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Rates and Work Drill Sheet

Academic Support For Math 1105 (Cornell University)

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Rates and Work Drill

Warm-up Problems

Problem 1
A water purifier takes 4 minutes to purify 100 cubic feet of water. How long will
it take to purify the contents of a 15 foot  15 foot  10 foot tank that is filled exactly
halfway with water?

Problem 2
A baker has 8 hours to make 150 pies for a party. If after the first 5 hours he has
made 60 pies, by how many pies per hour does he need to increase his rate to finish the
order on time?

Problem 3
A stockbroker works 10 hours a day on Monday, Wednesday, and Friday, 11
hours a day on Tuesday and Thursday, and 8 hours on Saturday. If she earns $600 each
week day, and $300 on Saturday, what are her average earnings per hour over this 6-day
period?

Problem 4
Two coal carts are approaching each other from opposite ends of a 400-yard track.
If Cart A is traveling at 40 feet per second, and Cart B is traveling at 56 feet per second,
how long will it take before they collide?

A) 75 seconds
B) 48 seconds
C) 23 1/3 seconds
D) 12 1/2 seconds
E) 4 1/6 seconds

Problem 5
9 identical machines, each working at the same constant rate, can stitch 27 jerseys
in 4 minutes. How long would it take 4 such machines to stitch 60 jerseys?

Problem 6

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It takes Brenda 3 hours to walk a 12 mile scenic loop. If she wanted to walk the
loop twice without stopping, and she cut her walking speed in half, how much time
would she need?

Problem 7
A gang of criminals hijacks a train heading due south. The police are contacted,
and a police car stationed 50 miles north of the train starts driving toward the train on a
roadway alongside the track. If the train is traveling at 50 miles per hour, and the police
car is traveling at 80 miles per hour, how long does it take the police car to catch up with
the train?

Problem 8
Rachel and Terry are assembling brochures. Rachel can assemble a parcel of
brochures in 10 minutes, and Terry can assemble a parcel of brochures in 8 minutes.
How long will it take the two working together to assemble 9 parcels of brochures?

Problem 9
With 4 identical servers, a new Internet search provider can process 9,600 search
requests per hour. If the search provider adds 2 more such servers, how long will it take
all of the servers, working at full capacity, to process 216,000 search requests?

Problem 10
A pipe siphons ink from an 800-liter drum at a rate of r liters per minute. If two
such pipes were used, the drum could be emptied 100 minutes faster than when one pipe
is used. What is the value of r?

Problem 11
Sabrina and Janis are assembling tanks. If Sabrina can assemble a tank in 8
hours, and Janis can assemble a tank in 13 hours, approximately how many hours will it
take the two of them working together to assemble a tank?

A) 21
B) 18
C) 7
D) 5
E) 2

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Problem 12
Etienne is sitting down to eat a batch of 20 cookies, just as Jacques begins making
more cookies at a rate of 16 cookies per hour. If Etienne eats 20 cookies per hour, how
long will it take before there are no cookies?

Problem 13
Phil collects virtual gold in an online computer game, and then sells the virtual
gold for real dollars. After playing 10 hours a day for 6 days, he collects 540,000 gold
pieces. If he sells this virtual gold at a rate of $1 per 1,000 gold pieces, what are his
average earnings per hour, in real dollars?

Problem 14
After completing a speed training, Alyosha can translate Russian literature into
English at a rate of 10 more than twice as many words per hour as he was able to
translate before the training. If he was previously able to translate 10 words per minute,
how many words can he now translate in an hour?

Problem 15
Jenny takes 3 hours to sand a picnic table. Laila can do the same job in 1/2 hour.
How many hours will it take the two women to sand a picnic table if they work together
at their constant rates?

A) 1/6
B) 2/9
C) 1/3
D) 3/7
E) 5/6

Problem 16
Jurgen and Mott are participating in a cashew-eating contest. What is Jurgen’s
eating speed, in cashews consumed per minute?

1) Jurgen eats one pound of cashews per minute.

2) Together, Jurgen and Mott eat 1,400 cashews in 5 minutes.

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Problem 17
If five average workers in a workshop take 24 minutes to string 80 violins, how
long will it take twelve average workers to string 720 violins?

Problem 18
Riders board the Jelly Coaster in groups of 4 every 15 seconds. If there are 200
people in front of Kurt in line, approximately how long will it take before he is aboard the
Jelly Coaster?

