The denominator of a fraction is 3 more than its numerator.
Question:

The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is $2 \frac{9}{10}$. Find the fraction.

 

Solution:

Let the numerator be $x$.

$\therefore$ Denominator $=x+3$

$\therefore$ Original number $=\frac{x}{x+3}$

According to the question:

$\frac{x}{x+3}+\frac{1}{\left(\frac{x}{x+3}\right)}=2 \frac{9}{10}$

$\Rightarrow \frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}$

$\Rightarrow \frac{x^{2}+(x+3)^{2}}{x(x+3)}=\frac{29}{10}$

$\Rightarrow \frac{x^{2}+x^{2}+6 x+9}{x^{2}+3 x}=\frac{29}{10}$

$\Rightarrow \frac{2 x^{2}+6 x+9}{x^{2}+3 x}=\frac{29}{10}$

$\Rightarrow 29 x^{2}+87 x=20 x^{2}+60 x+90$

$\Rightarrow 9 x^{2}+27 x-90=0$

$\Rightarrow 9\left(x^{2}+3 x-10\right)=0$

$\Rightarrow x^{2}+3 x-10=0$

$\Rightarrow x^{2}+5 x-2 x-10=0$

$\Rightarrow x(x+5)-2(x+5)=0$

$\Rightarrow(x-2)(x+5)=0$

$\Rightarrow x-2=0$ or $x+5=0$

$\Rightarrow x=2$ or $x=-5$ (rejected)

So, numerator $=x=2$

denominator $=x+3=2+3=5$

So, required fraction $=\frac{2}{5}$

 

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