Let the number of machines be \( m \). The number of toys assembled by one machine in one hour is: \[ \frac{60}{3} = 20 \, toys per hour \] The total number of toys assembled by \( m \) machines in one hour is: \[ 20m \] After replacing 40% of the machines, the number of new machines is \( 0.4m \), and the old machines remain \( 0.6m \). The number of toys assembled by the new machines is: \[ \frac{60}{2} \times 0.4m = 30m \] The total number of toys assembled by all the machines in one hour is: \[ 30m + 20 \times 0.6m = 30m + 12m = 42m \] The percent increase in the number of toys is: \[ \frac{42m - 20m}{20m} \times 100 = \frac{22m}{20m} \times 100 = 20% \] To calculate the percent increase, first find the initial and final amounts, then use the formula \( \frac{final - initial}{initial} \times 100 \).