A) 5 minutes
B) 8 minutes
C) 10 minutes
D) 13 minutes
E) 20 minutes

Problem 19
Machines A and B both shrinkwrap CDs continuously, but Machine B works 50%
faster than Machine A. If, in a 24-hour period, Machine B shrinkwraps 48,000 more CDs
than Machine A, what is Machine A’s rate in CDs per hour?

Problem 20
It takes a team of 8 pastry chefs 5 days to produce 3,200 fruit tarts. If 3,600 fruit
tarts are needed in exactly 3 working days, and all pastry chefs work at the same rate,
how many pastry chefs will be needed to fill the order on time?

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Medium Problems

Problem 21
Two robots, working together at their respective constant rates, take 6 minutes to
polish 88 pounds of gemstones. If Robot A’s rate of polishing is 3/5 that of Robot B,
how long would it take Robot A alone to polish 165 pounds of gemstones?

A) 15.75 minutes
B) 18 minutes
C) 18.75 minutes
D) 27.5 minutes
E) 30 minutes

Problem 22
Car A starts driving north at 2:00 p.m., traveling at a constant rate of 40 miles per
hour. Car B leaves from the same starting point in direct pursuit at 3:00 p.m., driving at a
constant rate of 30 miles per hour. If each car holds 8 gallons of fuel, which is consumed
at a rate of 30 miles per gallon, how far apart are the two cars when Car A runs out of
fuel?

A) 30 miles
B) 60 miles
C) 90 miles
D) 120 miles
E) 150 miles

Problem 23
A population of bacteria doubles at a constant rate, increasing from 50 to 3,200
bacteria in exactly two days. How large will the population of bacteria be after 16 more
hours?

A) 6,400
B) 12,800
C) 25,600
D) 51,200
E) 102,400

Problem 24
A pipe is filling a 120-gallon drum. What is its average fill rate, in gallons per
second?

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1) Once the drum is halfway full, the pipe increases its speed to 6 gallons per
second.

2) The pipe fills the first half of the drum in 20 seconds, then works at double
speed until the pipe is full.

Problem 25
One robot, working independently at a constant rate, can assemble a doghouse in
12 minutes. What is the maximum number of complete doghouses that can be assembled
by 10 such robots, each working on separate doghouses at the same rate for 2 1/2 hours?

A) 20
B) 25
C) 120
D) 125
E) 150

Problem 26
A semiconductor company predicts that it will be able to double the density of
transistors on its circuits (measured in transistors per square mm) every 18 months. If
this prediction holds true, and the company’s current generation of circuits have a density
of 5 million transistors per square mm, what will be the density of the company’s circuits,
measured in millions of transistors per square mm, exactly 30 years from now?

A) 5  218
B) 5  220
C) 5  226
D) 5  236
E) 5  245

Problem 27
Soda Q is bottled at a rate of 500 liters/second, 24 hours a day. Soda V is bottled
at a rate of 300 liters/second, 24 hours a day. If twice as many bottles of Soda V as of
Soda Q are filled in a day, what is the ratio of the volume of a bottle of Soda Q to a bottle
of Soda V?

A) 3/10
B) 5/6
C) 6/5

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D) 8/3
E) 10/3

Problem 28
The superintendent of a school district uses a phone tree system to contact parents
in case of an emergency. The district office contacts 5 families, each of which then
contacts 5 families, until all the families have been reached. If all of the contacts are
successful, and if it takes 15 minutes for a family to receive the emergency information
and contact the next family, how many families will have been contacted after 1 hour?

A) 3905
B) 3125
C) 3000
D) 780
E) 625

Problem 29
Today, Samson’s Train traveled at a constant rate along a straight North-South
route that included the famous 10-mile long Chasm Bridge. Was Samson’s Train on the
Chasm Bridge at 2:00 p.m.?

1) At 9:00 a.m., Samson’s Train was 200 miles north of the Chasm Bridge, and at
4:00 p.m, the train was 140 miles south of Chasm Bridge.

2) Samson’s Train started its trip at 6:00 a.m., and traveled at a rate of less than 70
miles per hour.

Problem 30
It takes Audrey 4 hours to complete a certain job. Ferris can do the same job in 3
hours. Audrey and Ferris decided to collaborate on the job, working at their respective
rates. While Audrey worked continuously, Ferris took 3 breaks of equal length. If the
two completed the job together in 2 hours, how many minutes long was each of Ferris’s
breaks?

A) 5
B) 10
C) 15
D) 20
E) 25

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Challenge Problems

Problem 31
A turtle climbs to the top of a plateau at a rate of 4 miles an hour, crosses the
plateau at a rate of x miles per hour, and descends the other side of the plateau at a rate of
x2 miles per hour. If each portion of the journey is equal in distance, what is the turtle’s
average speed for the entire trip, in terms of x?

