Question

A fraction is equal to 5/9 when three is added to the numerator and to the denominator. Then when 5 is subtracted from numerator and denominator it is 1/5. Find the fraction

Answer 1

$\mathfrak{\huge{Answer:}}$

Let's assume the fraction to be = $\tt{\frac{x}{y}}$

Now, start doing the operations given.
Condition 1 : When 3 is added to the numerator and denominator ( The new fraction becomes = $\tt{\frac{x+3}{y+3}}$ ), the fraction is equal to $\tt{\frac{5}{9}}$. In algebraic form :

$\tt{\frac{x+3}{y+3}= \frac{5}{9}}$

=》 9x + 27 = 5y + 15

=》 9x + 27 - 15 = 5y

=》 y = $\tt{\frac{9x + 12}{5}}$

Condition 2 : When 5 is subtracted from the numerator and the denominator ( The new fraction becomes = $\tt{\frac{x-5}{y-5}}$ ), the fraction is equal to $\tt{\frac{1}{5}}$. In algebraic form :

$\tt{\frac{x-5}{y-5} = \frac{1}{5}}$

=》 5x - 25 = y - 5

=》 5x - 25 + 5 = y

=》 y = 5x - 20

We know that : y = y

=》 $\tt{\frac{9x + 12}{5}}$ = 5x - 20

=》 9x + 12 = 25x - 100

=》 12 + 100 = 25x - 9x

=》 x = $\tt{\frac{112}{16}}$

=》 x = 7

y = 5 (7) - 20

=》 y = 15

Fraction becomes : $\tt{\frac{7}{15}}$

Answer 2

Answer:

Step-by-step explanation:

Let the Original Fraction be x/y.

Case |,

( x+3) / (y+3) = 5 / 9.

=> 9 ( x +3) = 5 ( y+3)

=> 9x + 27 = 5y + 15

=> 9x - 5y = 15 - 27

=> 9x - 5y = -12 _____________(1).

Case || ,

( x-5 )/ ( y -5) = 1 / 5

=> 5 ( x-5) = y - 5

=> 5x - 25 = y - 5

=> 5x - y = 20

=> 5x - 20 = y _____________(2)

Putting y = 5x - 20 in eq (1).

=> 9x - 5 ( 5x - 20) = -12

=> 9x - 25 x + 100 = - 12

=> - 16 x = -112

=> x = 112 / 16

=> x = 7.

Putting x = 7 in eq (2). we get,

=> 35 - 20 = y

=> y = 15.

Hence,

The Original Fraction = x/y = 7 / 15.