Many problems involve three quantities, such as the number of workers, the time taken and the amount of work. Hold one quantity fixed at a time, or measure the work in a single unit such as worker-days or machine-hours. **Quantities compared** | **What stays fixed** | **Relationship** Workers and time | the amount of work | inverse proportion Work done and time | the number of workers | direct proportion Work done and workers | the time | direct proportion Why: if every worker does the same amount each day, the total work equals workers x days x amount per worker per day. Fixing the job fixes the product workers x days, which is inverse proportion. Fixing the team makes the work proportional to the number of days. Method - Find the work needed for one unit of output, such as one room or one page, in worker-days or machine-hours. - Multiply by the number of units required to find the total work. - Divide the total work by the new number of workers, or by the new number of days. - If the team changes part-way, work done so far = workers x time already worked, and work remaining = total work - work done. Divide the remaining work by the new team. - Check the direction of each change: more output needs more time; more workers need less time. Example Six cleaners take 5 hours to clean 4 identical classrooms. How many hours will 10 cleaners take to clean 12 such classrooms? Work for 4 classrooms: 6 x 5=30 cleaner-hours Work for 1 classroom: 304=7.5 cleaner-hours Work for 12 classrooms: 12 x 7.5=90 cleaner-hours Time for 10 cleaners: 9010=9 hours Check one step at a time: 1 cleaner cleans 4 classrooms in 30 hours; 10 cleaners take 3 hours; 12 classrooms is three times the work, so 9 hours. Assume that every cleaner works at the same steady rate and that all classrooms need the same work. When the team changes: 10 workers can finish a job in 12 days. After 4 days, 2 workers leave. Total work =10 x 12=120 worker-days; work done =10 x 4=40 worker-days; work remaining =120-40=80 worker-days. The 8 workers left need 808=10 more days, so the job takes 4+10=14 days in all.

Direct and inverse proportion together

Secondary 2

Many problems involve three quantities, such as the number of workers, the time taken and the amount of work. Hold one quantity fixed at a time, or measure the work in a single unit such as worker-days or machine-hours.

Quantities compared	What stays fixed	Relationship
Workers and time	the amount of work	inverse proportion
Work done and time	the number of workers	direct proportion
Work done and workers	the time	direct proportion

Why: if every worker does the same amount each day, the total work equals workers $\times$ days $\times$ amount per worker per day. Fixing the job fixes the product workers $\times$ days, which is inverse proportion. Fixing the team makes the work proportional to the number of days.

Method

Find the work needed for one unit of output, such as one room or one page, in worker-days or machine-hours.
Multiply by the number of units required to find the total work.
Divide the total work by the new number of workers, or by the new number of days.
If the team changes part-way, work done so far $=$ workers $\times$ time already worked, and work remaining $=$ total work $-$ work done. Divide the remaining work by the new team.
Check the direction of each change: more output needs more time; more workers need less time.

Example

Six cleaners take $5$ hours to clean $4$ identical classrooms. How many hours will $10$ cleaners take to clean $12$ such classrooms?

Work for $4$ classrooms: $6\times5=30$ cleaner-hours

Work for $1$ classroom: $\dfrac{30}{4}=7.5$ cleaner-hours

Work for $12$ classrooms: $12\times7.5=90$ cleaner-hours

Time for $10$ cleaners: $\dfrac{90}{10}=9$ hours

Check one step at a time: $1$ cleaner cleans $4$ classrooms in $30$ hours; $10$ cleaners take $3$ hours; $12$ classrooms is three times the work, so $9$ hours. Assume that every cleaner works at the same steady rate and that all classrooms need the same work.

When the team changes: $10$ workers can finish a job in $12$ days. After $4$ days, $2$ workers leave. Total work $=10\times12=120$ worker-days; work done $=10\times4=40$ worker-days; work remaining $=120-40=80$ worker-days. The $8$ workers left need $\dfrac{80}{8}=10$ more days, so the job takes $4+10=14$ days in all.

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