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?

by Gmat_mission » Wed Apr 17, 2019 12:33 am

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

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

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

[spoiler]OA=C[/spoiler]

Source: Veritas Prep
Last edited by Gmat_mission on Sat Apr 27, 2019 4:22 am, edited 1 time in total.
Source: — Data Sufficiency |

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by Vincen » Sat Apr 27, 2019 4:21 am

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Gmat_mission wrote:Is \(\left(\frac{a}{p}\right)\left(p^2+r^2+s^2\right)=ap+br+cs?\)

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

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

[spoiler]OA=C[/spoiler]

Source: Veritas Prep
Hi Gmat_mission.

Let's start by solving the parenthesis: $$\left(\frac{a}{p}\right)\left(p^2+r^2+s^2\right)=ap+\frac{a}{p}\cdot r^2+\frac{a}{p}\cdot s^2.$$ Now, let's take a loot at the given statements.

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

Replacing it into the expression above we get $$ap+\frac{c}{s}\cdot r^2+\frac{c}{s}\cdot s^2=ap+\frac{a}{p}\cdot r^2+cs$$ Since we don't know if \(\frac{a}{p}\cdot r^2\) is equal to \(br\) or not, this statment is NOT SUFFICIENT.

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

Again, if we replace this into the expression at the beginning we will get $$ap+\frac{a}{p}\cdot r^2+\frac{a}{p}\cdot s^2=ap+br+\frac{a}{p}\cdot s^2.$$ Since we don't know if \(\frac{a}{p}\cdot s^2\) is equal to \(cs\) or not, this statment is NOT SUFFICIENT.

Statement 1 + Statement 2:

In this case we get $$ap+\frac{a}{p}\cdot r^2+\frac{a}{p}\cdot s^2=ap+br+cs.$$ An this is what we needed to answer. So, the answer is YES.

Therefore, using both statements together is SUFFICIENT.

Hence, the correct answer is the option _C_.

I hope it helps you. <i class="em em---1"></i>