The following table shows the amount of work done by each pump during the hour when they are both working.

The total work done by both pumps in an hour is:

\(\displaystyle \frac{1}{4}+\frac{1}{6}=\frac{5}{12}\)

The remaining work to be completed by the slowest pump is:

\(\displaystyle 1-\frac{5}{12}=\frac{7}{12}\)

The time taken by the slowest pump to complete filling the pool is the quotient of the remaining work by the work rate of the slowest pump:

\(\displaystyle \frac{7}{12}\div\frac{1}{6}=\frac{7}{12}\times6=\frac{7}{2}\)

It will take the slowest pump 7/2 hours to complete filling the pool.

All five workers have the same work rate, therefore they all complete an equal portion of the work done. Each worker's rate is:

\(\displaystyle \frac{1}{5}\times250=50\)

Each worker can then classify 50 files in a day.

To classify 1250 files, the company therefore needs the following number of temporary workers:

\(\displaystyle \frac{1250}{50}=25\)

Twenty-five workers are needed to classify the 1250 files in a day.

Let \(\displaystyle T\) be the number of hours it took to fill the tank.

The inlet pipe takes three hours to fill the tank, so it can fill \(\displaystyle \frac{1}{3}\) tank in one hour, and \(\displaystyle \frac{1}{3} T\) tank in \(\displaystyle T\) hours.

The drain can empty the tank in five hours, so it can remove \(\displaystyle \frac{1}{5}\) tank in one hour; since it started two hours after the filling started, it worked for \(\displaystyle T-2\) hours to empty \(\displaystyle \frac{1}{5}\left (T-2 \right )\) tank worth of water.

The work performed by the drain was against that performed by the inlet pipe, so the difference of their results is one tank of water. Therefore, the equation to solve for \(\displaystyle T\) is

\(\displaystyle \frac{1}{3} T- \frac{1}{5}\left (T-2 \right ) = 1\)

\(\displaystyle \frac{1}{3} T- \frac{1}{5} T+ \frac{2}{5}\right ) = 1\)

\(\displaystyle \left (\frac{1}{3} - \frac{1}{5} \right )T = 1- \frac{2}{5}\right )\)

\(\displaystyle \left (\frac{5}{15} - \frac{3}{15} \right )T = \frac{3}{5}\)

\(\displaystyle \frac{2}{15} T = \frac{3}{5}\)

\(\displaystyle T = \frac{3}{5} \cdot \frac{15} {2} = \frac{45}{10} = \frac{9}{2} = 4 \frac{1}{2}\)

Let \(\displaystyle T\) be the amount of time it would take for Phillip to paint the house by himself. Then he can paint \(\displaystyle \frac{1}{T}\) of a house per hour. Similarly, since Bryan can paint the house by himself in 25 hours, he can paint \(\displaystyle \frac{1}{25}\) of the house per hour. 

Since the two brothers together paint one house in 16 hours, Bryan's share of the work is to paint \(\displaystyle \frac{1}{25} \cdot 16 = \frac{16}{25}\) of one house. Phillip's share of the work is to paint \(\displaystyle \frac{1}{T} \cdot 16 = \frac{16}{T}\) of the house. Their shares together add up to one house, so the problem to be solved is

\(\displaystyle \frac{16}{T}+ \frac{16}{25}= 1\)

\(\displaystyle \frac{16}{T}= 1 - \frac{16}{25}\)

\(\displaystyle \frac{16}{T}= \frac{9}{25}\)

Cross-multiply and solve:

\(\displaystyle 9T = 16 \cdot 25 = 400\)

\(\displaystyle 9T \div 9 = 400 \div 9\)

\(\displaystyle T = 44 \frac{4}{9}\) hours.

Of the given choices, 45 hours comes closest.

Let \(\displaystyle T\) be the number of hours it took to fill the tank.

The inlet pipe takes \(\displaystyle \frac{5}{2}\) hours to fill the tank, so it fills \(\displaystyle \frac{2}{5}\) tank per hour. The drain empties the tank in \(\displaystyle \frac{7}{2}\) hours, so it empties \(\displaystyle \frac{2}{7}\) tank per hour. In \(\displaystyle T\) hours, the inlet pipe filled \(\displaystyle \frac{2}{5}T\) tanks of water, but the drain let out \(\displaystyle \frac{2}{7}T\) tanks of water; the one tank of water was the difference of these amounts, so

\(\displaystyle \frac{2}{5}T- \frac{2}{7}T = 1\)

\(\displaystyle \left (\frac{2}{5} - \frac{2}{7} \right )T = 1\)

\(\displaystyle \left (\frac{14}{35} - \frac{10}{35} \right )T = 1\)

\(\displaystyle \frac{4}{35} T = 1\)

\(\displaystyle T = \frac{35}{4} = 8\frac{3}{4}\)

This makes 9 hours the correct response.

