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The average marks of boys in class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is

Option 1)
80

Option 2)
60

Option 3)
40

Option 4)
20

Answers (1)

best_answer

As we learnt in

ARITHMETIC Mean -

For the values x₁, x₂, ....x_n of the variant x the arithmetic mean is given by

$\bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}$

in case of discrete data.

let us donote sum of marks scored by boys as $S_{b}$ and that by girls as $S_{g}$ ,the no. of boys and girls be $n_{b}$ and $n_{g}$ respectively,

Then average $\left ( boys \right )=\frac{S_{b}}{n_b}=52 - - - - - -\left ( 1 \right )$

average $\left ( girls \right )=\frac{S_{g}}{n_g}=42 - - - - - -\left ( 2 \right )$

average $\left ( boys\ and\ girls \right )=\frac{S_{b}+S_{g}}{n_{b}+n_{g}}=50- - - - - -\left ( 3\right )$

From (1), (2) and (3), we have

$S_{b}=52n_{b}$ & $S_{g}=42n_{g}$ playing in 3

$\frac{52n_{b}+42n_{g}}{n_{b}+n_{g}}=50$

$\Rightarrow \frac{52\frac{n_{b}}{n_{g}}+42}{\frac{n_{b}}{n_{g}}+1}=50$

Let us denote $n_b/n_g=x$ then

$\Rightarrow \frac{52x+42}{x+1}=50$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

Hence, $n_b/n_g=4$

$n_b=4n_{g}$

Hence, $\frac{n_{b}}{n_{b+n_{g}}}$

$=\frac{4n_{g}}{5n_{s}}$ $=\frac{4}{5}$

$\Rightarrow$ 80%

Option 1)

80

Correct option

Option 2)

60

Inorrect option

Option 3)

40

Inorrect option

Option 4)

20

Inorrect option

Posted by

prateek

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