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12. Find the sub – triplicate ratio of the following:(iii) $125{{a}^{3}}:343{{b}^{6}}$(iv)${64{{m}^{3}}}/{729{{n}^{3}}}\;:{216{{m}^{3}}}/{27{{n}^{3}}}\;$

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,$125{{a}^{3}}:343{{b}^{6}}$

$={}^{3}\sqrt{\left( 125{{a}^{3}} \right)}:{}^{3}\sqrt{\left( 343{{b}^{6}}
\right)}$

$={{\left( 125{{a}^{3}} \right)}^{{1}/{3}\;}}:{{\left( 343{{b}^{6}}
\right)}^{{1}/{3}\;}}$

$=5a:7{{b}^{2}}$

Therefore, sub – triplicate ratio is $5a:7{{b}^{2}}$

Solution:

Given,${64{{m}^{3}}}/{729n{}^{3}}\;:{216{{m}^{3}}}/{27{{n}^{3}}}\;$

${{=}^{3}}\sqrt{\left( 64{{m}^{3}}/729{{n}^{3}}
\right)}{{:}^{3}}\sqrt{\left( 216{{m}^{3}}/27{{n}^{3}} \right)\overset{\grave{\
}}{\mathop{\ }}\,}$

$={{\left( 64{{m}^{3}}/729{{n}^{3}} \right)}^{1/3}}:{{\left(
216{{m}^{3}}/27{{n}^{3}} \right)}^{1/3}}$

By simplification we get,

$=4m/9n:6m/3n$

$=\left( 4m/9n \right)\times \left( 3n/6m \right)$

$=2/9$

Therefore, sub – triplicate ratio is $2:9.$