Test Index

Ratio and Proportion

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Question : 21 of 112

Marks: +1, -0

If

2 s: 3 t

is the duplicate ratio of

2 s-p: 3 t-p,

then

$p^{2}=6 s t$
$p=6 s t$
$2 p=3 s t$
none

Solution:

Concept:

The duplicate ratio of

a:b

is

a^{2}:b^{2}

.

Explanation:

Given duplicate ratio of

(2s-p):(3t-p)

equals

2s:3t

.

So

\frac{(2s-p)^{2}}{(3t-p)^{2}} = \frac{2s}{3t}

.

Cross multiply:

3t(2s-p)^{2} = 2s(3t-p)^{2}

.

Expand:

3t(4s^{2} - 4sp + p^{2}) = 2s(9t^{2} - 6tp + p^{2})

.

Simplify:

12s^{2}t - 12stp + 3tp^{2} = 18st^{2} - 12stp + 2sp^{2}

.

Cancel

-12stp

on both sides:

12s^{2}t + 3tp^{2} = 18st^{2} + 2sp^{2}

.

Rearrange:

12s^{2}t - 18st^{2} = 2sp^{2} - 3tp^{2}

.

Factor:

6st(2s - 3t) = p^{2}(2s - 3t)

.

Assuming

2s - 3t \neq 0

, cancel to get

p^{2} = 6st

.

Answer:

p^{2} = 6st

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