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(6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7

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(6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7 [#permalink] New post 06 Oct 2010, 10:05
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A
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Difficulty:

  45% (medium)

Question Stats:

62% (02:42) correct 38% (01:50) wrong based on 232 sessions
\(\frac{6.804^6*1.701^{-13}}{2^{19}*3.402^{-7}}\).

A. .8502
B. 1.000
C. 1.7010
D 3.402
E. 6.804
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Oct 2013, 01:08, edited 1 time in total.
Edited the question.
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Re: MGMAT CAT exponents problem [#permalink] New post 06 Oct 2010, 10:18
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tonebeeze wrote:
Can you please walk me through simplifying this exponents problem. Thanks

(6.804)^6 (1.701)^-13
_________________________ =

(2)^19 (3.402)^-7


a. .8502
b. 1.000
b. 1.7010
c. 3.402
d. 6.804


\(\frac{6.804^6*1.701^{-13}}{2^{19}*3.402^{-7}}=\frac{(2^2*1.701)^6}{2^{19}*(2*1.701)^{-7}*1.701^{13}}=\frac{2^{12}*1.701^6*(2*1.701)^{7}}{2^{19}*1.701^{13}}=\frac{2^{12}*1.701^6*2^7*1.701^{7}}{2^{19}*1.701^{13}}=\frac{2^{19}*1.701^{13}}{2^{19}*1.701^{13}}=1\).

Answer: B.

Please do check the questions when posting and use formating for such formulas (writing-mathematical-symbols-in-posts-72468.html).
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Re: MGMAT CAT exponents problem [#permalink] New post 06 Oct 2010, 10:21
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tonebeeze wrote:
Can you please walk me through simplifying this exponents problem. Thanks

(6.806)^6 (1.701)^-13
_________________________ =

(2)^9 (3.402)^-7


a. .8502
b. 1.000
b. 1.7010
c. 3.402
d. 6.804


I'm sure the question must have \((6.804)^6\) in the numerator; otherwise you won't have much simplification. I also suspect that the denominator should have a 2^19 in it, instead of 2^9; otherwise you won't get one of the five answer choices. With those changes, then if you notice that 3.402 = 2(1.701), and that 6.804 = (2^2)(1.701), we can factor out 2's in order to cancel:

\(\frac{(6.804)^6 (1.701)^{-13}}{(2^{19})(3.402)^{-7}} = \frac{(2^2)^6(1.701)^6 (1.701)^{-13}}{(2^{19})(2^{-7})(1.701)^{-7}} = \frac{(2^{12})(1.701)^{-7}}{(2^{12})(1.701)^{-7}} = 1\)
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Re: MGMAT CAT exponents problem [#permalink] New post 11 Mar 2012, 01:45
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the hardest thing here is to discern that:

2*1.701=3.402
4*1.701=6.804

In order to save time and avoid making mistakes by re-writing decimals, 3.402 can be substituted with '2x' and 6.804 with 4x. Therefore, the initial formula is quite simplified:

\(\frac{(4x)^6*x^-13}{2^19*(2x)^-7}\)

Afterwards, it takes 20 seconds to arrive to the final result
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Re: (6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7 [#permalink] New post 08 Oct 2013, 11:39
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tonebeeze wrote:
Can you please walk me through simplifying this exponents problem. Thanks

(6.806)^6 (1.701)^-13
_________________________ =

(2)^9 (3.402)^-7


a. .8502
b. 1.000
b. 1.7010
c. 3.402
d. 6.804


Question is wrong. Please correct. Thanks
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Re: (6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7 [#permalink] New post 09 Oct 2013, 01:08
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Re: (6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7 [#permalink] New post 26 Oct 2013, 12:03
I realized something when looking at this one: they would never have you doing the kind of calculations that would lead to those long decimals. I figured it had to be B based simply on the fact that they rarely if ever have you calculating out that far past the decimal point.
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Re: (6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7 [#permalink] New post 26 Feb 2014, 01:37
Re-writing the equation in the powers of 1.701 & 2 :

2^12 . 1.701^6 . 1.701^ -13

Divide by

2^19 . 2^ -7 . 1.701^-7

Answer = 1 = B
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Re: (6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7 [#permalink] New post 30 Jul 2015, 14:09
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Re: (6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7 [#permalink] New post 30 Aug 2015, 07:00
igormkd wrote:
the hardest thing here is to discern that:

2*1.701=3.402
4*1.701=6.804

In order to save time and avoid making mistakes by re-writing decimals, 3.402 can be substituted with '2x' and 6.804 with 4x. Therefore, the initial formula is quite simplified:

\(\frac{(4x)^6*x^-13}{2^19*(2x)^-7}\)

Afterwards, it takes 20 seconds to arrive to the final result


How are you even supposed to know that 2*1.701 = 3.402??? I'm supposed to hand-multiply that out on the test?
Re: (6.804)^6*(1.701)^-13/(2)^19*(3.402)^-7   [#permalink] 30 Aug 2015, 07:00
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