Question

the numerator of a fraction is 3 less than its denominator if 1 is added to the denominator the fraction is decreased by 1 by 15 find the fraction ​

Answer 1

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QueStI0N -

The numerator of a fraction is 3 less than its denominator.

If 1 is added to the denominator the fraction is decreased by (1 / 15) .

Find the fraction .

S0lUti0N -

Let the denominator of the required Fraction be x.

Therefore the numerator becomes ( x - 3 ) .

So, the fraction becomes [ ( x - 3 ) / x ] .

According to the question -

New Fraction - [ ( x - 3 ) / ( x - 1 ) ]

$=> \dfrac{ x-3}{x-1} = \dfrac{x-3}{x} - \dfrac{1}{15} \\ \\ => \dfrac{ x-3}{x-1} = \dfrac{ 15(x-3)-x} { 15x} \\ \\ => Expanding \: And \: Cancelling \: \\ \\ => { x } ^ 2 - 14x + 45 = 0 \\ \\ => { x } ^ 2 - 9 x - 5x + 45 = 0 \\ \\ => x( x - 9 ) - 5 ( x - 9 ) = 0 \\ \\ => ( x - 5)(x-9) = 0 \\ \\ => x = 5 \: Or \: x = 9 \\ \\ If \: x = 5, \: the \: required \: fraction \:: - \dfrac{2}{5} \\ \\ If \: x = 9, \: the \: required \: fraction \:: - \dfrac{6}{9} </p><p></p><p>$

Answer 2

$\underline{\huge\mathbb{GIVEN:-}}$

✥ The numerator of a fraction is 3 less than its denominator.

✥ If 1 is added to the denominator, the fraction is decreased by 1 by 15.

$\underline{\huge\mathbb{TO\ FIND:-}}$

➟ The fraction.

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

Let the denominator be x, and the numerator be (x - 3).

The fraction is :

$\displaystyle{\sf{\frac{x-3}{x} }}$

A/q,

$\bold{\displaystyle{\sf{\frac{x-3}{x+1}=\frac{x-3}{x} -\frac{1}{15} }}}$

⇾ x - 3 / x+1 = [15 ( x- 3 ) - 1 × x] / 15x

⇾ (x - 3 / x +1) =[ 15x - 45 - x] /15x

⇾ 15x (x-3/x+1) = [ 14x - 45 ]

⇾ 15x² - 45x = (14x - 45) (x +1)

⇾ 15x² - 45x = x(14x - 45) + 1 (14x - 45)

⇾ 15x² - 45x = 14x² - 45x + 14x - 45

⇾ 15x² - 14x² - 45x + 45x - 14x + 45

⇾ x² - 14 x + 45 = 0

⇾ x² - 9x - 5x + 45 = 0

➼ [By Middle term splitting]

⇾ x(x- 9) - 5(x - 9) = 0

⇾ (x - 9) or (x-5)= 0

⇾ x= 9 or x = 5

⇰ If x = 9 ,

Fraction = (x - 3) / x = (9-3)/9 = 6/9

⇰ If x = 5,

Fraction = (x - 3) / x = (5-3)/5 = 2/5