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Find the reciprocal ratio of the following:(iii) ${{a}^{3}}{{b}^{2}}:{{a}^{2}}{{b}^{3}}$(iv) $81p{{q}^{2}}:54{{p}^{2}}q$

The ratio is used for comparing two quantities of the sane kind.

The ratio formula for two numbers says a and b is given by a:b or a/b. When two or more such ratios are equal, they are said to be in proportion.

The concept of ratio and proportion is majorly based on ratios and fractions.  

Solution:

Given ratio,${{a}^{3}}{{b}^{2}}:{{a}^{2}}{{b}^{2}}$

The reciprocal of the given ratio is

$1/{{a}^{3}}{{b}^{2}}:1/{{a}^{2}}{{b}^{3}}$

$=\left( 1/{{a}^{3}}{{b}^{2}} \right)\times \left( {{a}^{2}}{{b}^{3}}/1 \right)$

$=b:a$

Therefore, reciprocal of the ratio is $b:a$

Solution:

Given ratio,

$81p{{q}^{2}}:54{{p}^{2}}q$

The reciprocal of the given ratio

$\left( 1/81p{{q}^{2}} \right):1/54{{p}^{2}}q$

$=\left( 1/81p{{q}^{2}} \right)\times \left( {{54}^{2}}q/1 \right)$

By simplification we get,

$=2p/3q$

Therefore, reciprocal of the ratio is $2p:3q$