Test Index

Ratio and Proportion

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Question : 4 of 112

Marks: +1, -0

If

(a^{2}+b^{2}):(a^{2}-b^{2})=41: 9

(a^{3}+b^{3}):(a^{3}-

b^{3})

is

$5: 4$
$125: 64$
$189: 61$
$41: 9$

Solution:

Concept:

Find the ratio

\frac{a}{b}

from the given squares ratio, then compute the cubes ratio.

Explanation:

Given

\frac{a^2+b^2}{a^2-b^2} = \frac{41}{9}

.

Cross-multiplying:

9(a^2+b^2) = 41(a^2-b^2)

.

Expand:

9a^2+9b^2 = 41a^2-41b^2

.

Bring terms:

9b^2+41b^2 = 41a^2-9a^2

, giving

50b^2 = 32a^2

.

Thus

\frac{a^2}{b^2} = \frac{32}{50} = \frac{16}{25}

.

Taking positive root:

\frac{a}{b} = \frac{5}{4}

(since ratio is positive).

Let

a=5k

and

b=4k

.

Then

a^3+b^3 = 125k^3+64k^3 = 189k^3

.

And

a^3-b^3 = 125k^3-64k^3 = 61k^3

.

So the required ratio

(a^3+b^3):(a^3-b^3) = 189:61

.

Answer:

189:61

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