Question

19. The sum of the numerator and denominator of a fraction is 8. If 3 is
added to both of the numerator and the denominator, the fraction
3/4 becomes.Find the fraction.
​

Answer 1

$\Large{\underline{\underline{\tt{\orange{Given}}}}}$

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

$\Large{\underline{\underline{\tt{\orange{Find\:out}}}}}$

Find the fraction

$\Large{\underline{\underline{\tt{\orange{Solution}}}}}$

❍Let the required fraction be x/y

★ Sum of the numerator and denominator of a fraction is 8.

• x + y = 8 -----(i)

★ 3 is added to both of the numerator and the denominator, the fraction 3/4 becomes.

• x + 3/y + 3 = 3/4

➞ 4(x + 3) = 3(y + 3)

➞ 4x + 12 = 3y + 9

➞ 4x - 3y = 9 - 12

➞ 4x - 3y = - 3 -----------(ii)

Multiply (i) by 3 and (ii) by 1

• 3x + 3y = 24
• 4x - 3y = -3

Add both the equations

➞ (3x + 3y) + (4x - 3y) = 24 - 3

➞ 3x + 3y + 4x - 3y = 21

➞ 7x = 21

➞ x = 21/7 = 3

Put the value of x in eqⁿ (i)

➞ x + y = 8

➞ 3 + y = 8

➞ y = 8 - 3 = 5

$\large{\underline{\boxed{\tt{Therefore,\:x = 3\:and\:y = 5}}}}$

$\large{\underline{\boxed{\tt{Required\: fraction=\dfrac{x}{y}=\dfrac{3}{5}}}}}$

Answer 2

$\huge\underline\mathrm{SOLUTION:-}$

AnswEr:

• The required fraction is 3/5 .

Given:

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

Need To Find:

• The required fraction = ?

$\huge \underline \mathsf \red {ExPlanation:-}$

Let the required fraction be x/y .

Then, we have:

x + y = 8 ……(i)

And, x + 3/ y + 3 = 3/4

➠ 4(x + 3) = 3(y + 3)

➠ 4x + 12 = 3y + 9

➠ 4x - 3y = -3 ……(ii)

On multiplying (i) by 3, we get:

3x + 3y = 24

On adding (ii) and (iii), we get:

7x = 21

➠ x = 3

On substituting x = 3 in (i), we get:

3 + y = 8

➠ y = (8 - 3) = 5

∴ x = 3 and y = 5

Hence:

• $\dag$The required fraction is 3/5 .

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$