Question

If twice the numerator of a fraction is decreased by 50% and thrice the denominator is increased by 200%, the resultant fraction is \( \frac{121}{150} \). What was the original fraction ?

• A. \( \frac{1100}{150} \)
• B. \( \frac{1098}{150} \)
• C. \( \frac{1089}{150} \)
• D. \( \frac{9810}{150} \)

Hint:

Always trust your algebraic steps and double-check the arithmetic. Sometimes the initial approach is correct.

Answer 1

The Correct Option is C

Step 1: Represent the original fraction.
Let the original fraction be \( \frac{n}{d} \).
Step 2: Apply the changes described in the problem.
Twice the numerator decreased by 50\%: \( 2n - 0.5(2n) = 2n - n = n \).
Thrice the denominator increased by 200\%: \( 3d + 2(3d) = 3d + 6d = 9d \).
Step 3: Set up the equation based on the resultant fraction.
The resultant fraction is \( \frac{n}{9d} = \frac{121}{150} \).
Step 4: Solve for the original fraction \( \frac{n}{d} \).
Multiply both sides by 9:
$$\frac{n}{d} = 9 \times \frac{121}{150} = \frac{1089}{150}$$ This matches option (3).