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.

4 feet

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9 feet

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7 feet

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2 feet

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Solution

The correct option is A 9 feet
For balance condition
$40 y = 80 x$ ........... (1)
$60 y = 80 (x + 1 \frac{1}{2})$ ....... (2)
Solving eq. (1) and (2) by substitution method
$20 y = 120 \Rightarrow y = 6; x = y / 2 = 3$

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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 112 feet further away from the fulcrum to obtain a balance. Find the original distance between the two weights.

The correct option is A 9 feetFor balance condition 40y=80x ........... (1)60y=80(x+112) ....... (2)Solving eq. (1) and (2) by substitution method20y=120⇒y=6;x=y/2=3

The correct option is A 9 feet
For balance condition
$40 y = 80 x$ ........... (1)
$60 y = 80 (x + 1 \frac{1}{2})$ ....... (2)
Solving eq. (1) and (2) by substitution method
$20 y = 120 \Rightarrow y = 6; x = y / 2 = 3$