2x
A)
x2
 x  2
2

B)
3
 x  2
2
C)
4 x2
D)
 x  2
2

12 x 2
E)
 x  2
2

Problem 32
Melted chocolate is continuously pumped into a large mixing vat by a fill tube,
and continuously drained out by a dispenser tube. The fill tube pumps at a rate of z
gallons per day. The dispenser tube operates twice as quickly, but does not operate on
Saturdays and Sundays. If the vat is 2/3 full on a Tuesday, and is 1/4 full exactly 3 weeks
later, what is the capacity of the vat in terms of z?

A) 36z
B) 108z/5
C) 12z
D) 12z/5
E) 5z/4

Problem 33
A kangaroo and a dingo are having a race from Bush A to Bush B. The dingo
runs in a straight line, while the kangaroo jumps forward in a single semicircular arc.
They travel along their respective trajectories at equal speeds, until the kangaroo reaches
the height of its arc. At this point the kangaroo’s speed begins to increase, while the
dingo’s speed remains constant. If the two animals reach Bush B simultaneously, what is
the proportion of the kangaroo’s average downward speed to the dingo’s speed?

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
A)
 2
4  
2

B)
4

C)
2

D)
4 
 2
E)


Problem 34
Two flocks of geese are migrating toward each other from opposite ends of a
1,200-mile, straight-line migratory flyway. Flock X is flying north at a rate 20% less
than that at which Flock Y is flying south, and the two flocks pass each other after 13
hours and 20 minutes. If Flock Y had started along the pathway 6 hours before Flock X,
how many miles would Flock Y have flown when the two flocks passed each other?

A) 680 miles
B) 720 miles
C) 800 miles
D) 850 miles
E) 950 miles

Problem 35
Ahmet and Nesuhi were painting their shop. Did either of the two brothers paint
at least 100 square feet?

1) Ahmet painted at a rate of 40 square feet/hour, and Nesuhi painted half again as
fast as Ahmet.

2) Nesuhi painted for an hour longer than Ahmet.

Problem 36
A team of 3 runners is competing in a 3-stage relay race. The first runner travels
a third of the racetrack before handing the baton to the second runner, who runs a third of
the racetrack before handing off to the third runner, who runs to the finish line. The
runner’s respective rates, in feet per second, are 10, 15, and z, and there is no lag time

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between one stage and the next. In terms of z, what is the average rate, in feet per
second, of the three runners over the course of the race?

25  z
A)
3
12.5  z
B)
2
18 z
C)
z6
6z
D)
z6
36 z
E)
z  25

Problem 37
Grace and Sergei are walking toward each other down a straight road that is d
miles in length. It takes them t hours to meet each other. If Sergei walks 1 mile an hour
faster than Grace, how long would it have taken Grace to walk the entire d miles herself?

d t
A)
2
d t
B)
2t
2dt
C)
d t
dt
D)
d t
t 1
E)
d

Problem 38
If Trains A and B traveled directly toward each other, how long did it take them
to meet?

1) If Train A had traveled twice as fast and Train B had traveled 3 times as fast,
the two trains would have taken 1 hour to meet.

2) Train B traveled half as fast as Train A.

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Problem 39
Three brothers are hired for a painting job. The first brother shows up at 10:00
a.m. and begins painting at a steady rate of x square feet per hour. After t hours, the
second brother shows up and starts painting at a rate of 2x square feet per hour. After an
additional t hours, the third brother finally appears and starts painting at a rate of 4x
square feet per hour. The three brothers work without breaks until they finish the job at
6:00 p.m. In terms of x and t, what is their average rate, in square feet per hour, for the
entire 8-hour job?

x(28  5t )
A)
12
x(28  5t )
B)
4
x(4  5t )
C)
4
7 x  3t
D)
3
7x
E)
3

Problem 40
Spy X is chasing Spy Y along a deserted highway. When Spy X is exactly one
mile behind, Spy Y engages his car’s smokescreen, causing Spy X to drop to half her
maximum speed. 5 minutes later, the smokescreen clears, and Spy X begins driving at
maximum speed, catching up with Spy Y in 4 minutes. If Spy Y drives at a constant rate
of 80 miles per hour, what is Spy X’s maximum speed, in miles per hour?

