Problem 32.31 Easy Free 4 variations

Combined Work Rates

A kiddie pool is being filled by two hoses while its drain is open. The first hose alone would fill the empty pool in $20$ minutes, the second hose alone in $10$ minutes, and the drain alone would empty a full pool in $15$ minutes. With both hoses running and the drain open, how many minutes does it take to fill the pool from empty?

Work it out, then compare with the answer.

SolutionShow the worked solution

Measure work in pools and time in minutes, so each device has a rate in pools per minute. The two hoses add water at rates $\frac{1}{20}$ and $\frac{1}{10}$ . The drain removes water at rate $\frac{1}{15}$ .

Everything runs at once, so the rates add, with the drain counted negatively:

\frac{1}{20}+\frac{1}{10}-\frac{1}{15}=\frac{3+6-4}{60}=\frac{5}{60}=\frac{1}{12}.

The net rate is positive, so the pool does fill. One full pool at $\frac{1}{12}$ pool per minute takes

\boxed{12 \text{ minutes}}

Key idea. Times do not add, but rates do. Convert each worker to a rate, add the rates, then invert the total.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 32.31a

Two gardeners mow a lawn. Working alone, the first finishes in $1$ hour and the second in $1$ hour $15$ minutes. Working together at their respective constant rates, how many minutes do they need to mow the lawn once?

Work it out, then compare with the answer.

Solution

Work in minutes. The first gardener mows $\frac{1}{60}$ of the lawn per minute and the second $\frac{1}{75}$ . Together they mow

\frac{1}{60}+\frac{1}{75}=\frac{5+4}{300}=\frac{3}{100}

of the lawn per minute. Hence they need

\boxed{\tfrac{100}{3}\approx 33.3 \text{ minutes}}

Variation 32.31b

Alice and Bob can each finish building a pipe alone in $120$ minutes and $100$ minutes respectively. They start building together, and after $40$ minutes Charlie joins them, working at his own constant rate. The three of them finish the pipe together $10$ minutes after Charlie joins. How many minutes would Charlie need to build the pipe alone?

Work it out, then compare with the answer.

Solution

Together Alice and Bob build $\frac{1}{120}+\frac{1}{100}=\frac{11}{600}$ of the pipe per minute. In $40$ minutes they complete $\frac{440}{600}$ , which leaves $\frac{160}{600}$ .

The three finish this remainder in $10$ minutes, so their combined rate is $\frac{16}{600}$ . Charlie’s rate is therefore $\frac{16-11}{600}=\frac{1}{120}$ . Alone, he would need

\boxed{120 \text{ minutes}}

Variation 32.31c

A farmer’s cow and goat together would eat all the grass in a field in $45$ days; the cow and the goose together would eat it in $60$ days; and the goat and the goose together would eat it in $90$ days. Each animal grazes at its own constant rate. If the cow, the goat and the goose grazed the field together, how many days would it take them to eat all the grass?

Work it out, then compare with the answer.

Solution

Let $a,b,c$ be the rates of the cow, goat and goose, in fields per day. The three pairs give

a+b=\frac{1}{45},\qquad a+c=\frac{1}{60},\qquad b+c=\frac{1}{90}.

Adding the three equations counts each animal twice:

2(a+b+c)=\frac{4+3+2}{180}=\frac{1}{20}.

So $a+b+c=\frac{1}{40}$ .

The individual rates are $a=\frac{5}{360}$ , $b=\frac{1}{120}$ and $c=\frac{1}{360}$ . All are positive, so the data are consistent. Grazing together, the three animals need

\boxed{40 \text{ days}}

Variation 32.31d

A farming company must till $11$ identical plots of land. Four farmers, each working at the same constant rate, start the job and together till $2$ plots in $5$ hours. The company wants to hire additional farmers, each working at that same rate, so that the remaining $9$ plots are tilled within the next $12$ hours. What is the minimum number of additional farmers needed?

Work it out, then compare with the answer.

Solution

Four farmers working $5$ hours supply $20$ farmer-hours and till $2$ plots. So one farmer tills $\frac{1}{10}$ plot per hour.

With $n$ additional farmers, $4+n$ farmers working $12$ hours till $\frac{12(4+n)}{10}$ plots. This must be at least $9$ :

\frac{12(4+n)}{10}\ge 9 \iff 4+n\ge 7.5 \iff n\ge 3.5 .

The bound is sharp. With $n=3$ the farmers till only $8.4<9$ plots, while with $n=4$ they till $9.6\ge 9$ plots. The minimum is therefore

\boxed{4 \text{ additional farmers}}

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