Question

The domination of fraction exceeds its numinator by 4. If the numinator and dinominator are both incered by 3, the new fraction become 4/5 find the original fraction. Plezz tell the answer ​

Answer 1

let numerator be x

$as \: per \: conditions \\ \frac{x}{4 + x} \: and \\ \\ \: \frac{x + 3}{x + 7} = \frac{4}{5} \\ 5x + 15 = 4x + 28 \\ x = 13 \\ original \: fraction = \frac{13}{17}$

Answer 2

$\bf\blue{\underbrace{AnsweEr\implies \frac{13}{17}}}$

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

$\sf{\implies The\: numerator\:be\:x}$

$\sf{\implies The\: denominator\:be\:y}$

$\sf{\implies The\: orginal\: fraction\:be\:\frac{x}{y}}$

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

$\sf{\implies The\: orginal\: fraction=?}$

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

• By setting equation and calculate the value of x and y than calculate orginal fraction after putting the value of x and y]

$\sf\small\underline{Given\:in\:case\:(i):}$

• The denominator of fraction exceeds its numinator by 4]

$\rm{\implies y=x+4-------(i)}$

$\sf\small\underline{Given\:in\:case\:(ii):}$

• The numerator and denominator are both increased by 3, the new fraction become 4/5]

$\tt{\implies \dfrac{x+3}{y+3}=\dfrac{4}{5}}$

$\rm{\implies 5(x+3)=4(y+3)}$

$\rm{\implies 5x+15=4y+12}$

$\rm{\implies 5x-4y=12-15}$

$\rm{\implies 5x-4y=-3-----(ii)}$

• Putting the value of y=x+4 here]

$\rm{\implies 5x-4(x+4)=-3}$

$\rm{\implies 5x-4x-16=-3}$

$\rm{\implies x=-3+16\implies x=13}$

• Now putting the value of x=13 in eq (i)

$\rm{\implies y=x+4}$

$\rm{\implies y=13+4\implies y=17}$

$\sf\large{Hence,}$

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

$\sf{\implies The\: orginal\: fraction=\frac{13}{17}}$