A) 120
B) 115
C) 110
D) 105
E) 100

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Problem 1 Solution: 45 minutes

Since the problem asks for time, we know that we need two pieces in order to
solve: the rate and the amount of work. To find the rate, we can use the RTW chart to
interpret the first sentence, or simply divide the total work the purifier can do by the time
required. 100 cubic feet ÷ 4 minutes = 25 cubic feet/minute.

Rate  Time = Work

25 ft3/min  4 minutes = 100 ft3

Now we need to find the amount of work required. Since we are filling a tank, we want
to find the volume of the tank. Make sure not to overlook the condition that we are
filling halfway. The easiest way to avoid skipping this kind of detail is to build it into
your calculations from the start. 15  15  10  1/2 = 225  5 = 1125 cubic feet. Once
we have found the rate and the work, we simply plug these numbers back into the
formula and solve. 1125 cubic feet ÷ 25 cubic feet/minute = 45 minutes.

Rate  Time = Work

25 ft3/min  45 minutes = 1125 ft3

The job will take 45 minutes to complete.

Problem 2 Solution: 18 pies per hour

To answer this problem, we first need to identify what we are looking for. We
have been asked to find the amount by which the baker’s rate of pie-making must
increase, so we need to find his rate for the first 5 hours, and the rate needed to finish the
job over the next 3 hours. The “skeleton” equation for our solution looks like this:

Rate for last 3 hours – Rate for first 5 hours = Increase

The rate for the first 5 hours is a straight RTD equation: 60 ÷ 5 = 12 pies per hour.

Rate  Time = Work

First 5 hours 12 pies/hour  5 hours = 60 pies

The new rate is similarly straightforward, once we have established that there are 90 pies
left to make (150 pies needed – 60 pies made in the first 5 hours): 90 ÷ 3 = 30 pies per
hour.

Rate  Time = Work

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Last 3 hours 30 pies/hour  3 hours = 90 pies

From there we can plug the two rates into our solution equation to find the increase:

30 pies per hour – 12 pies per hour = 18 pies per hour

Problem 3 Solution: 55

First off, it’s important to realize that although this problem is asking for an
average, we don’t want to start averaging a bunch of numbers. Since we have been asked
for the average rate across the entire week, we can simply use our RTW formula. Dollars
per hour is a rate, so we want to use the equation R = W ÷ T. All we need to do is add up
the total work (in this case, earnings) and the total time, and we are all set.

5 week days  $600 = $3,000

1 Saturday  $300 = $300
Total earnings = $3,300

3 days  10 hours = 30
2 days  11 hours = 22
1 day  8 hours =8
Total hours = 60

The broker’s average earnings per hour = $3,300 ÷ 60 = $55.

Rate  Time = Work (Earnings)

$55/hr  60 hours = $3,300

Problem 4 Solution: D

This is a classic combined rates problem. Since the carts are moving directly
toward each other, we can add their rates together. Remember that when two objects are
moving in opposite directions—either toward each other or away from each other—we
generally add their rates. Watch for the unit conversion trap (answer choice E). Again,
the best way to avoid these is to do the conversion (or at least write down the units) up
front. In this case, it is probably easiest to convert the total distance to feet: 400 yards 
3 ft/yard = 1200 feet. The combined rate of the two carts is 96 ft/sec. Therefore the time
it takes for them to meet is 1200 feet ÷ 96 feet/second = 12 1/2 seconds.

Rate  Time = Distance

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Cart A 40 ft/sec  t = 40t

Cart B 56 ft/sec  t = 56t
Combined Rate 96 ft/sec  12 1/2 sec = 1200 feet

If you weren’t in the mood to do the division, you could also note that the answer should
be awfully close to 1200 ÷ 100 = 12 seconds. D is the only answer choice in that
neighborhood.

Problem 5 Solution: 20 minutes

The first place to start on one of these “identical machines” problems is to find the
rate for one machine. Since each of the 9 machines works at the same rate, we can
simply divide their total output by 9. 1 machine can stitch 3 jerseys in 4 minutes, which
gives us a rate of 3/4 jersey per minute.

Rate  Time = Work

1 machine 3/4 jersey/min  4 min = 27 ÷ 9 = 3 jerseys

From there, we can plug in the numbers specified in the question: 4 machines and 60
jerseys. To find the rate for 4 machines, we simply multiply the individual rate by 4,
which conveniently cancels out the denominator of our fractional rate. Together, the
machines produce 3 jerseys per minute. We get the time by dividing 60 jerseys by the
rate of 3 jerseys per minute.

