Question

Two numbers are in the ratio 4: 5. If the smaller is divided by 3 and the larger is increased by 1, the ratio becomes 1: 4. Find the numbers.​

Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf \implies Two \: numbers \: are \: in \: the \: ratio \: 4: 5$

$\sf If \: the \: smaller \: (4) \: is \: divided \: by \: 3 \: and \: the \: larger \: (5) \: is \: increased \: by \: 1.$

$\sf \implies The \: ratio \: becomes \: 1: 4$

$\bf \underline{ \underline{\maltese \: To \: find } }$

$\sf \implies Both \: the \: numbers = \: ?$

$\bf \underline{ \underline{\maltese \: Solution } }$

$\sf Let \: us \: assume \: that,$

$\sf \implies First \: number \: or \: smaller \: number \: be \: 4x.$

$\sf \implies Second \: number \: or \: larger \: number \: be \: 5x.$

$\bf \underline{Now},$

$\sf According \: to \: conditions \: given \: in \: the \: question,$

$\sf \implies Equation = \red{ \bf \dfrac{(4x \div 3)}{(5x + 1)} = \dfrac{1}{4} }$

$\sf \underline{Solving \: the \: above \: equation},$

$\sf \implies 4(4x \div 3) = 1(5x + 1)$

$\sf \implies \dfrac{ 4(4x)}{3} = 1(5x + 1)$

$\sf \implies \dfrac{16x}{3} = 1(5x + 1)$

$\sf \implies \dfrac{16x}{3} = 5x + 1$

$\sf \implies \dfrac{16x}{3} - \dfrac{5x}{1} = 1$

$\sf \implies \dfrac{16x - 5x \times 3}{3} = 1$

$\sf \implies \dfrac{x(16 - 5 \times 3)}{3} = 1$

$\sf \implies \dfrac{x }{3} = 1$

$\sf \implies x = 3 \times 1$

$\sf \implies x = 3$

$\bf \underline{Therefore},$

$\sf \implies First \: number \: or \: smaller \: number = \bf 4x = 4 \times 3 = 12$

$\sf \implies Second \: number \: or \: larger \: number = \bf5x = 5 \times 3 = 15$