Question

The denominator of a fraction is 1 more than 9 times the numerator. If the sum of the fraction and its reciprocal is \(\frac {101}{10}\) then, what will be the fraction?

(a) \(\frac {89}{10}\)

(b) \(\frac {10}{89}\)

(c) \(\frac {1}{10}\)

(d) 10

This question was addressed to me in examination.

The doubt is from Solution of Quadratic Equation by Factorisation in division Quadratic Equation of Mathematics – Class 10

Answer 1

The correct option is (c) \(\frac {1}{10}\)

To elaborate: Let the numerator be x.

Denominator of a fraction is 1 more than 9 times the numerator.

Denominator = 1+9x

The fraction is \(\frac {x}{11+9x}\)

Fraction + Reciprocal = \(\frac {101}{10}\)

\(\frac {x}{1+9x}+\frac {1+9x}{x}=\frac {101}{10}\)

\(\frac {x^2+(1+9x)^2}{x(1+9x)}=\frac {101}{10}\)

10(x^2+1+81x^2+18x)=101x(1+9x)

10(82x^2+18x+1)=101x+909x^2

820x^2+180x+10=101x+909x^2

89x^2-79x-10=0

89x^2-89x+10x-10x=0

89x(x-1)+10(x-1)=0

(x-1)(89x+10)=0

x=1, \(\frac {-10}{89}\)

The fraction is \(\frac {1}{10}\)