Rate  Time = Work

1 machine 3/4 jersey/min  4 min = 3 jerseys
4 machines 3/4  4 = 3 jerseys/min  t= 60/3 = 20 min = 60 jerseys

The 4 machines will take 20 minutes to produce 60 jerseys.

Problem 6 Solution: 12 hours

This problem asks us to compare an actual scenario with a hypothetical one.

Let’s start by setting up our chart for Brenda’s actual walk. This allows us to find her
rate: 4 miles per hour.

Rate  Time = Distance

4 mi/hr  3 hours = 12 miles

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Now we can set up a chart for the hypothetical situation, applying the changes as
described to give us 2 miles per hour and 24 miles. The new time is 24 ÷ 2 = 12 hours.

Rate  Time = Distance

2 mi/hr  12 hours = 24 miles

Alternately, we might note that both of the changes—doubling the distance and halving
the rate—have the same effect: they make the trip take twice as long as it would have
before. So the time required for this hypothetical situation is multiplied by four: 3 2 2
= 12 hours.

Problem 7 Solution: 1 hour, 40 minutes

This is a chase problem. The two vehicles are moving in the same direction, and
we want to know how long it will take the rear vehicle to catch up, so we subtract rates to
find out how quickly the rear vehicle is gaining. The difference between the rates is 80
mi/hr – 50 mi/hr = 30 mi/hr, so the police car gains on the train at a rate of 30 mi/hr.
Since the police car needs to gain 50 miles, the time it takes to catch up is 50 ÷ 30 = 5/3
hours, or 1 hour, 40 minutes.

Rate  Time = Distance

Police Car 80 mi/hr  =
Train 50 mi/hr  =
Police Car’s Gain 80 – 50 = 30 mi/hr  50÷30 = 5/3 = 50 miles

Problem 8 Solution: 40 minutes

Rachel and Terry are working together, so we can add their rates. To do so, we
need to find their individual rates, using the given times and the work quantity of 1
1 1 4 5 9
parcel. Adding the rates, together, we get a combined rate of    
10 8 40 40 40
parcels/minute.

Rate  Time = Work

Rachel 1/10 parcel/min  10 minutes = 1 parcel
Terry 1/8 parcel/min  8 minutes = 1 parcel
Rachel & Terry 9/40 parcel/min

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Now we can plug our numbers into the RTW formula. 9 parcels ÷ 9/40 parcels/min = 40
minutes.

Rate  Time = Work

Rachel & Terry 9/40 parcel/min  40 minutes = 9 parcels

Problem 9 Solution: 15 hours

We have been given the work, as well as the rate, for a smaller number of servers.
Since we are simply increasing the number of servers by 50%, we do not need to find the
rate for one server. We can simply multiply the rate for 4 servers by 1.5.

Rate  Time = Work

4 servers 9,600 searches/hr  1 hr = 9,600 searches
6 servers 9,600 x 1.5 =  1 hr = 14,400 searches
14,400 searches/hr

Now we can apply this rate to the given amount of work. 216,000 searches ÷ 14,400
searches/hr = 15 hours.

Problem 10 Solution: 4 liters/minute

Let’s start by filling in our chart with the pieces we know. Remember, two pipes
will double the rate at which ink is siphoned.

Rate  Time = Work

1 pipe r L/min  800/r = 800 L
2 pipes 2r L/min  800/2r = 800 L

Now let’s use the information we haven’t yet applied. We have been given the
relationship between the time for 1 pipe (800/r) and the time for 2 pipes (800/2r). We
can express this as a skeleton equation:

Time for 2 pipes + 100 = Time for 1 pipe

If we weren’t sure that we were adding 100 to the correct side, a good double-check
would be to ask which number is smaller. The time for 2 pipes should be shorter, so we
add 100 to that side to make it equal the time for 1 pipe. We can now plug in the times
that we found:

800/2r + 100 = 800/r

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(800 + 200r)/2r = 800/r

800 + 200r = 1600
200r = 800
r=4

1 pipe pumps 4 liters of ink per minute.

Problem 11 Solution: D

Since Sabrina and Janis are working together, we can add their rates.
Unfortunately, we are faced with some slightly ugly arithmetic:

Rate  Time = Work

Sabrina 1/8 tank/hr  8 hours = 1 tank
Janis 1/13 tank/hr  13 hours = 1 tank
Sabrina & Janis 1/8 + 1/13 =
13/104 + 8/104 =
21/104 tank/hr

Now we can plug this combined rate into the RTW formula to find the time. We might
also notice that since the work is equal to 1, the time will just be the reciprocal of the rate.

