Work and Combined-Rate Problems flashcards

$\frac{1}{h}$ of the job per hour, with the whole job counted as 1. A mechanic who rebuilds an engine in 5 hours does $\frac{1}{5}$ of it each hour.

$W=R\times t$: work done (fraction of the job) equals rate (jobs per hour) times time (hours). At $\frac{1}{9}$ per hour, 4 hours completes $\frac{4}{9}$, leaving $\frac{5}{9}$.

Add their rates, then invert the sum. For 6 and 10 hours: $\frac{1}{6}+\frac{1}{10}=\frac{4}{15}$, so the time is $\frac{15}{4}$ hours, or 3 hours 45 minutes.

Subtract the known rate from the team rate: $\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$, so B alone needs 15 hours.

The drain's rate is subtracted: $\frac{1}{6}-\frac{1}{15}=\frac{1}{10}$ tank per hour, so the tank fills in 10 hours.

Done: $\frac{5}{15}=\frac{1}{3}$, so $\frac{2}{3}$ remains. Team rate: $\frac{1}{15}+\frac{1}{10}=\frac{1}{6}$. Rest: $\frac{2}{3}÷\frac{1}{6}=4$ hours. Total: 9 hours.

Let the job be the LCM of the solo times. For 6 and 9 hours, call it 18 units: the rates are 3 and 2 units per hour, 5 together, so $18÷5=3.6$ hours.

Never add or average times. Add rates: $\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$, so 2 hours. The team time must fall between half the faster worker's time and that worker's time.