Question

The sum of the numerator and denominator of fraction is 8. If 3 is added to both the numerator and the denominator the fraction become 3/4. find the fraction

Answer 1

Given:-

• Sum of numerator & denominator = 8
• 3 added to both numerator & denominator fraction = 3/4

To find:-

• Original fraction = ?

Solution:-

→ Numerator + Denominator = 8

→ Numerator = 8 - denominator

Let the denominator be D & numerator be

(8 - D)

A/q,

→ (8 - D + 3)/(D + 3) = 3/4

→ (11 - D)/(D + 3) = 3/4

→ 3(D + 3) = 4(11 - D)

→ 3D + 9 = 44 - 4D

→ 3D + 4D = 44 - 9

→ 7D = 35

→ D = 35/7

→ D = 5

So, denominator = 5

→ Now, numerator = 8 - D

→ Numerator = 8 - 5

→ Numerator = 3

Therefore,

→ Fraction = Numerator/Denominator

→ Original fraction = 3/5

Hence,

$\therefore\underline{\boxed{\textsf{Original fraction obtained = {\textbf{ \dfrac{\text{3}}{\text{5}} }}}}}$

Answer 2

Given,

• The sum of the numerator and denominator of a fraction is 8 .
• If 3 is added to both the numerator and denominator then the fraction becomes 3/4 .

To Find,

• The fraction

Solution,

⇒Suppose the numerator be x

And , Suppose the denominator be y

According to the First Condition :-

• The sum of the numerator and denominator of a fraction is 8

$\implies \sf{ x + y = 8}$

$\implies \boxed{\sf{ x = 8 - y}}$

According to the Second Question :-

• If 3 is added to both the numerator and denominator then the fraction becomes 3/4 .

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

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

$\implies \sf{4x + 12 = 3y + 9}$

$\implies \sf{4x - 3y = 9 - 12}$

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

$\implies \sf{4(8 -y) - 3y = -3}$

(Put the value of  x from the First Condition)

$\implies \sf{32 - 4y -3y = -3}$

$\implies \sf{-7y = -3 - 32}$

$\implies \sf{ -7y = -35}$

$\implies \sf{y = \dfrac{35}{7}}$

$\implies \boxed{\sf{ y = 5}}$

Now Put the value of y in First Condition :-

$\implies \sf{ x = 8 -y}$

$\implies \sf{ x = 8 - 5}$

$\implies \boxed{\sf{ x = 3}}$

Therefore ,

$\boxed{\sf{\red{The \ Fraction = \dfrac{3}{5}}}}$

$\rule{200}1$