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In a Mathematics test, the average marks of boys is $x \%$ and the average marks of girls is $y \%$ with $x \neq y$. If the average marks of all students is $z \%$, the ration of the number of girls to the total number of students is
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Verified Answer
The correct answer is:
$\frac{z-x}{y-x}$
Given
Let no. of Boys $=B \&$ no of girls $=G$
$\therefore$ Sum of marks obtained by boys $=\mathrm{B} \cdot \mathrm{x}$
$\therefore$ Sum of marks obtained by girls = G.y
Now, given
$\begin{array}{c}
\frac{B x+G y}{B+G}=z \\
B(x-z)=G(z-y) \\
\frac{B}{G}=\frac{z-y}{x-z}
\end{array}$
Add 1
$\begin{array}{l}
\frac{B}{G}+1=\frac{z-y}{x-z}+1 \\
\Rightarrow \frac{B+G}{G}=\frac{x-y}{x-z} \\
\Rightarrow \frac{G}{B+G}=\frac{z-x}{y-x}
\end{array}$
Let no. of Boys $=B \&$ no of girls $=G$
$\therefore$ Sum of marks obtained by boys $=\mathrm{B} \cdot \mathrm{x}$
$\therefore$ Sum of marks obtained by girls = G.y
Now, given
$\begin{array}{c}
\frac{B x+G y}{B+G}=z \\
B(x-z)=G(z-y) \\
\frac{B}{G}=\frac{z-y}{x-z}
\end{array}$
Add 1
$\begin{array}{l}
\frac{B}{G}+1=\frac{z-y}{x-z}+1 \\
\Rightarrow \frac{B+G}{G}=\frac{x-y}{x-z} \\
\Rightarrow \frac{G}{B+G}=\frac{z-x}{y-x}
\end{array}$
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