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GMAT Tutor G
Joined: 24 Jun 2008
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Re: odds and evens!  [#permalink]

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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]

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1
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]

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

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let the number be x, now x is halved => x/2..the question asks what is x?

x(number) + x/2 (half of that number) = something

=> 3x/2 = something
=> x = 2 (something) / 3

so the answer choice must be divisible by 3

eliminate C, E right away

check the rest

D satisfies

3x/ 2 = 129

=> x = 86

Verify. 86 + 43 = 129

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

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1
u2lover wrote:
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

please explain

My thinking about this problem (detailed explanation):
use this flow chart
Attachments two digit.png [ 33.34 KiB | Viewed 12078 times ]

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

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Bunuel wrote:
u2lover wrote:
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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 57236
Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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ygdrasil24 wrote:
Bunuel wrote:
u2lover wrote:
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.

Hope it's clear.
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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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.

Hope it's clear.[/quote] Yes it is thanks So basically G ranges from 20 to 198 for all G >0
Math Expert V
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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ygdrasil24 wrote:
Yes it is thanks So basically G ranges from 20 to 198 for all G >0

No. G must also be a two digit number, so it ranges from 20 to 88.
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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Bunuel wrote:
ygdrasil24 wrote:
Yes it is thanks So basically G ranges from 20 to 198 for all G >0

No. G must also be a two digit number, so it ranges from 20 to 88.

Hmm... blunder as always By the way why cant G(max) be 98 , H(max) be 49 in that case
Math Expert V
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Posts: 57236
Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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ygdrasil24 wrote:
Bunuel wrote:
ygdrasil24 wrote:
Yes it is thanks So basically G ranges from 20 to 198 for all G >0

No. G must also be a two digit number, so it ranges from 20 to 88.

Hmm... blunder as always By the way why cant G(max) be 98 , H(max) be 49 in that case

We are told that EACH digit in the two-digit number G is halved, thus both digits of G must be even.
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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By the way why cant G(max) be 98 , H(max) be 49 in that case[/quote]

We are told that EACH digit in the two-digit number G is halved, thus both digits of G must be even.[/quote]

Hmmm..Okay under even constraint G max =88, Thanks
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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I followed the approach as :-

Multiplied each number with 2/3 and saw only 129 gives a 2 digit number i.e 43+86 which is possible,

for all of the rest number it gives a 3 digit number or is not multiple of 3.

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

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

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

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Maximum largest number = 88 + 44 = 132, so options A,B & C are eliminated

Number should be divisible by 3 (for ex a + a/2 = 3a/2)

129 > Divisible by 3 >>>>>>>>>>>> Answer = D
89 > Not divisible by 3
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Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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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.

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

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u2lover wrote:
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

let G = 10a+b
=> H = (10a+b)/2
G+H = (3/2)*(10a+b)

out of the options C and E go out as they are not divisible by 3.

substitute other options and try...

1) 153 = 3/2*G
=> G = a three digit number (goes out)

2) G = a three digit number (goes out)

4) no need to check but just to be sure...
G = 86
H = 43
G+H = 129
correct.

takes less than a minute to solve....
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GMAT 1: 690 Q49 V34 GMAT 2: 700 Q49 V35 Re: Each digit in the two-digit number G is halved to form a new  [#permalink]

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

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so, the digits of G must be even.
suppose G = 10A+B where A- tens digit, and B units digit.
now we are told that 5A+0.5B=H.

we are asked for the sum of G and H, or 15A+1.5B

knowing that A and B must be even, there aren't that many ways it can be arranged in that way. 4*4 = 16 possibilities.
nevertheless, since the results are over 100, we can definitely rule out 2, 4, and 6 as A. Thus, A must be 8.

Now, we have 15*8 = 120.
We're getting closer to the answer choice.
B can be 2 = so the min digit of H can be 1.
B can be 8 = so the max digit of H can be 4.

suppose B = 8 => 1.5B = 12, and the sum of G+H=132.
suppose B = 2 => 1.5B = 3, and the sum of G+H=123.

now we know for sure that G+H must be a number between 123 and 132. only one answer choice satisfies this condition.
_________________ Re: Each digit in the two-digit number G is halved to form a new   [#permalink] 23 Dec 2015, 19:08

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