Question

The numerator of a fraction is one less than the denominator. If 3 is added to both numerator
and denominator, the fraction increases to 3/28. Find the fraction.​

Answer 1

Answer:

3/4

Step-by-step explanation:

Let the denominator be x.

A.T.Q., the numerator will be (x - 1).

Original fraction = ${\sf{{\dfrac{x - 1}{x}}}}$

Now, as given in the question, 3 is added to both numerator and denominator.

So,

Numerator = x - 1 + 3 = x + 2

Denominator = x + 3

A.T.Q.

${\sf{{\dfrac{x + 2}{x + 3}} = {\dfrac{x - 1}{x}} + {\dfrac{3}{28}} }}$

On further solving, we get

$\Rightarrow{\sf{ {\dfrac{x + 2}{x + 3}} = {\dfrac{28(x - 1) + 3(x)}{(28)(x)}} }}$

$\Rightarrow{\sf{ {\dfrac{x + 2}{x + 3}} = {\dfrac{28x - 28 + 3x}{28x}} }}$

$\Rightarrow{\sf{ {\dfrac{x + 2}{x + 3}} = {\dfrac{31x - 28}{28x}} }}$

Cross - multiplication.

$\Rightarrow$ 28x(x + 2) = (x + 3)(31x - 28)

$\Rightarrow$ 28x(x + 2) = x(31x - 28) + 3(31x - 28)

$\Rightarrow$ 28x² + 56x = 31x² - 28x + 93x - 84

$\Rightarrow$ 28x² + 56x = 31x² + 65x - 84

Transposing the terms.

$\Rightarrow$ 28x² - 31x² = 65x - 56x - 84

$\Rightarrow$ - 3x² = 9x - 84

$\Rightarrow$ 3x² + 9x - 84 = 0

$\Rightarrow$ 3(x² + 3x - 28) = 0

$\Rightarrow$ x² + 3x - 28 = 0

Using Middle Term Factorisation, we get

$\Rightarrow$ x² - 4x + 7x - 28 = 0

$\Rightarrow$ x(x - 4) + 7(x - 4) = 0

$\Rightarrow$ (x + 7)(x - 4) = 0

Using zero product rule, we get

$\Rightarrow$ x + 7 = 0 and x - 4 = 0

$\Rightarrow$ x = - 7 and x = 4

Here, x is a natural number, where x > 0.

Hence, x = 4

The required fraction is :

$\Rightarrow{\sf{ {\dfrac{x - 1}{x}} }}$

$\Rightarrow{\sf{ {\dfrac{4 - 1}{4}} }}$

$\Rightarrow{\boxed{\sf{\red{ {\dfrac{3}{4}} }}}}$

Answer 2

$\huge\bold{Question}$

The numerator of a fraction is one less than the denominator. If 3 is added to both numerator

and denominator, the fraction increases to 3/28. Find the fraction.

$\huge\bold{AnsWer}$

Let the denominator be → x

According to the question numerator will be → (x - 1)

Hence the equation will be = ${\tt{{\dfrac{x - 1}{x}}}}$

Now, According to the question, 3 is added to both numerator and denominator.

Hence the equation will be

•Numerator = x - 1 + 3 = x + 2

•Denominator = x + 3

=${\tt{{\dfrac{x + 2}{x + 3}} = {\dfrac{x - 1}{x}} + {\dfrac{3}{28}} }}$

By solving this equation we get ,

=${\tt{ {\dfrac{x + 2}{x + 3}} = {\dfrac{28(x - 1) + 3(x)}{(28)(x)}} }}$

=${\tt{ {\dfrac{x + 2}{x + 3}} = {\dfrac{28x - 28 + 3x}{28x}} }}$

= ${\tt{ {\dfrac{x + 2}{x + 3}} = {\dfrac{31x - 28}{28x}} }}$

By Cross - multiplication we get ,

= 28x(x + 2) = (x + 3)(31x - 28)

= 28x(x + 2) = x(31x - 28) + 3(31x - 28)

= 28x² + 56x = 31x² - 28x + 93x - 84

= 28x² + 56x = 31x² + 65x - 84

By simply Transposing the terms.

= 28x² - 31x² = 65x - 56x - 84

= - 3x² = 9x - 84

= 3x² + 9x - 84 = 0

= 3(x² + 3x - 28) = 0

= x² + 3x - 28 = 0

By doing Middle Term Factorisation, we get

= x² - 4x + 7x - 28 = 0

= x(x - 4) + 7(x - 4) = 0

= (x + 7)(x - 4) = 0

By using zero product rule, we get

= x + 7 = 0 and x - 4 = 0

= x = - 7 and x = 4

Here, x is a natural number, where x > 0.

So, x = 4

The required fraction is :

=$\bold\red{ {\dfrac{x - 1}{x}} }$

=$\bold\red{\dfrac{4 - 1}{4}} }$

=$\bold\red{\dfrac{3}{4}} }$