Question

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 1/15. Find the fraction.

2 points

4/7 or 2/5

5/8 or 6/9

4/7 or 7/10

2/5 or 6/9

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Answer 1

AnSwer :

• Option 4 is the correct answer!!

$\bigstar\:\:{\underline{ \boxed{\large{ \red{ \sf{ \: \: \dfrac{2}{5} \: or \: \dfrac{6}{9} \: \: }}}}}}$

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SoluTion :

• Let us assume the denominator of the required fraction be x.

So,

• Numerator of the required fraction = x-3
• Original fraction = $\sf\dfrac{(x - 3)}{x}$

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Also,

If 1 is added to the denominator, then the new fraction obtained is $\sf\dfrac{x - 3}{x + 1}$

Now, According to the given conditions,

$: \implies \sf \: \: \dfrac{x - 3}{x + 1} = \dfrac{x - 3}{x - 1} - \dfrac{1}{15} \\ \\ \\ : \implies \sf \: \: \dfrac{x - 3}{x} - \dfrac{x - 3}{x + 1} = \dfrac{1}{15} \\ \\ \\ : \implies \sf \: \: \dfrac{(x - 3)(x + 1) - x(x - 3)}{x(x + 1)} = \dfrac{1}{15} \\ \\ \\ : \implies \sf \: \: \dfrac{ {x}^{2} - 2x - 3 - {x}^{2} + 3x }{ {x}^{2} + x } = \dfrac{1}{15} \\ \\ \\ : \implies \sf \: \: \dfrac{x - 3}{ {x}^{2} + x } = \dfrac{1}{15} \\ \\ \\ : \implies \sf \: \: {x}^{2} + x = 15( x - 3) \\ \\ \\ : \implies \sf \: \: {x}^{2} + x = 15x - 45 \\ \\ \\ : \implies \sf \: \: {x}^{2} - 14x + 45 = 0 \\ \\ \\ : \implies \sf \: \: {x}^{2} - 9x - 5x + 45 = 0 \\ \\ \\ : \implies \sf \: \:x(x - 9) - 5(x - 9) = 0 \\ \\ \\ : \implies \sf \: \:(x - 5)(x - 9) = 0 \\ \\ \\ : \implies \sf \: \:x = 0 + 5 \: \: or \: \: x = 0 + 9 \\ \\ \\ : \implies \sf \: \: { \boxed{ \bf{ \orange{ \: x = 5 \: }}}} \: \bigstar \: \: or \: \: { \boxed{ \bf{ \blue{ \: x = 9 \: }}}} \: \bigstar$

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Now,

• If x = 5

$: \implies \sf \: \: \dfrac{x - 3}{x} \\ \\ : \implies \sf \: \: \dfrac{5 - 3}{5} \\ \\ : \implies \sf \: \: { \boxed{ \bf{ \pink{\dfrac{ \: 2 \: }{ \: 5 \: }}}}} \: \bigstar$

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• If x = 9

$:\implies \sf \: \: \dfrac{x - 3}{x} \\ \\ :\implies \sf \: \: \dfrac{9 - 3}{9} \\ \\ :\implies \sf \: \: { \boxed{ \bf{ \purple{\dfrac{ \: 6 \: }{ \: 9 \: } }}}} \: \bigstar$

Therefore, required fraction is ${ \sf{ \: \: \dfrac{2}{5} \: or \: \dfrac{6}{9} \: \: }}$

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Answer 2

Refers to attachment ♡♡~