Question

A fraction will be equal to 6/8 if 1 is added to both numerator and denominator. If however, 5 is subtracted from both numerator and denominator the fraction will be equal to 1/3.What is the fraction

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:The\:fraction\:=\:\dfrac{6.2}{8.6}}}}$

Step-by-step-explanation:

Let the numerator of the fraction be x.

And the denominator of the fraction be y.

$\displaystyle{\therefore\:\sf\:The\:fraction\:=\:\dfrac{x}{y}}$

From the first condition,

$\displaystyle{\sf\:\dfrac{x\:+\:1}{y\:+\:1}\:=\:\dfrac{\cancel{6}}{\cancel{8}}}$

$\displaystyle{\implies\sf\:\dfrac{x\:+\:1}{y\:+\:1}\:=\:\dfrac{3}{4}}$

$\displaystyle{\implies\sf\:4\:(\:x\:+\:1\:)\:=\:3\:(\:y\:+\:1\:)}$

$\displaystyle{\implies\sf\:4x\:+\:4\:=\:3y\:+\:3}$

$\displaystyle{\implies\sf\:4x\:=\:3y\:+\:3\:-\:4}$

$\displaystyle{\implies\sf\:4x\:=\:3y\:-\:1}$

$\displaystyle{\implies\:\boxed{\sf\:x\:=\:\dfrac{3y\:-\:1}{4}}\sf\:\qquad\cdots\:(\:1\:)\:}$

From the second condition,

$\displaystyle{\sf\:\dfrac{x\:-\:5}{y\:-\:5}\:=\:\dfrac{1}{3}}$

$\displaystyle{\implies\sf\:3\:(\:x\:-\:5\:)\:=\:1\:(\:y\:-\:5\:)}$

$\displaystyle{\implies\sf\:3x\:-\:15\:=\:y\:-\:5}$

$\displaystyle{\implies\sf\:3x\:-\:y\:=\:-\:5\:+\:15}$

$\displaystyle{\implies\sf\:3x\:-\:y\:=\:10}$

$\displaystyle{\implies\sf\:3\:\left(\:\dfrac{3y\:-\:1}{4}\:\right)\:-\:y\:=\:10\:\qquad\cdots\:[\:From\:(\:1\:)\:]}$

$\displaystyle{\implies\sf\:\dfrac{3\:(\:3y\:-\:1\:)}{4}\:-\:y\:=\:10}$

$\displaystyle{\implies\sf\:\dfrac{9y\:-\:3}{4}\:-\:y\:=\:10}$

$\displaystyle{\implies\sf\:\dfrac{9y\:-\:3\:-\:4y}{4}\:=\:10}$

$\displaystyle{\implies\sf\:9y\:-\:4y\:-\:3\:=\:10\:\times\:4}$

$\displaystyle{\implies\sf\:5y\:-\:3\:=\:40}$

$\displaystyle{\implies\sf\:5y\:=\:40\:+\:3}$

$\displaystyle{\implies\sf\:5y\:=\:43}$

$\displaystyle{\implies\:\boxed{\blue{\sf\:y\:=\:\dfrac{43}{5}}}}$

By substituting this value of y in equation ( 1 ), we get,

$\displaystyle{\sf\:x\:=\:\dfrac{3y\:-\:1}{4}\:\qquad\cdots\:(\:1\:)\:}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{3\:\times\:\dfrac{43}{5}\:-\:1}{4}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{\dfrac{129}{5}\:-\:1}{4}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{\dfrac{129\:-\:5}{5}}{4}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{\dfrac{124}{5}}{4}}$

$\displaystyle{\implies\sf\:x\:=\:\dfrac{\cancel{124}}{5}\:\times\:\dfrac{1}{\cancel{4}}}$

$\displaystyle{\implies\:\boxed{\pink{\sf\:x\:=\:\dfrac{31}{5}}}}$

Now,

$\displaystyle{\sf\:The\:fraction\:=\:\dfrac{x}{y}}$

$\displaystyle{\implies\sf\:The\:fraction\:=\:\dfrac{\cancel{\dfrac{31}{5}}}{\cancel{\dfrac{43}{5}}}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:The\:fraction\:=\:\dfrac{6.2}{8.6}}}}}$

Answer 2

Answer:

$\rm \dfrac{6.2}{8.6}$

Step-by-step explanation:

Let numerator = x

and denominator = y

$\bf \underline{\underline{\red {Case-1}}}$

$\rm \dfrac{x+1}{y+1}= \dfrac{6}{8} = \dfrac{3}{4}$

$\rm 4(x+1) = 3(y+1)$

$\rm 4x+4=3y+3$

$\rm 4x-3y+1=0$

$\rm 4x=3y-1$

$\rm x=\dfrac{3y-1}{4}$

$\bf \underline{\underline{\red{Case-2}}}$

$\rm \dfrac{x-5}{y-5}= \dfrac{1}{3}$

$\rm 3(x-5) = y-5$

$\rm 3x-15 = y-5$

$\rm 3x = y-5+15$

$\rm 3x = y+10$

Put the value of x,

$\rm 3 \bigg( \dfrac{3y-1}{4} \bigg) = y+10$

$\rm \dfrac{9y-3}{4} = y+10$

$\rm 9y-3 = 4y+40$

$\rm 9y-4y=40+3$

$\rm 5y=43$

$\rm y=\dfrac{43}{5} = 8.6$

Put the value of y,

$\rm x = \dfrac{3y-1}{4}$

$\rm x = \dfrac{3×8.6-1}{4}$

$\rm x = \dfrac{25.8-1}{4}$

$\rm x = \dfrac{24.8}{4} = 6.2$

So, The fraction is,

$\rm \dfrac{x}{y} = \dfrac{6.2}{8.6}$