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

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

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 .

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

Answer 2

➲ ɢɪᴠᴇɴ :-

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.

➲ ᴛᴏ ғɪɴᴅ :-

• Fraction

➲ sᴏʟᴜᴛɪᴏɴ :-

Let numerator of fraction be x and denominator be y

then,

✒ According to 1st condition :-

• Numerator + Denominator = 8

• x + y = 8.

• x = 8 - y --(1)

✒ According to 2nd condition :-

• Num + 3 / Den + 3 = 3/4

$\implies \: \tt \frac{x + 3}{y + 3} = \frac{3}{4} \\ \\ \\ \implies \: \tt 4(x + 3) = 3(y + 3) \\ \\ \\ \implies \: \tt 4x + 12 = 3y + 9 \\ \\ \\ \implies \: \tt 4x - 3y = - 12 + 9 \\ \\ \\ \implies \: \tt 4x - 3y = - 3$

We get,

➦ 4x - 3y = -3. ---(2)

Put the value of (1) in (2) , we get,

➭ 4x - 3y = -3

➭ 4(8 - y) - 3y = -3

➭ 32 - 4y - 3y = -3

➭ 32 - 7y = -3

➭ -7y = -3 - 32

➭ -7y = -35

➭ y = -35/-7

➭ y = 5

Put y = 5 in (1) , we get,

➭ x = 8 - y

➭ x = 8 - 5

➭ x = 3

Hence,

• Fraction is x/y = 3/5