Work Problems - GMAT Quantitative

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Question

If 2 machines working at the same rate create 88 widgets in 4 minutes, how many widgets can 5 machines make in 2 minutes, working at the same rate?

Answer

To find the number of widgets created by each machine separately, divide 88 by 2:

$88/2=44$

This is the number of widgets created by 1 machine in 4 minutes.

To find the number of widgets in 1 minute, divide 44 by 4: $44/4minutes = 11widgets/min$

Use this rate to find the answer:

$5 machines\cdot 11widgets/min \cdot 2min = 110 widgets$ .

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Question

The inlet pipe leading into a water tank can fill the tank in 45 minutes; the drain can empty the tank in 25 minutes.

One day while draining the tank, someone left the inlet pipe on.

To the nearest minute, how long did it take for the tank to drain completely?

Answer

Let be the number of minutes that it takes to drain the tank.

Think of emptying the tank as one job. Then the drain can do one job in 25 minutes, or $\small \frac{1}{25}$ jobs per minute.

Now, think of the inlet pipe as doing a "negative" job - it is doing the opposite of emptying the tank, working against the drain. It is doing "negative one" job in 45 minutes, or $\small -\small \frac{1}{45}$ jobs per minute.

Now, think of this as a rate problem. In minutes, the drain does

$\small \small \frac{1}{25} x$ jobs

and the pipe does

$\small \small -\small \frac{1}{45} x$ jobs.

Together they do 1 job, the draining of the whole tank.

Set up an equation to solve for x:

$\small \small \small \frac{1}{25}x + \left (- \frac{1}{45} \right )x = 1$

$\small \small \small \small \frac{1}{25}x - \frac{1}{45} \right )x = 1$

$\small \small \left (\small \small \frac{1}{25} - \frac{1}{45} \right ) \right )x = 1$

$\small \small \small \left (\small \small \frac{9}{225} - \frac{5}{225} \right ) \right )x = 1$

$\small \small \small \small \left (\small \small \frac{4}{225} \right )x = 1$

$\small \small \small \small \small \left (\small \small \frac{225}{4} \right )\small \small \small \small \left (\small \small \frac{4}{225} \right )x = \small \small \small \small \left (\small \small \frac{225}{4} \right )1$

, which rounds to 56 minutes.

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Question

A tank has three inlet pipes. One, by itself, can fill the tank in one hour; another, by itself, can fill the tank in one hour and twenty minutes ; a third, by itself, can fill the tank in fifty minutes alone. If all three are on, then, to the nearest tenth of a minute, how long does it take to fill the tank?

Answer

Think of this as a rate problem, with rate being measured in "tanks per minute".

$Work = rate \times time$

The pipes can fill the tank up at rates of $\frac{1}{60}$ , $\frac{1}{80}$ , and $\frac{1}{50}$ tanks per minute, respectively.

In the time it takes to fill the tank, the three pipes fill up $\frac{1}{60}t$ of the tank, $\frac{1}{80}t$ of the tank, and $\frac{1}{50}t$ of the tank, respectively. Add the amounts of work done by the three tanks to get the total amount of work - one job.

$\frac{1}{60}t +\frac{1}{80}t +\frac{1}{50}t = 1$

$\left (\frac{1}{60} +\frac{1}{80} +\frac{1}{50} \right ) t= 1$

$\left (\frac{20}{1,200} +\frac{15}{1,200} +\frac{24}{1,200} \right ) t= 1$

$\frac{59}{1,200}t= 1$

$t= \frac{1,200}{59} \approx 20.3 ; min$

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Question

In a subdivision, several houses of uniform size and design need to be painted; there are two crews working together to paint them. Without the second crew, the first crew painted the first house in twelve hours; together, the two crews painted the second house in five hours. The second crew will paint the third house without the first crew; how long should it take them?

Answer

Let 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 $\frac{1}{12}$ house per hour; the second alone can paint $\frac{1}{t}$ house per hour; both together can paint $\frac{1}{5}$ 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.

