Question

Suppose that a fulcrum is placed so that weights of 40 kg. and 80 kg. are in balance. Furthermore, suppose that when 20 kg is added to the 40 kg. weight, the 80 kg weight must be moved $1\frac{1}{2}$ feet further away from the fulcrum to obtain a balance. Find the original distance between the two weights.

• A. 4 feet
• B. 9 feet
• C. 7 feet
• D. 2 feet

Answer 1

Answer: (B)

9 feet

For balance condition
$40y=80x$ ........... (1)
$60y=80(x+1\frac{1}{2})$ ....... (2)
Solving eq. (1) and (2) by substitution method
$20y=120\Rightarrow y=6;x=y/2=3$