Question

In the half-yearly examination of class IX of a school, the mean marks scored
by the boys is 52 and the mean marks scored by the girls is 48. If on the
whole, the mean marks of the class is 50.5, find the ratio of the number of
boys to the number of girls in the class.​

Answer 1

Step-by-step explanation:

Let the number of boys and girls be x and y

According to the question,

⇒ 52x+42y=50(x+y)

⇒ 52x+42y=50x+50y

⇒ 52x−50x=50y−42y

⇒ 2x=8y

∴ x=4y ------ ( 1 )

Total number of students in the class =x+y

=4y+y [ From ( 1 ) ]

=5y

Percentage of boys =

5y

4y

×100

=80%

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• In the half-yearly examination of class IX of a school, the mean marks scored by the boys is 52 and the mean marks scored by the girls is 48. If on the whole, the mean marks of the class is 50.5, find the ratio of the number of boys to the number of girls in the class.

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$\huge{AηsωeR} ✍$

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$\begin{gathered}\begin{gathered}\bf Given : - \begin{cases} &\sf{mean \: marks \: of \: boys = 52} \\ &\sf{mean \: marks \: of \: girls \: = 48}\\ &\sf{mean \: marks \: of \: the \: class = 50.5} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf Let :- \begin{cases} &\sf{number \: of \: boys \: = x} \\ &\sf{number \: of \: girls = y}\end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf \: To Find :- \begin{cases} &\sf{x : y} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\bf \:Combined Arithmetic Mean </p><p>$

$\bf \: {X_c} = \dfrac{{{n_1}{{\overline X }_1} + {n_2}{{\overline X }_2}}}{{{n_1} + {n_2}}}$

where,

$\sf \:  ⟼{\overline X _1} = mean \: of \: boys \: = 52$

$\sf \:  ⟼{\overline X _2} = mean \: of \: girls \: = 48$

$\sf \:  ⟼{\overline X _c} = mean \: of \: class \: = 50.5$

$\sf \:  ⟼{n _1} = number\: of \: boys \: = x$

$\sf \:  ⟼{n _2} = number\: of \: girls \: = y$

So, on substituting the values, we get

$\sf \:  ⟼50.5 = \dfrac{52 \times x + 48 \times y}{x + y}$

$\sf \:  ⟼50.5(x + y) = 52x + 48y$

$\sf \:  ⟼50.5x + 50.5y = 52x + 48y$

$\sf \:  ⟼50.5y - 48y = 52x - 50.5y$

$\sf \:  ⟼1.5y = 1.5x$

$\sf \:  ⟼x = y$

$\bf\implies \:x : y = 1 : 1$

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$\large{\boxed{\boxed{\sf{Hence, \sf \: {number _{boys}}: {number _{girls}}= 1 :1 }}}}$