# 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

As we learnt in

ARITHMETIC Mean -

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

$\dpi{100} \bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}$

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

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-