$\mathrm{Work }= \mathrm{rate}; \times ;\mathrm{ time}$

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

$\frac{1}{12} \cdot 5 + \frac{1}{t} \cdot 5 = \frac{1}{5} \cdot 5$

or

$\frac{5}{12}+ \frac{5}{t} = 1$

$\frac{5}{t} = \frac{7}{12}$

$t=60\div7\approx 8.6$

The third house will be painted in about 8.6 hours.

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Question

Which of the following is not a prime number?

Answer

By definition, a prime number is any number that is greater than and is only divisible by and itself. Therefore, by definition is not a prime number.

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Question

What is the sum of the first seven prime numbers?

Answer

The first seven prime numbers are:

To find the sum, all numbers must be added:

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Question

If a triangular field has a base of and a height of , and a garden is taking up 40% of the field. What is the area of the garden?

Answer

First, find the area of the triangular field:

The garden takes up 40% of the field, therefore:

Area of garden= $40%\cdot 25$

Area of garden=

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Question

What happens to the volume of a rectangular prism if the length, width, and height are doubled?

Answer

$V=l\cdot w\cdot h$

$New\ V=2l\cdot 2w\cdot 2h$

$New\ V=8\cdot l\cdot w\cdot h$

Then, the new volume is times the old volume.

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Question

Two inlet pipes lead into a large water tank. One pipe can fill the tank in 45 minutes; the second pipe can fill it in 40 minutes. At 8:00 AM, the first pipe is opened; at 8:10 AM, the second one is opened. To the nearest minute, at what time is the tank full?

Answer

Look at the work rates as "tanks per minute", not "minutes per tank".

The two pipes can fill the tank up at $\frac{1}{45}$ tanks per minute and $\frac{1}{40}$ tanks per minute.

Let be the time it took, in minutes, to fill the tank up. Then this is the amount of time that the first pipe had to let in water; the amount of time that the second pipe had, in minutes, is .

Since rate multiplied by time is equal to work, then the two pipes fill up $\frac{1}{45} t$ and $\frac{1}{40} \left ( t-10\right )$ tanks; together, they filled up $\frac{1}{45} t + \frac{1}{40} \left ( t-10\right )$ tank - one tank. This sets up the equation to be solved:

$\frac{1}{45} t + \frac{1}{40} \left ( t-10\right ) =1$

$360 \cdot \left [ \frac{1}{45} t + \frac{1}{40} \left ( t-10\right ) \right ] =360 \cdot 1$

$\frac{360}{45} t + \frac{360}{40} \left ( t-10\right ) =360$

$8 t + 9 \left ( t-10\right ) =360$

$t = 450 \div17 \approx 26.47$

This rounds to 26 minutes after the first pipe is opened, or 8:26 AM.

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Question

Philip, his wife Sharon, and their son Greg are planning to paint a greenhouse together. Philip can paint the greenhouse alone in four hours; Sharon can paint it alone in four and a half hours; Greg can paint it alone in three and a half hours. If they start at noon and don't stop, when, to the nearest minute, will they finish painting the greenhouse?

Answer

This can be solved by looking at their work rates in terms of "greenhouses per hour".

Philip can paint one greenhouse in four hours, or $\frac{1}{4}$ greenhouse per hour.

Sharon can paint one greenhouse in four and a half hours, or $\frac{1}{4 \frac{1}{2}} = \frac{1}{ \left ( \frac{9}{2} \right )} = \frac{2} {9}$ greenhouses per hour. Grag can paint one greenhouse in three and a half hours, or $\frac{1}{3 \frac{1}{2}} = \frac{1}{ \left ( \frac{7}{2} \right )} = \frac{2} {7}$ greenhouses per hour. If is the number of hours that it takes for the three to paint the greenhouse, then Philip, Sharon, and Grag will paint $\frac{1}{4} t$ , $\frac{2}{9} t$ , and $\frac{2}{7} t$ of the greenhouse, respectively; these three shares add up to one greenhouse, so we can set up and solve this equation:

