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2^3 * 17 * 5^2

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2^3 * 17 * 5^2 [#permalink] New post 27 Jul 2018, 09:28
Expert's post
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Question Stats:

100% (00:56) correct 0% (00:00) wrong based on 30 sessions
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Re: 2^3 * 17 * 5^2 [#permalink] New post 27 Jul 2018, 09:44
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Carcass wrote:
Quantity A
Quantity B
\(\frac{2^3 * 17 * 5^2}{60}\)
\(\frac{255}{2}\)


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

Kudos for R.A.E


QTY A = \(\frac{2^3 * 17 * 5^2}{60} = \frac{8 * 17 * 25}{60} = \frac{170}{3}\)

As we compare we find the numerator of QTY A is less than the numerator of QTY B and

the denominator of QTY A is greater than the denominator of QTY B

Hence QTY A < QTY B

Option B
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Re: 2^3 * 17 * 5^2 [#permalink] New post 07 Jun 2019, 07:54
Expert's post
Carcass wrote:
Quantity A
Quantity B
\(\frac{2^3 * 17 * 5^2}{60}\)
\(\frac{255}{2}\)


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

Kudos for R.A.E


Let's do a little prime factorization

Given:
Quantity A: \(\frac{2^3 * 17 * 5^2}{60}=\frac{(2)(2)(2)(17)(5)(5)}{(2)(2)(3)(5)}\)

Quantity B: \(\frac{255}{2}=\frac{(3)(5)(17)}{2}\)


Simplify Quantity A to get:
Quantity A: \(\frac{(2)(17)(5)}{(3)}\)

Quantity B: \(\frac{(3)(5)(17)}{2}\)


Divide both quantities by 5 and by 17 to get:
Quantity A: \(\frac{2}{3}\)

Quantity B: \(\frac{3}{2}\)

Answer: B

Cheers,
Brent
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Re: 2^3 * 17 * 5^2   [#permalink] 07 Jun 2019, 07:54
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