SSC GD Constable 6 Feb 2025 Shift 1 Paper

Section: Elementary Mathematics

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Question : 54 of 80

Marks: +1, -0

The mean proportion of

\;{a^{2}}/{b^{3}}

and

\;{9b^{2}}/{4a^{3}}

is _________.

$\;{3}/{2(ab)}$
$\;{9}/{4(ab)}$
$\;{3}/{2{√{ab}}}$
$\;{9}/{4{√{ab}}}$

Solution:

The mean proportion of

\;{ a ^2}/{b^3}

and

\;{9 b^2}/{4 a ^3}

.
Formula used:
Mean proportion between two values x and

y ={√{ }} x × y

Calculations:
Let

x =\;{ a ^2}/{b^3}

and

y =\;{9 b^2}/{4 a ^3}

.

\;⇒ \;\text " Mean proportion "\;=√{\;{a^2}/{b^3} × \;{9 b^2}/{4 a^3}}

\;⇒ \;\text " Mean proportion "\;=√{\;{a^2 × 9 b^2}/{b^3 × 4 a^3}}

\;⇒ \;\text " Mean proportion "\;=√{\;{9 × b^2 × a^2}/{4 × b^3 × a^3}}

\;⇒ \;\text " Mean proportion "\;=√{\;{9}/{4} × \;{b^2}/{b^3} × \;{a^2}/{a^3}}

\;⇒ \;\text " Mean proportion "\;=√{\;{9}/{4} × \;{1}/{b} × \;{1}/{a}}

\;⇒ \;\text " Mean proportion "\;=√{\;{9}/{4 × a × b}}

\;⇒ \;\text " Mean proportion "\;=\;{{√{ }} 9}/{{√{4}} × a × b}

\;⇒ \;\text " Mean proportion "\;=\;{3}/{2 {√{ }} a b}

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