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Each digit in the twodigit number G is halved to form a new
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Updated on: 13 Aug 2013, 23:40
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61% (01:43) correct 39% (01:10) wrong based on 1023 sessions
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Each digit in the twodigit number G is halved to form a new twodigit number H. Which of the following could be the sum of G and H? A. 153 B. 150 C. 137 D. 129 E. 89
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Originally posted by u2lover on 08 Aug 2006, 12:49.
Last edited by Bunuel on 13 Aug 2013, 23:40, edited 1 time in total.
Edited the question and added the OA.




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Re: Each digit in the twodigit number G is halved to form a new
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13 Aug 2013, 23:50




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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 G is a two gidit number
B  150 then H = 50, G = 100 G is a two gidit number
D  129 then H = 43, G = 86 ANSWER
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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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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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Can you enlighten me why not E? what am I missing?
46 and 23 is 89



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Re: PS: sum of 2 digit numbers
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08 Aug 2006, 13:16
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
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08 Aug 2006, 13:21
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 so, why are we diving these by 3?



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paddyboy wrote: u2lover wrote: Can you enlighten me why not E? what am I missing? 46 and 23 is 89 69, not 89
talking about feeling stupid on some days... can't stop laughing now... I need lunch no carbs



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Eat some fruits...
u2lover wrote: I need lunch no carbs



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Re: PS: sum of 2 digit numbers
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08 Aug 2006, 14:09
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 so, why are we diving these by 3?
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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haas_mba07 wrote: Eat some fruits... u2lover wrote: I need lunch no carbs
yup, take some fruit. i just finished 3 bowls cuz i had to divide every answer choice by 3.



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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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I got it in 30 seconds by picking numbers
I started out around 80 as the answer choices are about the 120130 mark..



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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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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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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!
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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!
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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!
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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! &nbs
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