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3 ⋅ 3 ⋅ 3 ⋅ 3/9 ⋅ 9 ⋅ 9 ⋅ 9=

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3 ⋅ 3 ⋅ 3 ⋅ 3/9 ⋅ 9 ⋅ 9 ⋅ 9= [#permalink]

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Question Stats:

88% (00:26) correct 12% (00:03) wrong based on 25 sessions

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\(\frac{3 ⋅ 3 ⋅ 3 ⋅ 3}{9 ⋅ 9 ⋅ 9 ⋅ 9}\)=

(A) \(\frac{1}{3}\)^4
(B) \(\frac{1}{3}\)^3
(C) 1/3
(D) 4/9
(E) 4/3
[Reveal] Spoiler: OA

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Re: 3 ⋅ 3 ⋅ 3 ⋅ 3/9 ⋅ 9 ⋅ 9 ⋅ 9= [#permalink]

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We should understand that (a.a/b.b) can be written as (a/b).(a/b) ----- (1)

Here "." represents multiplication and the above property is valid for multiplication.

Now applying (1) in the problem.

We can write the question as (3/9).(3/9).(3/9).(3/9) which on further simplification gives us (1/3).(1/3).(1/3).(1/3) = (1/3)^4

Hence the answer is A.
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3 ⋅ 3 ⋅ 3 ⋅ 3/9 ⋅ 9 ⋅ 9 ⋅ 9= [#permalink]

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New post 14 Oct 2017, 14:01
SajjadAhmad wrote:
\(\frac{3 ⋅ 3 ⋅ 3 ⋅ 3}{9 ⋅ 9 ⋅ 9 ⋅ 9}\)=

(A) \(\frac{1}{3}\)^4
(B) \(\frac{1}{3}\)^3
(C) 1/3
(D) 4/9
(E) 4/3

Exponents

\(\frac{3 ⋅ 3 ⋅ 3 ⋅ 3}{9 ⋅ 9 ⋅ 9 ⋅ 9}=\)

\(\frac{3^4}{(3^2)^4}=\frac{3^4}{3^8}= \frac{1}{3^{(8-4)}}= \frac{1}{3^4}=(\frac{1}{3})^4\)

Cancel factors

Factor out four 3s from both numerator and denominator, to yield

\(\frac{1 ⋅ 1 ⋅ 1 ⋅ 1}{3 ⋅ 3 ⋅ 3 ⋅ 3}=\frac{1}{3^4} =
(\frac{1}{3})^4\)

Answer A

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3 ⋅ 3 ⋅ 3 ⋅ 3/9 ⋅ 9 ⋅ 9 ⋅ 9=   [#permalink] 14 Oct 2017, 14:01
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