$\frac{1}{4} t + \frac{2}{9} t + \frac{2}{7} t = 1$

$\frac{ 252}{1} \cdot \left ( \frac{1}{4} t + \frac{2}{9} t + \frac{2}{7} t \right )=252 \cdot 1$

$\frac{ 252}{1} \cdot \frac{1}{4} t +\frac{ 252}{1} \cdot \frac{2}{9} t +\frac{ 252}{1} \cdot \frac{2}{7} t =252$

$191 t \div 191 =252 \div 191$

$t = \frac{252}{191 }$

Let's convert this to minutes by multiplying by 60:

$\frac{252}{191 } \times 60 \approx 79.2$

This rounds to one hour and 19 minutes, so the three finish at $1:19 \textrm{ PM}$ .

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Question

A shoe factory has two pieces of equipment to package the shoes: and .

is a better performer and makes packages an hour while produces only packages an hour.

The company has an order to ship shoes. How many hours will it take for the factory to complete the packages necessary to ship the order?

Answer

If in an hour produces packages and produces packages respectively, then both machines produce packages in an hour altogether.

Since the order requires packages, the factory will take:

$\frac{250}{50}=5 \text{ hours}$

Therefore, it will take 5 hours for the factory to complete the packages necessary for the shipment.

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Question

Two pumps are used to fill a pool. One pump can fill the pool by itself in hours while the oher can fill the pool by itself in hours. Both pumps are open for an hour when the fastest pump stops working. How long will it take the slower pump to complete filling the pool?

Answer

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:

$\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

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

$1-\frac{5}{12}=\frac{7}{12}$

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:

$\frac{7}{12}\div\frac{1}{6}=\frac{7}{12}\times6=\frac{7}{2}$

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

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Question

When working at the same constant rate, temporary workers at a company can classify files in a day. How many temporary workers would be needed to classify files in a day?

Answer

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

$\frac{1}{5}\times250=50$

Each worker can then classify 50 files in a day.

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

$\frac{1250}{50}=25$

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

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Question

Bryan and his brother Philip, working together, can paint their father's house in a total of sixteen hours. Bryan, working alone, would take twenty-five hours. Which of the following is closest to the number of hours it would take Philip to paint the house, working alone?

Answer

Let be the amount of time it would take for Phillip to paint the house by himself. Then he can paint $\frac{1}{T}$ of a house per hour. Similarly, since Bryan can paint the house by himself in 25 hours, he can paint $\frac{1}{25}$ 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 $\frac{1}{25} \cdot 16 = \frac{16}{25}$ of one house. Phillip's share of the work is to paint $\frac{1}{T} \cdot 16 = \frac{16}{T}$ of the house. Their shares together add up to one house, so the problem to be solved is

$\frac{16}{T}+ \frac{16}{25}= 1$

$\frac{16}{T}= 1 - \frac{16}{25}$

$\frac{16}{T}= \frac{9}{25}$

Cross-multiply and solve:

$9T = 16 \cdot 25 = 400$

$9T \div 9 = 400 \div 9$

$T = 44 \frac{4}{9}$ hours.

Of the given choices, 45 hours comes closest.

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Question

A large water tank has an inlet pipe that can fill the tank completely in two and one-half hours, and a drain that can empty it completely in three and one-half hours. On one occasion, when the tank was being filled, the drain was left open; the drain was not closed until the tank was completely full. Which of the following answers comes closest to the number of hours it took to fill the tank?

Answer

Let be the number of hours it took to fill the tank.

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

$\frac{2}{5}T- \frac{2}{7}T = 1$

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

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

$\frac{4}{35} T = 1$

$T = \frac{35}{4} = 8\frac{3}{4}$

This makes 9 hours the correct response.

