Question

A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get 18/11, but if the numerator is increased by 8 and the denominator is doubled, we get 2/5. Find the fraction. ​

Answer 1

Answer:

$\huge \rm \pink{(answer)}$

The required fraction will be 12/25

Let us assume that the numerator of the fraction be x.

And the denominator of the fraction be y.

Given that:

A fractions such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get 18/11.

→ 3x/(y - 3) = 18/11

→ 3x × 11 = 18(y - 3)

→ 33x = 18y - 54

→ 3 × 11x = 3(6y - 18)

→ x = (6y - 18)/11

Given that:

But if the numerator is increased by 8 and the denominator is doubled, we get 2/5.

→ (x + 8)/2y = 2/5

→ 5(x + 8) = 2(2y)

→ 5(x + 8) = 4y

→ 5x + 40 = 4y

→ 5x = 4y - 40

→ x = (4y - 40)/5

Comparing both the equation:

(6y - 18)/11 = (4y - 40)/5

→ 5(6y - 18) = 11(4y - 40)

→ 30y - 90 = 44y - 440

→ 44y - 30y = - 90 + 440

→ 14y = 350

→ y = 350/14

→ y = 25

Here we get: Denominator = 25

Substituting the value of y in any equation:

→ x = (6 × 25 - 18)/11 or x = (4 × 25 - 40)/5

→ x = (150 - 18)/11 or x = (100 - 40)/5

→ x = 132/11 or x = 60/5

→ x = 12 or x = 12

Here we get: Numerator = 12

We know that:

Fraction = Numerator / Denominator

Fraction = 12/25

Answer 2

Answer:

Let the fraction be

y

x

Then, according to the given conditions, we have

y−3

3x

=

11

18

and

2y

x+8

=

5

2

⇒11x−6y−18 and 5x+40=4y

⇒11x−6y+18=0 and 5x−4y+40=0

By cross-multiplication, we have

(−6)×40−(−4)×18

x

=

11×40−5×18

−y

=

11×(−4)−5×(−6)

1

⇒

−240+72

x

=

440−90

−y

=

−44+30

..1

⇒−168

... ... x =−350

y= −14

...1

⇒x= −14 and y= −14

... .... −168 .−350

⇒x=12 and y=25

Hence, the fraction is 25

.... .... .... . ... ... ... ... .... ... .12

$\huge\mathfrak\red{Sanjeet ⚡}$