, 08:35

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