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Kiana has two older twin brothers. The product of their three ages is

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Math Expert
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Kiana has two older twin brothers. The product of their three ages is  [#permalink]

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New post 28 Mar 2019, 23:35
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

72% (02:05) correct 28% (02:31) wrong based on 29 sessions

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Intern
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Joined: 06 Dec 2017
Posts: 14
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Re: Kiana has two older twin brothers. The product of their three ages is  [#permalink]

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New post 28 Mar 2019, 23:43
1
The product of ages = 128 = \(2^7 ;i.e. 2^3 * 2^3 * 2\) ;

8+8+2 = 18
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Re: Kiana has two older twin brothers. The product of their three ages is  [#permalink]

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New post 29 Mar 2019, 00:27
Bunuel wrote:
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?

(A) 10
(B) 12
(C) 16
(D) 18
(E) 24


prime factorization will help to determine the age of the twins.

128 = 64*2 = 32 * 2 * 2 = 4 * 8 * 2 * 2 = \(2^2 *2^3 *2 *2\) = \(2^3 2^3 * 2\)

age of the twins = 2^3.

8 + 8 + 2 = 18.

D is the correct answer.
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Re: Kiana has two older twin brothers. The product of their three ages is  [#permalink]

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New post 29 Mar 2019, 02:48
Bunuel wrote:
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?

(A) 10
(B) 12
(C) 16
(D) 18
(E) 24


a*b^2= 128
128 = 2*8^2
so sum = 18
IMO D
GMAT Club Bot
Re: Kiana has two older twin brothers. The product of their three ages is   [#permalink] 29 Mar 2019, 02:48
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