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6^85^34^2/3^68^2

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6^85^34^2/3^68^2 [#permalink] New post 06 Aug 2017, 12:42
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Question Stats:

73% (01:19) correct 26% (01:00) wrong based on 26 sessions


Quantity A
Quantity B
\(\frac{6^85^34^2}{3^68^2}\)
\(\frac{6^95^34^3}{3^82^8}\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: 6^85^34^2/3^68^2 [#permalink] New post 21 Sep 2017, 09:01
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If we compare the two quantities we see we can simplify them until we get 1/64 and 1/96 so that A is larger!
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Re: 6^85^34^2/3^68^2 [#permalink] New post 02 Apr 2018, 12:21
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Answer: A
We just compare the two values and try to simplify them.
A ------------------------------------------------B
6^8 * 5^3 * 4^2 / 3^6 * 8^2 ------------------ 6^9 * 5^3 * 4^3 / 3^8 * 2^8
Multiply them by the minimum power of each element in the two values. 3^6 /6^8 * 5^3 * 4^2
The result will be a simple form
1/ 8^2 -------------------------------------------6 * 4 /3^2 * 2^8

We convert 8^2 and 2^ 8 to one single form. 8^2 = (2^3)^2 = 2^2*3 = 2^ 6
1/ 2^6--------------------------------------------6* 4 / 2^ 8 *3^2
Multiply both is 2^6:
1-------------------------------------------------6 * 4 / 2^2 * 3^2 = 2/3
So B is bigger than A.[/b]
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Re: 6^85^34^2/3^68^2 [#permalink] New post 04 Apr 2018, 07:25
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Instead of simplifying the whole terms, we can remove the terms which are common in both of the quantities and then compare the remaining parts
Re: 6^85^34^2/3^68^2   [#permalink] 04 Apr 2018, 07:25
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