Question

9
A fraction becomes , if 2 is added to both the numerator and the denominator
11
5
If, 3 is added to both the numerator and the denominator it becomes Find the
fraction.
6
Ro​

Answer 1

Correct Question :

A fraction becomes ⁹/₁₁ , if 2 is added to both the numerator and denominator . If 3 is added to both numerator and denominator , it becomes  ⁵/₆ . Find the fraction .

Solution :

Let fraction be " ˣ/ᵧ "

where ,

• x denotes numerator
• y denotes denominator

A fraction becomes ⁹/₁₁ , if 2 is added to both the numerator and denominator

$:\implies \sf \dfrac{x+2}{y+2}=\dfrac{9}{11}$

$:\implies \sf 11(x+2)=9(y+2)$

$:\implies \sf 11x+22=9y+18$

$:\implies \sf x=\dfrac{9y-4}{11}...(1)$

If 3 is added to both numerator and denominator , it becomes  ⁵/₆

$:\implies \sf \dfrac{x+3}{y+3}=\dfrac{5}{6}$

$:\implies \sf 6(x+3)=5(y+3)$

$:\implies \sf 6x+18=5y+15$

$:\implies \sf 6x-5y+3=0$

Sub. (1) here ,

$:\implies \sf 6\left(\dfrac{9y-4}{11}\right)-5y+3=0$

Multiply both sides of the eq. by " 11 " ,

$:\implies \sf 11\times 6\left(\dfrac{9y-4}{11}\right)-11\times 5y+11\times 3=11\times 0$

$:\implies \sf 6(9y-4)-55y+33=0$

$:\implies \sf 54y-24-55y+33=0$

$:\implies \sf y=9\ \; \bigstar$

Sub. y value in (1) ,

$:\implies \sf x=\dfrac{9(9)-4}{11}$

$:\implies \sf x=\dfrac{81-4}{11}$

$:\implies \sf x=\dfrac{77}{11}$

$:\implies \sf x=7\ \; \bigstar$

So , Fraction =  $\sf \dfrac{x}{y}$  =  $\sf \dfrac{7}{9}\ \; \bigstar$

Answer 2

$\mathcal{\green{\underline{\underline{\red{QUESTION:-}}}}}$

✨ A fraction becomes ⁹/₁₁ , if 2 is added to both the numerator and denominator . If 3 is added to both numerator and denominator , it becomes ⁵/₆ . Find the fraction .

$\mathcal{\green{\underline{\underline{\red{ANSWER:-}}}}}$

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

• A fraction becomes ⁹/₁₁ , if 2 is added to both the numerator and denominator .

• If 3 is added to both numerator and denominator , it becomes ⁵/₆ .

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• The original fraction .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

Let,

• $\bf{{\red{Numerator\:be\:}}{X\:.}}$

• $\bf{{\red{Denominator\:be\:}}{Y\:.}}$

So,

$\red\checkmark$ The original fraction = $\bf\pink{\dfrac{X}{Y}}$

⍟ According to the question,

CASE - 1 :-

$\bf\blue{\dfrac{X\:+\:2}{Y\:+\:2}\:=\:\dfrac{9}{11}\:}$

$\rm{:\implies\:11X\:+\:22\:=\:9Y\:+\:18\:}$

$\rm{:\implies\:11X\:-\:9Y\:+\:22\:-\:18\:=\:0\:}$

$\bf{:\implies\:11X\:-\:9Y\:+\:4\:=\:0\:}$----(1)

CASE - 2 :-

$\bf\blue{\dfrac{X\:+\:3}{Y\:+\:3}\:=\:\dfrac{5}{6}\:}$

$\rm{:\implies\:6X\:+\:18\:=\:5Y\:+\:15\:}$

$\rm{:\implies\:6X\:-\:5Y\:+\:18\:-\:15\:=\:0\:}$

$\bf{:\implies\:6X\:-\:5Y\:+\:3\:=\:0\:}$----(2)

✨ Now, multiple 5 in the equation (1) .

$\bf\purple{:\implies\:55X\:-\:45Y\:+\:20\:=\:0\:}$----(a)

✨ And multiple 9 in the equation (2) .

$\bf\purple{:\implies\:54X\:-\:45Y\:+\:27\:=\:0\:}$----(b)

⍟ Substracting Equation (b) from Equation (a), we get

➳ 55X - 45Y + 20 - (54X - 45Y + 27) = 0

➳ 55X - 45Y + 20 - 54X + 45Y - 27 = 0

➳ X - 7 = 0

➳ X = 7

➪ Now, putting the value of X in the equation (1) .

➳ 11 × 7 - 9Y + 4 = 0

➳ 77 - 9Y + 4 = 0

➳ -9Y + 81 = 0

➳ -9Y = -81

➳ 9Y = 81

➳ Y = 81/9

➳ Y = 9

$\pink\therefore\:\bf{The\:original\:fraction\:is\:{\red{\dfrac{X}{Y}}},\:i.e.\:\green{\dfrac{7}{9}}\:.}$