Question

3. The sum of the numerator and denominator of a fraction is 12. If 1 is added to both numerator and denominator the fraction becomes 9/10. Find the fraction. ​

Answer 1

CORRECT QUESTION

The sum of the numerator and denominator of a fraction is 12. If 1 is added to both numerator and denominator the fraction becomes 3/4

TO FIND

Find the fraction.

SOLUTION

✞ Let the numerator be x and denominator be y

✒ According to the given condition

✯The sum of the numerator and denominator of a fraction is 12

• x + y = 12 ---(i)

✯If 1 is added to both numerator and denominator the fraction becomes 9/10.

• x + 1/y + 1 = 3/4

➠ 4(x + 1) = 3(y + 1)

➠ 4x + 4 = 3y + 3

➠ 4x - 3y = 3 - 4

➠ 4x - 3y = -1 ---(ii)

✞ Multiply (i) by 3 and (ii) by 1

• 3x + 3y = 36
• 4x - 3y = - 1

✞ Add both the equation

➠ (3x + 3y)+(4x - 3y) = 36+(-1)

➠ 3x + 3y + 4x - 3y = 36 - 1

➠ 7x = 35

➠ x = 35/7 = 5

✞ Put the value of x in equation (ii)

➠ 4x - 3y = -1

➠ 4*5 - 3y = -1

➠ 20 - 3y = -1

➠ 20 + 1 = 3y

➠ y = 21/3 = 7

Hence, the required fraction is 5/7

Answer 2

✰ Correction in question:-

The sum of the numerator and denominator of a fraction is 12. If 1 is added to both numerator and denominator the fraction becomes 3/4 Find the fraction.

✰ Given:-

• Let numerator :- x

• Let denominator :- y

$\implies \sf{ Sum \; of \; x + y = 12}$

★ So, we can say that y = 12 - x ------(1)

✰ To find :-

• Fraction of numerator and denominator

✰ Solution:-

★ When 1 is added to both x and y :-

• $\sf{ \dfrac{x + 1}{y + 1} = \dfrac{3}{4}}$

★ Put eq(1) in this:-

• $\sf{ \dfrac{x + 1}{ (12 - x) + 1} = \dfrac{3}{4}}$

• $\sf{ \dfrac{x + 1}{-x + 13} = \dfrac{3}{4}}$

★ Cross multiply:-

• $\sf{ 4(x + 1) = 3( -x + 13)}$

• $\sf{ 4x + 4 = -3x + 39}$

• $\sf{ 4x + 3x = 39 - 4}$

• $\sf{ 7x = 35}$

• $\sf{ x = \cancel\dfrac{35}{7}}}$

• $\sf{ x = 5}$

★ Put value of x in eq(1):-

y = 12 - 5

y = 7

★ Hence, Fraction of numerator and denominator:-

$\implies \sf\boxed{\dag \; \dfrac{x}{y} = \dfrac{5}{7}}$

$\rule{200}{2}$