Question

The numerator of a fraction is 9 less than the denominator If the numerator is increased by 2 and the denominator by 9, we again get the same fraction. Find the fraction.​

Answer 1

Answer:

2/9 is the answer.

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Answer 2

Answer:

The required fraction is 2/9

Step-by-step explanation:

Given :

• The numerator of a fraction is 9 less than the denominator.
• If the numerator is increased by 2 and the denominator by 9, we again get the same fraction.

To find :

the fraction

Solution :

Let the numerator of the fraction be x and denominator of the fraction be y.

As, fraction = numerator/denominator

fraction = x/y

Numerator = denominator - 9

x = y - 9  [eqn. 1]

⇒ Numerator is increased by 2.

new numerator = x + 2

⇒ Denominator is increased by 9.

new denominator = y + 9

$\sf New \ fraction=\dfrac{x+2}{y+9}$

It's given that we get the same fraction.

Therefore,

  $\sf \dfrac{x+2}{y+9}=\dfrac{x}{y}$

Put x = y - 9, [ ∵ eqn. 1 ]

$\sf \dfrac{x+2}{y+9}=\dfrac{x}{y} \\\\ \sf \sf \dfrac{y-9+2}{y+9}=\dfrac{y-9}{y} \\\\ \sf \dfrac{y-7}{y+9}=\dfrac{y-9}{y} \\\\ \sf (y)(y-7)=(y-9)(y+9) \\\\ \sf y^2-7y=y^2-9^2 \ [ applying \ (a+b)(a-b)=(a^2-b^2) \ identity] \\\\ -7y=-81 \\\\ \sf 7y=81 \\\\ \sf y=\dfrac{81}{7}$

Put y = 81/7 in eqn. 1,

$\sf x=y-9 \\\\ \sf x=\dfrac{81}{7}-9 \\\\ \sf x=\dfrac{81-63}{7} \\\\ \sf x=\dfrac{18}{7}$

The required fraction is :

$\implies \sf \dfrac{x}{y} \\\\\\ \implies \sf \dfrac{\dfrac{18}{7}}{\dfrac{81}{7}} \\\\\\ \implies \sf \dfrac{18}{81} \\\\\\ \implies \sf \dfrac{9 \times 2}{9 \times 9} \\\\\\ \implies \sf \dfrac{2}{9}$