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)

As we learnt in 

ARITHMETIC Mean -

For the values x1, x2, ....xn 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

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