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

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21 Mar 2017, 06:44
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Difficulty:

15% (low)

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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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.

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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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