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Home » The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is . Find the fraction.
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.
CBSE | Class 10 | Exercise 10e | Maths | Quadratic Equations | RS Aggarwal
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.

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}
\begin{array}{l} \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 \text { or } x+5=0 \\ \Rightarrow x=2 \text { or } x=-5 \text { (rejected) } \end{array}

So, number =x=2

\text { denominator }=x+3=2+3=5

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

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