Test Index

Ratio and Proportion

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Question : 26 of 112

Marks: +1, -0

What quantity must be added to the terms of the ratio

p+q: p-q

to make it equal to

(p+q)^{2}

:(p-q)^{2} ?

$\frac{(q+p)^{2}}{p}$
$(q-p) 2 p$
$\frac{q^{2}-p^{2}}{2 p}$
none

Solution:

Concept:

Adding the same quantity to both terms of a ratio yields a new ratio.

Explanation:

Let

x

be the quantity to be added.

The new ratio is

\frac{p+q+x}{p-q+x} = \frac{(p+q)^2}{(p-q)^2}

.

Cross-multiply:

(p+q+x)(p-q)^2 = (p-q+x)(p+q)^2

.

Expand and group

x

terms:

x[(p-q)^2 - (p+q)^2] = (p-q)(p+q)[(p+q) - (p-q)]

.

Simplify:

(p-q)^2 - (p+q)^2 = -4pq

and

(p+q)-(p-q)=2q

.

Thus

x(-4pq) = (p^2 - q^2)(2q)

.

Assuming

q \neq 0

, divide by

2q

:

x(-2p) = p^2 - q^2

.

Hence

x = -\frac{p^2 - q^2}{2p} = \frac{q^2 - p^2}{2p}

.

Answer:

\frac{q^{2}-p^{2}}{2p}

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