Question

If the numerator and denominator of a fraction, both are decreased by one the fraction becomes 2/3. If the numerator and denominator both are increased by one the fraction becomes 3/4. Find the fraction

Answer 1

Answer:

Step-by-step explanation:

Let the fraction is x/y

According to question

X-1/y-1=2/3

3x-3=2y-2

3x-2y=1 ______________(1)

Again

X+1/y+1=3/4

4x+4=3y+3

4x-3y=-1 ______________(2)

Equation (1)×4=> 12x-8y=4

Equation (2)×3=> 12x-9y=-3

___________________________

On Subtracting y=7

On putting value of "y" in equation (1)

we get ,

3x-2×7=1

=> 3x-14=1

=> 3x=15

=> x=5

So the fraction is 5/7

Answer 2

Answer:

Required fraction is $\frac{5}{7}$

Step-by-step explanation:

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

Where x is numerator and y is denominator.

Now numerator x is decreasing by 1 and become $x - 1$

and denominator y is decreasing by 1 and become $y - 1$

According to question,

$\frac{x - 1}{y - 1} = \frac{2}{3}$

By cross multiplication,

$3(x - 1) = 2(y - 1)$

$3x - 2y = - 2 + 3$

$3x - 2y = 1......(1)$

Again numerator x is increasing by 1 and become $x + 1$

and denominator y is increasing by 1 and become $y + 1$

According to question,

$\frac{x + 1}{y + 1} = \frac{3}{4}$

By cross multiplication,

$4x + 4 = 3y + 3 \\ 4x - 3y = - 1.....(2)$

From equation (1),

$3x = 1 + 2y \\ x = \frac{1 + 2y}{3}$

Putting value of x in equation (2),

$4 \times \frac{1 + 2y}{3} - 3y = - 1$

$4 + 8y - 9y = - 3$

$- y = - 7 \\ y = 7$

From equation (1),

$3x - 2 \times7 = 1 \\ 3x = 1 + 14 \\ 3x = 15 \\ x = 5$

Required fraction is $\frac{5}{7}$