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Question

To the nearest hour, how many hours would it take three koala bears - Stuffy, Fluffy, and Muffy - to eat all of the leaves on Mr. Meany's farm if:

Stuffy can eat all the leaves alone in three times as much time that the three together can eat them;

Fluffy can eat all the leaves alone in four times as much time that the three together can eat them; and,

Muffy can eat all the leaves in twenty-four hours?

Answer

Let be the number of hours it takes the three koala bears together to eat the leaves. Then Stuffy can eat the leaves in hours, Fluffy can eat them in hours, and Muffy can eat them in 24 hours. Therefore, in one hour, Stuffy, Fluffy, and Muffy can eat $\frac{1}{3t}$ , $\frac{1}{4t}$ , and $\frac{1}{24}$ of the leaves, respectively, and in hours, Stuffy can eat $\frac{1}{3t} \cdot t= \frac{1}{3}$ of the leaves, Fluffy can eat $\frac{1}{4t} \cdot t= \frac{1}{4}$ of the leaves, and Muffy can eat $\frac{1}{24} t$ 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 in the equation:

$\frac{1}{3}+\frac{1}{4}+ \frac{1}{24}t = 1$

$\frac{8}{24}+\frac{6}{24}+ \frac{1}{24}t = 1$

$\frac{14}{24}+ \frac{1}{24}t = 1$

$\frac{1}{24}t = 1 - \frac{14}{24}$

$\frac{1}{24}t = \frac{10}{24}$

$\frac{1}{24}t \cdot 24 =\frac{10}{24}\cdot 24$

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

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Question

A large water tank has an inlet pipe that can fill the tank completely in three hours, and a drain that can empty it completely in five hours. On one occasion, two hours after the filling of the tank started, the drain was accidentally opened. The error was not immediately discovered, and the drain was not closed until the tank was completely filled. Which of the following comes closest to the number of hours it took to fill the tank?

Answer

Let 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 $\frac{1}{3}$ tank in one hour, and $\frac{1}{3} T$ tank in hours.

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

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

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

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

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

$\frac{2}{15} T = \frac{3}{5}$

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

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Question

Together, Mary and I, can trade stocks at a rate of stocks every minutes. I alone on the other hand can only trade stock every minutes. How fast can Mary trade, alone?

Answer

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

$M+\frac{1}{2}=\frac{15}{4}$ ,.

is Mary's rate.

By simply manipulating the terms, we end up with

$M=\frac{13}{4}$ , which is the final answer.

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Question

It takes Samuel thirty minutes to make a notebook cover and forty-five minutes to make a book cover. How many hours will it take Samuel to make thirteen notebook covers and three times as many book covers?

Answer

To begin, convert minutes to hours for each project.

Notebook Cover: $30:min\cdot\frac{1 :hr}{60:min}=0.5:hr$

Samuel is making thirteen of these, so we need to multiply the result by thirteen. is the amount of time Samuel needs to make thirteen notebook covers.

Book Cover: $45:min\cdot \frac{1:hr}{60:min}=0.75:hr$

Samuel is making three times as many book covers as he is making notebook covers. , so he is making book covers.

It will take Samuel to make 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:

. That's almost a full work-week of work!

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Question

A doughnut factory has a machine that takes four hours and twenty minutes to make $\textup{1,000 doughnuts}$ . Another machine is brought in that can do the same job in $\textup{two hours and twenty minutes}$ . It is decided to go ahead and allow both machines to work alongside each other until the older machine is worn out.

How long does it take for the $\textup{two machines}$ working together to make $\textup{ 1,000 doughnuts? }$ (Choose the time that is closest to the actual time.)

Answer

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

$\frac{1,000}{260} = \frac{50}{13}$ doughnuts per minute.

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

$\frac{1,000}{140} = \frac{50}{7}$ doughnuts per minute.

Working together, the machines make

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

doughnuts per minute, or, equivalently, $\textup{1,000 doughnuts per 91 minutes}$ .

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

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