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Is (a/p)(p^2+r^2+s^2)=ap+br+cs?

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Is (a/p)(p^2+r^2+s^2)=ap+br+cs?  [#permalink]

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New post 21 Mar 2017, 06:44
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Question Stats:

86% (01:51) correct 14% (02:36) wrong based on 62 sessions

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Is \((\frac{a}{p})(p^2+r^2+s^2)=ap+br+cs\)?

(1) \(\frac{c}{s}=\frac{a}{p}\)

(2) \(\frac{a}{p}=\frac{b}{r}\)

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Re: Is (a/p)(p^2+r^2+s^2)=ap+br+cs?  [#permalink]

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New post 21 Mar 2017, 08:03
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ziyuen wrote:
Is \((\frac{a}{p})(p^2+r^2+s^2)=ap+br+cs\)?

(1) \(\frac{c}{s}=\frac{a}{p}\)

(2) \(\frac{a}{p}=\frac{b}{r}\)


\(\big ( \frac{a}{p} \big )(p^2+r^2+s^2) = ap +\frac{a }{p}\times r^2 + \frac{a }{p}\times s^2 \quad (*)\)

(1) \(\frac{c}{s}=\frac{a}{p}\)
\(\implies (*) = ap + \frac{a }{p}\times r^2 + \frac{c}{s}\times s^2 = ap + cs + \frac{a }{p}\times r^2\)

We can't know whether \(\frac{a }{p}\times r^2= br\) or not. Insufficient.

(2) \(\frac{a}{p}=\frac{b}{r}\)
\(\implies (*) = ap + \frac{b }{r}\times r^2 + \frac{a}{p}\times s^2 = ap + br + \frac{a }{p}\times s^2\)

We can't know whether \(\frac{a }{p}\times s^2= cs\) or not. Insufficient.

Combine (1) and (2) we have
\((1) = ap + \frac{b }{r}\times r^2 + \frac{c}{s}\times s^2 = ap + br + cs\). Sufficient.

The answer is C
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Re: Is (a/p)(p^2+r^2+s^2)=ap+br+cs?  [#permalink]

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New post 22 May 2017, 17:40
hazelnut wrote:
Is \((\frac{a}{p})(p^2+r^2+s^2)=ap+br+cs\)?

(1) \(\frac{c}{s}=\frac{a}{p}\)

(2) \(\frac{a}{p}=\frac{b}{r}\)


The key to solving this question is substitution and recognizing patterns
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Re: Is (a/p)(p^2+r^2+s^2)=ap+br+cs? &nbs [#permalink] 22 May 2017, 17:40
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