Question

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.​

Answer 1

$\bf\: \underline{\red{GIVEN }}$

• 1/4 of the boys and 1/6 of the girls are overweighted.
• 1/4 of boys + 1/6 of girls = 1/5.

$\bf\: \underline{\red{TO \: FIND}}$

• Ratio of number of boys and girls in class.

ㅤ

Let the total boys in class be x

and total girls in class be y.

Now,

$\sf \implies \: \dfrac{1}{4} \: of \: boys = \dfrac{1}{4} \: of \: \bf{x= \dfrac{x}{4} }$

$\sf \implies \: \dfrac{1}{6} of \: girls = \dfrac{1}{6} \: of \: y = \bf{\dfrac{y}{6} }$

According to question,

$\bf \implies \: (x + y) \times \dfrac{1}{5} = \dfrac{x}{4} + \dfrac{y}{6}$

$\sf \implies \: \dfrac{x + y}{5} = \dfrac{6x + 4y}{24}$

$\sf \implies \: 24(x + y) = 5(6x + 4y)$

$\sf \implies \: 24x + 24y = 30x + 20y$

$\sf \implies \: 30x - 24x = 24y - 2y$

$\sf \implies \: 6x = 4y$

$\large \implies \boxed {\boxed {\tt \blue { \dfrac{x}{y} = \dfrac{2}{3} }}}$

$\bf \therefore \: \underline{2/3 \: is \: the \: required \: ratio \: of \: boys \: and \: girls}$

Answer 2

$\bold{\large{\purple{\underbrace{Answer \implies 2:3}}}}$

$\sf\large{\underline{\underline{\red{Given:}}}}$

$\sf{\longrightarrow \dfrac{1}{4}th\:of\:the\:boys\:and\:\dfrac{1}{6}th \:of \:the\:girls}$$\sf{ \: \: \: \: \: in\:the\:class\:are\:overweight.}$

$\sf{And,}$

$\sf{\longrightarrow \dfrac{1}{5}th\:of\:the\: class \:is\:overweight}$

$\sf\large{\underline{\underline{\red{To\:Find:}}}}$

$\sf{\longrightarrow Ratio\:of\:boys\:with\:respect\:to\:number}$$\sf{of\:girls\:in\:class......?}$

$\sf\large{\underline{\underline{\red{Let:}}}}$

$\sf{Total\:number\:of\:boys = 'x'}$

$\sf{Total\:number\:of\:girls = 'y'}$

$\sf{Also,}$

$\sf{Overweight\:of\:Boys = \dfrac{1}{4}th\:of\:x = \dfrac{x}{4}}$

$\sf{Overweight\:of\:Girls = \dfrac{1}{6}th\:of\:x = \dfrac{x}{6}}$

$\sf{So,}$

$\sf{ \: \: \: \: \: Overweight\:of\:class = x + \dfrac{y}{5}}$

$\sf\large{\underline{\underline{\red{According\:To\:Question:}}}}$

$\sf{\fbox{\green{\implies Overweight\:(Boys) + Overweight\:(Girls) = Overweight \:(Class)}}}$

$\sf{\implies (\dfrac{x}{4} + \dfrac{y}{6}) = (x + \dfrac{y}{5})}$

$\sf{\implies (\dfrac{3x + 2y }{12}) = (x + \dfrac{y}{5})}$

$\sf{\implies (15x + 10y) = (12x + 12y)}$

$\sf{\implies (3x = 2y)}$

$\sf{\implies x = \dfrac{2y}{3}}$

$\sf{\implies \dfrac{x}{y} = \dfrac{2}{3}}$

$\sf{\implies x:y = 2:3}$

$\sf\large{\underline{\underline{\red{Hence:}}}}$

$\bold{\underline{Ratio\:of\:no.\:of\:boys\:to\:the\:no.}}$$\bold{\underline{of\:girls\:in\:the\:class}}$

$\bold{\underline{\fbox{= 2:3}}}$