Each machine at a toy factory assembles a certain kind of toy at a constant rate of one toy...

Question

Each machine at a toy factory assembles a certain kind of toy at a constant rate of one toy every 3 minutes. If 40 percent of the machines at the factory are to be replaced by new machines that assemble this kind of toy at a constant rate of one toy every 2 minutes, what will be the percent increase in the number of toys assembled in one hour by all the machines at the factory, working at their constant rates?

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20% We set up the variable for the machine count and turn "one toy every 3 minutes" and "one toy every 2 minutes" into hourly rates. M = total number of machines 60/3 = 20 toys per hour per old machine 60/2 = 30 toys per hour per new machine We write the original hourly output when all machines are old. Q_old = 20M We use "40 percent of the machines" to split the machines into old and new groups and then add their outputs. Q_new = (0.6M) 20 + (0.4M) 30 = 12M + 12M = 24M We compare the change to the original to get the percent increase, then turn the resulting fraction into a percent. Percent increase = Q_new - Q_oldQ_old 100% = 24M - 20M/20M 100% = 4M20M 100% = 1/5 100% = 20%

Explanation	Calculations	Help
$We set up the variable for the machine count and turn "one toy every 3 minutes" and "one toy every 2 minutes" into hourly rates.$	$M = total number of machines \frac{60}{3} = 20 toys per hour per old machine \frac{60}{2} = 30 toys per hour per new machine$
$We write the original hourly output when all machines are old.$	$Q_{old} = 20 M$
$We use "40 percent of the machines" to split the machines into old and new groups and then add their outputs.$	$Q_{new} = (0.6 M) \times 20 + (0.4 M) \times 30 = 12 M + 12 M = 24 M$
$We compare the change to the original to get the percent increase, then turn the resulting fraction into a percent.$	$Percent increase = \frac{Q _{new} - Q _{old}}{Q _{old}} \times 100% = \frac{24 M - 20 M}{20 M} \times 100% = \frac{4 M}{20 M} \times 100% = \frac{1}{5} \times 100% = 20%$

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