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# 75/4^2*3^2/45*2^4/45

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75/4^2*3^2/45*2^4/45 [#permalink]  10 Jul 2019, 23:54
Expert's post
00:00

Question Stats:

69% (01:07) correct 30% (01:52) wrong based on 23 sessions
 Quantity A Quantity B $$\frac{75}{4^2} \times \frac{3^2}{45} \times \frac{2^4}{45}$$ $$\frac{3^2}{4^2} \times \frac{2^2}{5^2} \times \frac{10}{3}$$

A)The quantity in Column A is greater.
B)The quantity in Column 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: 75/4^2*3^2/45*2^4/45 [#permalink]  11 Jul 2019, 04:46
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Expert's post
Carcass wrote:
 Quantity A Quantity B $$\frac{75}{4^2} \times \frac{3^2}{45} \times \frac{2^4}{45}$$ $$\frac{3^2}{4^2} \times \frac{2^2}{5^2} \times \frac{10}{3}$$

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Break every number into the product of prime numbers...

Quantity A: $$\frac{75}{4^2} \times \frac{3^2}{45} \times \frac{2^4}{45}=\frac{(3)(5^2)}{(2^2)^2} \times \frac{3^2}{(3^2)(5)} \times \frac{2^4}{(3^2)(5)}=\frac{(3)(5^2)}{2^4} \times \frac{1}{5} \times \frac{2^4}{(3^2)(5)}=\frac{(2^4)(3)(5^2)}{(2^4)(3^2)(5^2)}=\frac{1}{3}$$

Quantity B: $$\frac{3^2}{4^2} \times \frac{2^2}{5^2} \times \frac{10}{3}=\frac{3^2}{(2^2)^2} \times \frac{2^2}{5^2} \times \frac{(2)(5)}{3}=\frac{3^2}{2^4} \times \frac{2^2}{5^2} \times \frac{(2)(5)}{3}=\frac{(2^3)(3^2)(5)}{(2^4)(3)(5^2)}=\frac{3}{(2)(5)}=\frac{3}{10}$$

Since $$\frac{1}{3}>\frac{3}{10}$$, the correct answer is A.

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

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Kudos [?]: 1843 [0], given: 7861

Re: 75/4^2*3^2/45*2^4/45 [#permalink]  11 Jul 2019, 04:49
Expert's post
Hi Sir,

awesome explanation as usual from you.

Regards
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Re: 75/4^2*3^2/45*2^4/45 [#permalink]  11 Jul 2019, 04:52
Expert's post
Carcass wrote:
Hi Sir,

awesome explanation as usual from you.

Regards

Thanks Carcass!
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Brent Hanneson – Creator of greenlighttestprep.com

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Re: 75/4^2*3^2/45*2^4/45 [#permalink]  11 Jul 2019, 09:12
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hi- before calculating exact values of Quantity A and Quantity B, it is suggested that both the quantities are reduced by common numbers. it will make the calculation easier
Re: 75/4^2*3^2/45*2^4/45   [#permalink] 11 Jul 2019, 09:12
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