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:

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:

$Let \: \: the \: \: fraction \: \: be  \: \: yx</p><p></p><p> \\ \\ \\ </p><p>Then, \: \: according \: \: to \: \: the \: \: given \: \: conditions, \: \: we \: \: have</p><p></p><p>$

$3x \: = \: 1118 \:  and  \: 2yx \: + \: 8 \: =52</p><p> \\ \\ </p><p>⇒11x \: −6y \: −18  \: and  \: 5x \: + \: 40 \: =4y</p><p> \\ \\ </p><p>⇒11x \: − \: 6y \: + \: 18 \: =0  \: and  \: 5x \: − \: 4y \: +40 \: =0</p><p></p><p>$

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$By \: \: cross- \:multiplication, \: \: we \: \: have$

$(−6)×40−(−4)×18x \: = \: 11×40−5×18−y \: = \: 11×(−4)−5×(−6)1</p><p>$