Question

if 2 is added to the numerator and denominator.it become 9 upon 10 and if 3 is subtracted from the numerator and denominator. it become 4 upon 5 . find the fraction​

Answer 1

Answer:

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Answer 2

SOLUTION :-

Let,

Numerator = x

Denominator = y

According to the first condition,

$\longrightarrow \sf \dfrac{x+2}{y+2}=\dfrac{9}{10} \\\\\longrightarrow 10(x+2)=9(y+2) \\\\\longrightarrow 10x+20=9y+18 \\\\\longrightarrow 10x-9y = -2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .......................(1)$

According to the second condition,

$\longrightarrow\sf \dfrac{x-3}{y-3}=\dfrac{4}{5} \\\\\longrightarrow5(x-3)=4(y-3)\\\\\longrightarrow 5x-15=4y-12 \\\\\longrightarrow 5x-4y = 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(2)$

Equation(2) × 2,

$\longrightarrow \sf 10x-8y = 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(3)$

Equation(3) - Equation(1),

$\longrightarrow\sf -y = - 8 \\\\\longrightarrow \bf y = 8$

Substitute y = 8 in equation(2),

$\longrightarrow \sf 5x - 4 \times 8 = 3 \\\\\longrightarrow 5x- 32 = 3 \\\\\longrightarrow 5x= 35\\\\\longrightarrow \bf x = 7$

 $\bf FRACTION = \dfrac{7}{8}$