Question

As shown in the diagram above, a lever resting on a fulcrum has weights of \( w_1 \) pounds and \( w_2 \) pounds, located \( d_1 \) feet and \( d_2 \) feet from the fulcrum. The lever is balanced and \( w_1d_1 = w_2d_2 \). Suppose \( w_1 \) is 50 pounds and \( w_2 \) is 30 pounds. If \( d_1 \) is 4 feet less than \( d_2 \), what is \( d_2 \), in feet? [Official GMAT-2018]

Hint:

For a balanced lever, the product of the weight and the distance from the fulcrum must be equal on both sides. Use this principle to set up and solve the equation.

Answer 1

Step 1: Set up the equation using the balance condition.
The lever is balanced, so the condition \( w_1d_1 = w_2d_2 \) must hold. We are given the following information:
- \( w_1 = 50 \) pounds
- \( w_2 = 30 \) pounds
- \( d_1 = d_2 - 4 \) feet (since \( d_1 \) is 4 feet less than \( d_2 \))
The balance equation is: \[ 50 \times (d_2 - 4) = 30 \times d_2 \] Step 2: Solve for \( d_2 \).
Expand the equation: \[ 50d_2 - 200 = 30d_2 \] Now, subtract \( 30d_2 \) from both sides: \[ 50d_2 - 30d_2 = 200 \] \[ 20d_2 = 200 \] Solve for \( d_2 \): \[ d_2 = \frac{200}{20} = 10 \] Step 3: Conclusion.
Thus, \( d_2 = 10 \) feet.