Question

The numerator of a fraction is four less than its denominator. If 1 is added to both, its
numerator and denominator, the fraction becomes half
. Find the fraction.​

Answer 1

Question:

The numerator of a fraction is 4 less than its denominator .If 1 is added to both its numerator and denominator , the fraction becomes half.

Find the fraction.

Answer:

Required fraction is 3/7.

Solution:

Let the numerator and the denominator of the required fraction be N and D respectively.

Then ,

The required fraction will be N/D.

According to the question,

The numerator of the fraction is 4 less then its denominator.

Thus,

=> N = D - 4 ------(1)

Also,

It is given that, when 1 is added to both its numerator and denominator , then the fraction becomes half.

Thus,

=> (N+1)/(D+1) = 1/2

=> (D-4+1)/(D+1) = 1/2

=> (D-3)/(D+1) = 1/2

=> 2(D-3) = (D+1)

=> 2D - 6 = D + 1

=> 2D - D = 1 + 6

=> D = 7

Now,

Putting D = 7 in eq-1 ,we get;

=> N = D - 4

=> N = 7 - 4

=> N = 3

Hence,

The required fraction is N/D ie; 3/7.

Answer 2

Question:

The numerator of a fraction is 4 less than its denominator .If 1 is added to both its numerator and denominator , the fraction becomes half. Find the fraction. ?

Solution

Assumption

Numerator be p - 4 (Given)

Denominator be p

As per Question it is given that 1 is added to both , its numerator and denominator

Hence,

$\tt{\rightarrow\dfrac{p-4+1}{p+1}=\dfrac{1}{2}}}$

So,

$\tt{\rightarrow\dfrac{p-3}{p+1}=\dfrac{1}{2}}}$

So,

After cross multiplication we get,

⇒ 2(p - 3) = 1(p + 1)

⇒ 2p - 6 = p + 1

⇒ 2p - p = 1 + 6

⇒ p = 7

Therefore,

Fraction will be,

$\tt{\rightarrow\dfrac{p-4}{p}}}$

$\tt{\rightarrow\dfrac{7-4}{7}}}$

$\tt{\rightarrow\dfrac{3}{7}}}$

Hence,

Fraction

$\tt{\rightarrow\dfrac{3}{7}}}$