Question

(20) Two numbers are in the ratio of 1:3. If 5 is added to both the numbers, the ratio becomes 1:2. Find the numbers.

Answer 1

Let the numbers be x and y.

Given

Two numbers are in the ratio of 1:3

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

$\implies 3x=y$

$\implies 3x-y=0--(1)$

If 5 is added to both the numbers, the ratio becomes 1:2

$\implies \dfrac{x+5}{y+5} = \dfrac{1}{2}$

$\implies 2x+10=y+5$

$\implies 2x-y=5-10$

$\implies 2x-y=-5--(2)$

Solving (1) and (2), we get,

$3x-y=0\\\underline{2x-y=-5}\\\underline{\underline{x=5}}$

Substituting x = 5 in the second equation, we get

$\implies 2(5) - y = -5\\\implies 10-y=-5\\\implies -y=-5-10\\\implies -y=-15\\\implies y = 15$

$\boxed{\boxed{\bold{Therefore, \ the \ numbers \ are \ 5 \ and \ 15}}}}}}}}}}$

                                                       

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ numbers \ are \ 5 \ and \ 15 \ respectively.}$

$\sf\orange{Given:}$

$\sf{\implies{Two \ numbers \ are \ in \ the \ ratio }}$

$\sf{of \ 3:5}$

$\sf{\implies{If \ 5 \ is \ added \ to \ both \ the \ numbers,}}$

$\sf{the \ ratio \ becomes \ 1:2.}$

$\sf\pink{To \ find;}$

$\sf{The \ numbers.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ numbers \ be \ x \ and \ y.}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{\frac{x}{y}=\frac{1}{3}}$

$\sf{\therefore{3x=y}}$

$\sf{\therefore{3x-y=0...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{\frac{x+5}{y+5}=\frac{1}{2}}$

$\sf{\therefore{2(x+5)=y+5}}$

$\sf{\therefore{2x+10=y+5}}$

$\sf{\therefore{2x-y=-5...(2)}}$

$\sf{Subtract \ equation \ (2) \ from \ equation (1)}$

$\sf{3x-y=0}$

$\sf{-}$

$\sf{2x-y=-5}$

____________________

$\boxed{\sf{\therefore{x=5}}}$

$\sf{Substitute \ x=5 \ in \ equation \ (1)}$

$\sf{3(5)-y=0}$

$\sf{\therefore{15-y=0}}$

$\boxed{\sf{\therefore{y=15}}}$

$\sf\purple{\tt{\therefore{The \ numbers \ are \ 5 \ and \ 15 \ respectively.}}}$