Question

The denominator of a fraction exceeds its numerator by 3.If 1 is added to both numerator and denominator yhe difference between new and originalfractions is 1/24 .find the original fraction

Answer 1

let the denominator of a fraction be =x
let the numerator of a fraction be =x+3
required fraction=x+3/x
new fraction=x+3+1/x+1
=x+4/x+1
A/Q
x+3/x - x+4/x+1 =1/24
(x+3)(x+1)-(x)(x+4)/x(x+1)=1/24
x2+x+3x+3-x2-4x/x2+x=1/24
3/x2+x=1/24
3×24=x2+x
72=x2+x
0=x2+x-72
0=x2+9x-8x-72
0=x(x+9)-8(x+9)
0=(x+9)(x-8)
let x+9=0 and x-8=0
x= -9 and x=8
• •
• denominator can not be negative
•
• •denominator=x=8
and numerator=x+3=8+3=11
required fraction=11/8

Answer 2

Answer:

The fraction can be $\frac{5}{8}$  or $\frac{4}{3}$

Step-by-step explanation:

Let the numerator be x

The denominator of a fraction exceeds its numerator by 3.

Denominator = x+3

Fraction : $\frac{x}{x+3}$

1 is added to both numerator and denominator

So, Fraction becomes:  $\frac{x+1}{x+3+1}$

                                        $\frac{x+1}{x+4}$

Now we are given that difference between new and original fraction is 1/24

So, $\frac{x+1}{x+4}-\frac{x}{x+3}=\frac{1}{24}$

$\frac{3}{x^2+7x+12}=\frac{1}{24}$

$24 \times 3=x^2+7x+12$

$72=x^2+7x+12$

$x^2+7x+12-72=0$

$x^2+7x-60=0$

$x^2+12x-5x-60=0$

$x(x+12)-5(x+12)=0$

$(x-5)(x+12)=0$

$x=5,-12$

when x = 5

Fraction : $\frac{x}{x+3}$ =$\frac{5}{8}$

when x = -12

Fraction : $\frac{x}{x+3}$ =$\frac{-12}{-12+3}=\frac{-12}{-9}=\frac{4}{3}$

So, the fraction can be $\frac{5}{8}$  or $\frac{4}{3}$