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The numerator of a fraction is 5 less than the denominator. If 6 is added to the...


Question

The numerator of a fraction is 5 less than the denominator. If 6 is added to the numerator and 4 to the denominator the fraction is doubled. What is the fraction?

Options

A) \(\frac{2}{7}\)

B) \(\frac{5}{8}\)

C) \(\frac{3}{8}\)

D) \(-\frac{1}{4}\)

The correct answer is C.

Explanation:

If the numerator be \(x\), the fraction becomes \(\frac{x}{x+5}\)
Again numerator + 6 and denominator + 4 gives \(\frac{x+6}{x+5+4}\)
Therefore, \(\cfrac{x+6}{x+5+4}=\cfrac{2x}{x+5}\)
Cross multiplying, gives \(2x(x+9)=(x+6)(x+5)\)
\(2x^2+18x=x^2+11x+30\)
Collecting like terms
\(x^2+7x-30=0\)
\((x+10)(x-3)=0\)
Therefore, \(x+10=0\) and \(x-3=0\)
\(x=-10\) and \(x=3\)
At \(x=-10\), the fraction is \(-\frac{10}{5}\) and at \(x=3\), the fraction is \(\frac{3}{8}\)

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Dicussion (1)

  • If the numerator be \(x\), the fraction becomes \(\frac{x}{x+5}\)
    Again numerator + 6 and denominator + 4 gives \(\frac{x+6}{x+5+4}\)
    Therefore, \(\cfrac{x+6}{x+5+4}=\cfrac{2x}{x+5}\)
    Cross multiplying, gives \(2x(x+9)=(x+6)(x+5)\)
    \(2x^2+18x=x^2+11x+30\)
    Collecting like terms
    \(x^2+7x-30=0\)
    \((x+10)(x-3)=0\)
    Therefore, \(x+10=0\) and \(x-3=0\)
    \(x=-10\) and \(x=3\)
    At \(x=-10\), the fraction is \(-\frac{10}{5}\) and at \(x=3\), the fraction is \(\frac{3}{8}\)