Sabrina & Janis 21/104 tank/hr  104/21 minutes = 1 tank

At this point, we do not need to do long division! We have been invited to approximate,
and we don’t want to ignore that invitation. 104/21 is about 100/20 = 5. The answer is
D.
We might also use some intuition to work the answer choices and avoid setting up
this problem at all! We can immediately eliminate A and B, since these times exceed
either worker’s individual time. Also, since we know that Sabrina is the faster worker,
we know that Janis’s contribution will be less than Sabrina’s. The two together won’t
work twice as fast as Sabrina, but they will work more than twice as fast as Janis.
Therefore, the total time should be more than half of Sabrina’s individual time, and less
than half of Janis’s individual time. 4 < t < 6.5, which leaves D as the only possible
answer.

Problem 12 Solution: 5 hours

Etienne and Jacques are working at cross-purposes (although perhaps Etienne

doesn’t mind), so we can subtract their rates. 20 cookies/hour -16 cookies/hour = 4
cookies/hour, so the quantity of cookies decreases by 4 per hour. Since Etienne begins
with a pile of 20 cookies, it will take him 5 hours to eat all the cookies.

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Rate  Time = Work

Jacques & Etienne 4 cookies/hr  5 hours = 20 cookies

Note that Etienne eats a lot more than 20 cookies. In 5 hours he eats 100 cookies—the
initial 20, plus the 80 that Jacques makes in that 5 hours.

Problem 13 Solution: $9/hr

To solve for average earnings, we need to fill in this formula:

Total earnings / Total hours = Average earnings per hour

Since the gold-dollar exchange rate is 1,000:1, Phil’s dollar earnings for the 6 days are
540,000/1,000 = $540. His total hours worked is 10 hours/day  6 days = 60 hours.
Therefore, his average hourly earnings are $540/60 = $9/hour.

Problem 14 Solution: 1210

Since we’ve been asked to find the new rate in words per hour, let’s start by
setting up an equation to find this value:

New words/hr = 10 + 2(Old words/hr)

We have been given the old rate in words per minute, so we need to convert. 10
words/min  60 min/hr = 600 words/hr. Now we just have to plug into the equation:

New words/hr = 10 + 2(600) = 1,210

Note that we would not want to start by working with the rate per minute. If we did,
we’d get 10 + 2(10) = 30 words/minute, then 30  60 = 1,800 words/hr. We reached this
inflated number because we added an additional 10 words per minute instead of per hour.
This is another reason to perform your conversions right away!

Problem 15 Solution: D

Since the two women are working together, we can add their rates. To find their
individual rates, we need to divide work by time. (Be careful when dividing the work by
1/2. The rate is the reciprocal of 1/2, or 2 tables/hour.)

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Once we find Jenny and Laila’s combined rate, we simply divide the work
required by this rate. 1 table ÷ 7/3 tables/hour = 3/7 hour. The answer is D.

Rate  Time = Work

Jenny 1/3 table/hr  3 hours = 1 table
Laila 2 tables/hr  1/2 hour = 1 table
Jenny & Laila 1/3 + 2 = 7/3 tables/hr  3/7 hour = 1 table

Problem 16 Solution: E

To find Jurgen’s rate, we need work (in quantity of cashews) and time. Statement
1 gives us work in terms of weight, but not in terms of quantity. Without a conversion
factor between weight and quantity, we can’t determine Jurgen’s rate. Statement 1 is
insufficient.
Statement 2 gives us Jurgen and Mott’s combined work, in terms of quantity, for
5 minutes. But without knowing anything about the relationship between the two
contestants’ rates, we can’t determine how many of the cashews were consumed by
Jurgen, and therefore cannot determine his individual rate. Statement 2 is insufficient.
If we try to combine Statements 1 and 2, we can determine that in the 5 minutes
referred to in Statement 2, Jurgen eats 5 pounds of cashews. Therefore, this quantity
must be less than or equal to the 1,400 cashew total. However, the question is asking for
Jurgen’s exact rate, and we still don’t know the actual number of cashews he consumed.
The two statements combined are still insufficient. The answer is E.

Problem 17 Solution: 90 minutes

Since we have been given work and time, we can begin by finding the combined
rate for 5 workers.