Let \(\displaystyle t\) be the number of hours it takes the three koala bears together to eat the leaves. Then Stuffy can eat the leaves in \(\displaystyle 3t\) hours, Fluffy can eat them in \(\displaystyle 4t\) hours, and Muffy can eat them in 24 hours. Therefore, in one hour, Stuffy, Fluffy, and Muffy can eat \(\displaystyle \frac{1}{3t}\), \(\displaystyle \frac{1}{4t}\), and \(\displaystyle \frac{1}{24}\) of the leaves, respectively, and in \(\displaystyle t\) hours, Stuffy can eat \(\displaystyle \frac{1}{3t} \cdot t= \frac{1}{3}\) of the leaves, Fluffy can eat \(\displaystyle \frac{1}{4t} \cdot t= \frac{1}{4}\) of the leaves, and Muffy can eat \(\displaystyle \frac{1}{24} t\) of the leaves. Since together they are eating all of the leaves, the sum of the three amounts is one task, so we solve for \(\displaystyle t\) in the equation:

\(\displaystyle \frac{1}{3}+\frac{1}{4}+ \frac{1}{24}t = 1\)

\(\displaystyle \frac{8}{24}+\frac{6}{24}+ \frac{1}{24}t = 1\)

\(\displaystyle \frac{14}{24}+ \frac{1}{24}t = 1\)

\(\displaystyle \frac{1}{24}t = 1 - \frac{14}{24}\)

\(\displaystyle \frac{1}{24}t = \frac{10}{24}\)

\(\displaystyle \frac{1}{24}t \cdot 24 =\frac{10}{24}\cdot 24\)

\(\displaystyle t = 10\)

It takes 10 hours for all three koalas together to eat all the leaves on Mr. Meany's farm.

To begin, convert minutes to hours for each project.

Notebook Cover: \(\displaystyle 30\:min\cdot\frac{1 \:hr}{60\:min}=0.5\:hr\)

   Samuel is making thirteen of these, so we need to multiply the result by thirteen. \(\displaystyle 13*0.5\:hr=6.5\:hr\) is the amount of time Samuel needs to make thirteen notebook covers.

Book Cover: \(\displaystyle 45\:min\cdot \frac{1\:hr}{60\:min}=0.75\:hr\)

   Samuel is making three times as many book covers as he is making notebook covers. \(\displaystyle 13*3=39\), so he is making \(\displaystyle 39\) book covers.

\(\displaystyle 39*0.75\:hr=29.25\:hr\)

It will take Samuel \(\displaystyle 29.25\:hr\) to make \(\displaystyle 39\) book covers.

Add up the calculated times to get the total number of hours it will take Samuel to make the given number of notebook covers and book covers:

\(\displaystyle 6.5+29.25=35.75 \:hr\). That's almost a full work-week of work!

To solve this problem, we need to set up an equation as follows

\(\displaystyle M+\frac{1}{2}=\frac{15}{4}\),.

\(\displaystyle M\) is Mary's rate.

By simply manipulating the terms, we end up with

 \(\displaystyle M=\frac{13}{4}\), which is the final answer.

The first machine can make \(\displaystyle \textup{1,000 doughnuts in 4 hours and 20 minutes, or 260 minutes}\); this is

\(\displaystyle \frac{1,000}{260} = \frac{50}{13}\) doughnuts per minute.

Similarly, the second machine can make \(\displaystyle \textup{1,000 doughnuts in 2 hours and 20 minutes, or 140 minutes}\); this is

\(\displaystyle \frac{1,000}{140} = \frac{50}{7}\) doughnuts per minute.

Working together, the machines make 

\(\displaystyle \frac{50}{13} + \frac{50}{7} = \frac{50 \cdot 7 }{13 \cdot 7 } + \frac{13 \cdot 50}{13 \cdot 7} = \frac{350}{91}+ \frac{650}{91} = \frac{1,000}{91}\)

doughnuts per minute, or, equivalently, \(\displaystyle \textup{1,000 doughnuts per 91 minutes}\).

\(\displaystyle \textup{91 minutes is 1 hour and 31 minutes, making the closest choice 1 hour and 30 minutes}\).

Let \(\displaystyle t\) be the amount of time it takes for the second crew to paint a house without the first.

Think of this as a rate problem, with rate being measured in "houses per hour" rather than "hours per house". The first crew alone can paint \(\displaystyle \frac{1}{12}\) house per hour; the second alone can paint \(\displaystyle \frac{1}{t}\) house per hour; both together can paint \(\displaystyle \frac{1}{5}\) house per hour.

We can find the portion of the house each crew does in an amount of time by multiplying its rate in house per hour by the time in hourse elapsed.

\(\displaystyle \mathrm{Work }= \mathrm{rate}\; \times \;\mathrm{ time}\)

Let's look at what happens when the two crews are working together over five hours, adding their efforts:

\(\displaystyle \frac{1}{12} \cdot 5 + \frac{1}{t} \cdot 5 = \frac{1}{5} \cdot 5\)

or

\(\displaystyle \frac{5}{12}+ \frac{5}{t} = 1\)

\(\displaystyle \frac{5}{t} = \frac{7}{12}\)

\(\displaystyle 7t = 60\)

\(\displaystyle t=60\div7\approx 8.6\)

The third house will be painted in about 8.6 hours.