Question

the numerator of the fraction is 2 less then denominater. If 3 is added toh both the numerator the fraction becomes 3/4​

Answer 1

Solution :-

Let :-

Numerator = x

Denominator = y

According to the first condition :-

$\longrightarrow \sf x= y-2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(1)$

According to the second condition :-

$\longrightarrow \sf \dfrac{x+3}{y+3} = \dfrac{3}{4} \\\\\longrightarrow 4(x+3)= 3(y+3) \\\\\longrightarrow 4x+12 = 3y+9 \\\\\longrightarrow 4x-3y = -3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................(2)$

Substitute the value of x in eq(2) :-

$\longrightarrow \sf 4(x-2)-3y = -3 \\\\\longrightarrow 4x-8-3y = -3 \\\\\longrightarrow 4y-3y = -3+8 \\\\\longrightarrow \bf y = 5$

Substitute the value of y in eq(1) :-

$\longrightarrow \sf x=5-2 \\\\\longrightarrow \bf x = 3$

$\bf FRACTION = \dfrac{3}{5}$

Answer 2

SOLUTION :-

• The numerator of the fraction is 2 less then denominator

• If 3 is added to both the numerator and denominator the fraction becomes 3/4

TO DETERMINE :-

The fraction

EVALUATION :-

We know that

$\displaystyle \sf{Fraction = \frac{Numerator}{Denominator} }$

Now it is given that the numerator of the fraction is 2 less then denominator

Let denominator = x

Then numerator = x - 2

So the required fraction

$\displaystyle \sf{ = \frac{x - 2}{x} }$

Again it is also stated that if 3 is added to both the numerator and denominator the fraction becomes 3/4

Hence by the above condition

$\displaystyle \sf{ \frac{(x - 2) + 3}{x + 3} = \frac{3}{4} }$

$\implies \displaystyle \sf{ \frac{x + 1}{x + 3} = \frac{3}{4} }$

$\implies \displaystyle \sf{ 4x + 4 = 3x + 9 }$

$\implies \displaystyle \sf{ 4x - 3x = 9 - 4}$

$\implies \displaystyle \sf{ x = 5}$

Hence the required fraction

$\displaystyle \sf{ = \frac{5 - 2}{5} }$

$\displaystyle \sf{ = \frac{3}{5} }$

FINAL ANSWER :-

$\boxed{ \: \: \displaystyle \sf{ The \: required \: fraction= \frac{3}{5} } \: \: }$

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