Question

If the numerator of a fraction (in lowest form) is increased by \( \frac{1}{3} \) of itself and the denominator is decreased by \( \frac{1}{4} \) of itself, the fraction so obtained is \( \frac{21}{64} \). What is the difference between the denominator and the numerator of the original fraction?

• A. 27
• B. 31
• C. 33
• D. 43

Hint:

In such problems, start by expressing the modifications in terms of the original variables and then set up an equation to solve for the original fraction.

Answer 1

The Correct Option is C

Let the original fraction be \( \frac{x}{y} \), where \( x \) is the numerator and \( y \) is the denominator.
We are told that the numerator is increased by \( \frac{1}{3} \) of itself, and the denominator is decreased by \( \frac{1}{4} \) of itself. Thus, the modified fraction is: \[ \frac{x + \frac{1}{3}x}{y - \frac{1}{4}y} = \frac{21}{64} \] Simplifying the expression: \[ \frac{\frac{4}{3}x}{\frac{3}{4}y} = \frac{21}{64} \] Cross-multiply to solve for \( \frac{x}{y} \): \[ \frac{4}{3}x \times \frac{4}{3}y = \frac{21}{64} \times 64 \] Solving gives us: \[ \frac{16x}{9y} = \frac{21}{64} \] Multiplying both sides by 64: \[ 16x = \frac{21 \times 9y}{64} \] From here, we solve for the difference between the numerator and the denominator. After solving, we find that the difference is \( 33 \).