Question

If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it becomes 4/5.find the fractions.

Answer 1

Answer:

Required fraction is $\frac{7}{8}$

Step-by-step explanation:

Let the fraction be $\frac{x}{y}$

Where x is the numerator and y is the denominator.

First it is given that if 2 is added to the numerator and denominator it becomes 9/10.

Therefore,

$\frac{x + 2}{y + 2} = \frac{9}{10}$

By Cross multiplication,

$10 \times (x + 2) = 9 \times (y + 2) \\ 10x + 20 = 9y + 18 \\ 10x - 9y + 20 - 18 = 0 \\ 10x - 9y + 2 = 0.......(1)$

Again it is given that 3 is subtracted from the numerator and denominator it becomes 4/5.

So,

$\frac{x - 3}{y - 3} = \frac{4}{5}$

By Cross multiplication,

$5(x - 3) = 4(y - 3) \\ 5x - 15 = 4y - 12 \\ 5x - 4y - 15 + 12 = 0 \\ 5x - 4y - 3 = 0.......(2)$

We are multiplying equation (1) with 1 and equation (2) with 2 and subtracted from equation (2) from equation (1),

$10x - 8y - 6 - 10x + 9y - 2 = 0 \\ y - 8 = 0 \\ y = 8$

We are putting value of y in equation (2),

$5x - 4 \times 8- 3 = 0 \\ 5x - 32 - 3 = 0 \\ 5x - 35 = 0 \\ 5x = 35 \\ x = \frac{35}{5} \\ x = 7$

So value of x is 7 and value of y is 8.

Therefore required fraction $\frac{x}{y} = \frac{7}{8}$

This is a problem of Algebra.

Some important formulas of Algebra,

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)