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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] New post 10 Jul 2019, 23:54
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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] New post 11 Jul 2019, 04:46
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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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Re: 75/4^2*3^2/45*2^4/45 [#permalink] New post 11 Jul 2019, 04:49
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Hi Sir,

awesome explanation as usual from you.

Regards
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Re: 75/4^2*3^2/45*2^4/45 [#permalink] New post 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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Re: 75/4^2*3^2/45*2^4/45 [#permalink] New post 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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