Question

The numerator of a fraction is 1 less than the denominator. If 3 is added to each of the

numerator and denominator, the fraction is increased by 3/28. Find the fraction.

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Answer 1

GIVEN :–

• The numerator of a fraction is 1 less than the denominator.

• If 3 is added to each of the numerator and denominator, the fraction is increased by 3/28.

TO FIND :–

• Original fraction = ?

SOLUTION :–

• Let the denominator of fraction = x

• Then Numerator = (x - 1)

$\\ \implies \sf Fraction = \dfrac{x - 1}{x}$

▪︎ According to the question –

$\\ \implies \sf \dfrac{x - 1 + 3}{x + 3} =\dfrac{x - 1}{x} + \dfrac{3}{28}$

$\\ \implies \sf \dfrac{x + 2}{x + 3} =\dfrac{x - 1}{x} + \dfrac{3}{28}$

$\\ \implies \sf \dfrac{x + 2}{x + 3} =\dfrac{28(x - 1) + 3x}{28x}$

$\\ \implies \sf \dfrac{x + 2}{x + 3} =\dfrac{28x - 28 + 3x}{28x}$

$\\ \implies \sf \dfrac{x + 2}{x + 3} =\dfrac{ 31x - 28}{28x}$

$\\ \implies \sf (x + 2)(28x) = (x + 3)(31x - 28)$

$\\ \implies \sf 28 {x}^{2} + 56x= 31 {x}^{2} - 28x + 93x - 84$

$\\ \implies \sf - 3{x}^{2} - 9x + 84 = 0$

$\\ \implies \sf {x}^{2} + 3x - 28 = 0$

$\\ \implies \sf {x}^{2} + 7x - 4x - 28 = 0$

$\\ \implies \sf x(x + 7) - 4(x + 7) = 0$

$\\ \implies \sf (x - 4)(x + 7)= 0$

$\\ \implies \sf x = 4 \: , \: - 7$

• So that , Original fraction ${ \bold{ \: \: \dfrac{3}{4} \: \: or \: \: \dfrac{ - 8}{ - 7} (rejected)}}$

• Hence , Original fraction = $\: \sf \dfrac{3}{4}$

Answer 2

$~~~ \: Let \: the \: numerator \: be \: \red{x}. \\~~~ \:The \: dinomintor = \red{x + 1}. \\~~~ \: Fraction = \red{\frac{x}{x + 1}}.$

Now, Given that:-

$\frac{(x) + 3}{(x + 1) + 3} = \frac{x }{x + 1} + \frac{3}{28} \\\\ ↣\frac{x + 3}{x + 4} = \frac{x }{x + 1} + \frac{3}{28} \\\\ ↣ \frac{x + 3}{x + 4} - \frac{x }{x + 1} = \frac{3}{28} \\\\ ↣ \frac{(x + 3)(x + 1) - x(x + 4)}{(x + 4)(x + 1)} = \frac{3}{28} \\\\ ↣ \frac{ {x}^{2} + x + 3x + 3 - {x}^{2} - 4x }{ {x}^{2} + x + 4x + 4 } = \frac{3}{28} \\\\↣ \frac{3}{ {x}^{2} + 5x + 4 } = \frac{3}{28} \\\\ ↣ {x}^{2} + 5x + 4 = 28 \\\\ ↣{x}^{2} + 5x + 4 - 28 = 0 \\\\↣ {x}^{2} + 5x - 24 = 0 \\\\ ↣{x}^{2} + 8x - 3x - 24 = 0 \\\\ ↣x(x + 8) - 3(x + 8) = 0 \\\\↣ (x - 3)(x + 8) = 0 \\\\ ↣x = 3 \: or \: x = - 8$

The fraction $(\frac{x}{x + 1})= \red{\frac{3}{4}}$ or $\red{\frac{8}{7}}$.