1/4 of the boys and 1/6 of the girls in the class are overweight. Given that 1/5 of the class is overweight, find the ratio of the number of boys to the number of girls in the class.
Answers
Explanation:
\bf\: \underline{\red{GIVEN }}GIVEN
1/4 of the boys and 1/6 of the girls are overweighted.
1/4 of boys + 1/6 of girls = 1/5.
\bf\: \underline{\red{TO \: FIND}}TOFIND
Ratio of number of boys and girls in class.
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Let the total boys in class be x
and total girls in class be y.
Now,
\sf \implies \: \dfrac{1}{4} \: of \: boys = \dfrac{1}{4} \: of \: \bf{x= \dfrac{x}{4} }⟹41ofboys=41ofx=4x
\sf \implies \: \dfrac{1}{6} of \: girls = \dfrac{1}{6} \: of \: y = \bf{\dfrac{y}{6} }⟹61ofgirls=61ofy=6y
According to question,
\bf \implies \: (x + y) \times \dfrac{1}{5} = \dfrac{x}{4} + \dfrac{y}{6}⟹(x+y)×51=4x+6y
\sf \implies \: \dfrac{x + y}{5} = \dfrac{6x + 4y}{24}⟹5x+y=246x+4y
\sf \implies \: 24(x + y) = 5(6x + 4y)⟹24(x+y)=5(6x+4y)
\sf \implies \: 24x + 24y = 30x + 20y⟹24x+24y=30x+20y
\sf \implies \: 30x - 24x = 24y - 2y⟹30x−24x=24y−2y
\sf \implies \: 6x = 4y⟹6x=4y
\large \implies \boxed {\boxed {\tt \blue { \dfrac{x}{y} = \dfrac{2}{3} }}}⟹yx=32
\bf \therefore \: \underline{2/3 \: is \: the \: required \: ratio \: of \: boys \: and \: girls}∴2/3istherequiredratioofboysandgirls
Answer:
2:3
hope helps.......
