Question

08. In a fraction's numerator is increased by 1 and the denominator is increased by 2 then the fract
ion be comes 2/3. But when the numerator is increased by 5 and the denominator is increased by 1 then the fraction becomes 5/4. What is the value of the original fraction?*​

Answer 1

Answer:

Let the numerator and denominator be p and q respectively.

Then, (p + 2) / (q + 3) = 7/9

⇒ 9(p + 2) = 7(q +3)

⇒ 9p - 7q = 3 ....(i)

Again, (p - 1) / (q - 1) = 4/5

⇒ 5p - 4q = 1 ....(ii)

Solving (i) and (ii) we get,

p = 5, q = 6

Required fraction = 5/6

Answer 2

$\large \bf \clubs \: Given :-$

• In a fraction's numerator is increased by 1 and the denominator is increased by 2 then the fraction be comes 2/3.
• When the numerator is increased by 5 and the denominator is increased by 1 then the fraction becomes 5/4.

$\large \bf \clubs \: To \: Find :-$

• The Original Fraction

$\large \bf \clubs \: Solution :-$

• For Original Fraction Let,

• Numerator = x
• Denominator = decrease 1

So ,

$\text{Original Fraction} = \dfrac{ \text x}{ \text y}$

✏ When numerator is increased by 1 and the denominator is increased by 2 :

$\text{Fraction Becomes : } \dfrac{ \text x + 1}{ \text y + 2}$

According To Question :

$\dfrac{\text x + 1}{\text y + 2} = \dfrac{2}{3}$

:$\longmapsto3(\text x + 1) = 2(\text y + 2)$

:$\longmapsto3\text {x + 3 = 2y + 4}$

:$\longmapsto \bf{3x - 2y = 1} \: \: - - - (1)$

✏ When the numerator is increased by 5 and the denominator is increased by 1 :

$\text{Fraction Becomes : } \dfrac{ \text x + 5}{ \text y + 1}$

According To Question :

$\dfrac{\text x + 5}{\text y + 1} = \dfrac{5}{4}$

:$\longmapsto\text{4(x + 5) = 5(y + 1)}$

:$\longmapsto\text{4x + 20 = 5y + 5}$

:$\longmapsto \bf4x - 5 y = - 15 \: - - - (2)$

✏ Solving (1) and (2) :

$\purple{ \large :\longmapsto \underline {\boxed{{\bf x = 5 \: \: and \: \:y = 7 } }}}$

Hence ,

$\large \underline{ \pink{ \underline{ \pmb{ \frak {Original \: Fraction = \dfrac{5}{7} } }}}}$