Rate  Time = Work

5 workers 80/24= 10/3 violins/min  24 min = 80 violins

From here, we can find the rate for 12 workers through one of two equivalent methods.
First, we could find the rate for 1 worker by dividing the combined rate by 5. 10/3 ÷ 5 =
2/3 violin/min per worker. We could then multiply this rate by 12. 12 workers would
string 12  2/3 = 24/3 = 8 violins per minute.
We could also combine these two operations into one. In the above method, we
divided the rate by 5 and then multiplied by 12. We might also recognize that to convert
from 5 to 12, we can simply multiply by 12/5. 12/5  10/3 = 120/15 = 8 violins/minute.
Now that we have the rate for 12 workers, we simply plug this rate, along with the
new work required, into the RTW chart. 720 violins ÷ 8 violins/minute = 90 minutes.

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Rate  Time = Work

12 workers 8 violins/min  90 min = 720 violins

Problem 18 Solution: D

To find Kurt’s wait time, we need to know how long it will take for 200 people to
board the Jelly Coaster. We have been given a rate of 4 people per 15 seconds, which
converts to 16 people/minute. To find the time, we just need to divide the work by this
rate. 200 people ÷ 16 people/minute = 200/16 = 12.5 minutes. We have been asked for
an approximation, and this is now close enough to get us the answer of D, but if we want
to be very precise, we need to add in the 15-second wait for Kurt’s group to board, so
Kurt’s total wait time is 12.5 + .25 = 12.75 minutes.

Problem 19 Solution: 4,000 CDs/hour

This is a comparative rate question that turns out to be a straight algebra problem.
We have been given 2 equations about the machines’ rates. B is 50% faster than A, so
we can relate their rates with the equation b = 1.5a. We also know that B’s work for a
24-hour period exceeds A’s by 48,000. Let’s put our data into a chart to compare the
rates.

Rate  Time = Work

Machine A a  24 hours = 24a
Machine B b  24 hours = 24b

Now that we have expressions for the work of both machines, we can relate them in the
equation 24b = 24a + 48,000. From here, we can simplify:

24b – 24a = 48,000

24(b – a) = 48,000
b – a = 2,000

We can now combine this simplified equation with our previous equation, b = 1.5a.
Since we are solving for a, we should substitute for b.

1.5a – a = 2,000
.5a = 2,000
a = 4,000

Machine A shrinkwraps 4,000 CDs per hour.

An easier way to solve this problem would be to notice that since B is 50% faster
than A, the quantity by which its work exceeds A’s in an hour will equal 50%, or half, of

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A’s hourly rate. Since it shrinkwraps 48,000/24 = 2,000 more CDs per hour than A, we
know that A wraps 2,000  2 = 4,000 CDs per hour.

Problem 20 Solution: 15 chefs

We are given an initial RTW relationship, which we can use to find one chef’s
daily output. First, we divide total work by total time to get 3,200/5 = 640 tarts/day for
the team. Then we divide this rate by 8 to get one chef’s daily rate.

Rate  Time = Work

8 chefs 640 tarts  5 days = 3,200 tarts
1 chef 80 tarts  1 day = 80 tarts

From here, we can find one chef’s output for 3 days. We can then divide this into the
total work to find the number of chefs required. One chef produces 80  3 = 240 tarts in
3 days. To make a total of 3,600 tarts in 3 days, we will need 3,600 ÷ 240 = 15 chefs.
Alternately, we could apply fractions directly to the original numbers. The
quantity of work required is multiplied by 3,600/3,200 or 9/8. The number of days is
multiplied by 3/5. Since the number of workers determines the daily rate, we are looking
at the net effect on rate, or work divided by time. Since the work is multiplied by 9/8 and
the time is multiplied by 3/5, the net effect on rate is 9/8 ÷ 3/5 = 9/8  5/3 = 45/24. We
can multiply this by the number of workers to find the number needed. 8 chefs  45/24
= 45  8/24 = 45/3 = 15. 15 chefs will be needed.

Problem 21 Solution: E

When rate problems involve multiple situations, it can help to set up an initial
“skeleton” RTD chart for our solution. That way, we can easily determine what data is
needed, and fill in that data as we find it. Since we want to know how long Robot A will
take alone, our chart will look like this:

Rate  Time = Work

Robot A A’s rate  t min = 165 pounds

We know the work and we want to know the time, so we just need A’s rate. Let’s call the
rates a and b. Now let’s set up a chart representing what we know.

Rate  Time = Work

Robot A a lbs per min  6 min = 6a
Robot B b lbs per min  6 min = 6b
A & B together a+b lbs per min  6 min = 6(a+b) = 88

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At this point, we know that 6(a+b) = 88. We just need to apply the other piece of
information we know. A’s rate is 3/5 B’s. We can write this as a = (3/5)b. However,
since we are looking for a, let’s substitute for b in our rate chart:

a = (3/5)b
(5/3)a = b

6(a+(5/3)a) = 88
6(8/3)a = 88
48/3(a) = 88
a = 88(3/48)
a = 88(1/16) = 88/16 = 11/2

So A’s rate is 11/2 pounds per minute. Now we just plug into our original chart, and
we’re home free.

