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M26-03

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Math Expert
Joined: 02 Sep 2009
Posts: 52296

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16 Sep 2014, 00:24
00:00

Difficulty:

55% (hard)

Question Stats:

72% (02:11) correct 28% (02:39) wrong based on 29 sessions

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If $$5^{10x}=4,900$$ and $$2^{\sqrt{y}}=25$$ what is the value of $$\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}$$?

A. $$\frac{14}{5}$$
B. 5
C. $$\frac{28}{5}$$
D. 13
E. 14

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16 Sep 2014, 00:24
Official Solution:

If $$5^{10x}=4,900$$ and $$2^{\sqrt{y}}=25$$ what is the value of $$\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}$$?

A. $$\frac{14}{5}$$
B. 5
C. $$\frac{28}{5}$$
D. 13
E. 14

First thing one should notice here is that $$x$$ and $$y$$ must be some irrational numbers (4,900 has other primes than 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for $$x$$ and $$y$$.

$$5^{10x}=4,900$$

$$(5^{5x})^2=70^2$$

$$5^{5x}=70$$

Now, $$\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}} = 5^{5x}*5^{-5}*(2^{\sqrt{y}})^2$$

Since $$5^{5x}=70$$ then $$5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14$$

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Joined: 24 Jun 2015
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02 Jul 2015, 07:53
Hi,

This solution path is hard to visualize for me, is there a shortcut or an aproximation method viable to solve this question faster? If negative, what is your best advice with this kind of weird expressions?

Thanks a lot

Regards

Luis Navarro
Looking for 700
Math Expert
Joined: 02 Sep 2009
Posts: 52296

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02 Jul 2015, 08:09
2
luisnavarro wrote:
Hi,

This solution path is hard to visualize for me, is there a shortcut or an aproximation method viable to solve this question faster? If negative, what is your best advice with this kind of weird expressions?

Thanks a lot

Regards

Luis Navarro
Looking for 700

No shortcut there.

Study theory and practice questions from below links:
Theory on Exponents and Roots: math-number-theory-88376.html
Tips on Exponents and Roots: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

All DS roots problems to practice: search.php?search_id=tag&tag_id=49
All PS roots problems to practice: search.php?search_id=tag&tag_id=113

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

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Joined: 02 Jul 2012
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GMAT Date: 08-07-2012

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04 Jul 2015, 03:29
4√y=(2√y)2----- i dont get this part
Isnt (2√y)2== 2 (2√y)
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05 Jul 2015, 07:35
irus wrote:
4√y=(2√y)2----- i dont get this part
Isnt (2√y)2== 2 (2√y)

$$(2^{\sqrt{y}})^2=2^{2*\sqrt{y}}=4^{\sqrt{y}}$$
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Joined: 21 Jul 2017
Posts: 192
Location: India
GMAT 1: 660 Q47 V34
GPA: 4
WE: Project Management (Education)

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Updated on: 27 Oct 2018, 02:28
I think this is a high-quality question and I agree with explanation.

Originally posted by rever08 on 26 Oct 2018, 12:07.
Last edited by rever08 on 27 Oct 2018, 02:28, edited 1 time in total.
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27 Oct 2018, 01:35
rever08 wrote:
I think this is a poor-quality question and I agree with explanation. The question leaves room for us to assume 4 = 2 * 2^√y or 4= (2√y)^2. But, then yes 28/25 isn't available as an option so we need to assume the latter.

Not sure how you got that but the question is fine and there is no ambiguity there whatsoever.

Could yuu please show how you get these values? Thank you.
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GMAT 1: 660 Q47 V34
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27 Oct 2018, 02:30
Bunuel wrote:
rever08 wrote:
I think this is a poor-quality question and I agree with explanation. The question leaves room for us to assume 4 = 2 * 2^√y or 4= (2√y)^2. But, then yes 28/25 isn't available as an option so we need to assume the latter.

Not sure how you got that but the question is fine and there is no ambiguity there whatsoever.

Could yuu please show how you get these values? Thank you.

Ohh I made a brainless mistake . Thank you the question is clear now, have edited my post too.
Re: M26-03 &nbs [#permalink] 27 Oct 2018, 02:30
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M26-03

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