Rate  Time = Work

Robot A 11/2 lbs per min  30 min = 165 pounds

The time Robot A takes to polish 165 pounds of gems is 165/(11/2) = 330/11 = 30
minutes.

Problem 22 Solution: C

To begin with, we are not given any distance in this problem, so we need to
determine how far the two cars end up traveling before we can find the distance between
them. Our skeleton equation is as follows: Car A’s distance – Car B’s distance =
distance between cars.
Since the limiting factor in this case is A’s fuel supply, we must calculate how far
the car is able to drive before running out of fuel. This in itself is a rate problem of sorts:

30 miles per gallon  8 gallons = 240 miles

We now know that Car A will end up 240 miles north of its starting point. What about
Car B? Since we know Car B’s rate, we need to find its driving time to determine how
far it gets. We know Car B starts one hour later than Car A, and keeps going through the
end of the scenario, when Car A has driven 240 miles. So we need to know how long it
takes Car A to get to the end. Here comes the RTD chart:

Rate  Time = Distance

Car A 40 mi/hr  6 hours = 240 miles

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Car A drives for 6 hours. Therefore Car B, which leaves one hour later, drives for 5
hours. In this time it covers 30 miles/hour  5 hours = 150 miles. Therefore the two cars
end up 240 – 150 = 90 miles apart. The answer is C.
Alternately, since this is a chase problem, we could find the difference in the two
cars’ rates: 40 mi/hr – 30 mi/hr = 10 mi/hr. Once we have found A’s driving time (6
hours), we might reason this way: Car A spends 1 hour driving alone, for a gain in
distance of 40 miles. After that, Car A is chased for 5 hours by the slower Car B. Car A
pulls ahead at a rate of 10 miles per hour. 10 mi/hr  5 hr = 50 miles. Altogether, Car A
ends up 40 + 50 = 90 miles ahead of Car B.

Problem 23 Solution: B

The easiest way to approach this problem is to build a chart to see how many
doubling periods must occur to grow the population from 50 to 3,200. Most problems of
this sort rely on reducing any growth to a doubling period.

50
100
200
400
800
1,600
3,200

The population must double 6 times in 2 days. 48 hours / 6 doubling periods = 8 hours
per doubling period. Therefore the population doubles every 8 hours.
We are asked to determine the population after 16 more hours. 16 hours / 8 hours
per doubling period = 2 doubling periods. The population after 16 hours is 3,200  2 
2 = 12,800.

Problem 24 Solution: B

This is a classic combination trap. Statement 1 tells us that the rate for the second
60 gallons is 6 gallons per second. This is insufficient, as we are left wondering about
the rate at which the first half was filled.
Along comes Statement 2, which tells us that the drum was filled halfway in 20
seconds. Can we combine this with Statement 1 to solve? Absolutely! Do we need to do
so? No! Statement 2 is sufficient all by itself. The answer is B. (Always remember to
give Statement 2 its best shot before trying to combine statements.) Let’s see why
statement 2 is sufficient.

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For the first half, we know the work and time, so we can find the rate. Once we
have the rate for the first half, we can double it to find the rate for the second half. We
now have the rate and work for the second half, so we can find the time. Now we can
plug our data into the formula for average rate: Total Work / Total Time.
The quickest way to crunch these numbers is to determine that at double speed,
the second half will be filled in 20/2 = 10 seconds. This means total work is 120 gallons
and total time is 30 seconds. 120 gallons / 30 seconds = 4 gallons/second. Not
convinced? Try filling the two parts of Statement 2 into a chart:

Rate  Time = Work

First half 60 gal/20 sec =  20 sec = 60 gallons
3 gal/sec
Second half Double the rate =  10 sec = 60 gallons
6 gal/sec
Total 120 gallons/30 sec =  30 sec = 120 gallons
4 gal/sec

Problem 25 Solution: C

D is a trap. This doesn’t come up all the time, but it’s worthwhile to be able to
recognize it if you see it. Note that in this case, each robot is independently assembling
complete doghouses. Since the question asks us for the number of completed doghouses
after 2 1/2 hours, we need to remove any incomplete doghouses from our calculations.
Since one robot completes a doghouse in 12 minutes, we can determine that the
individual hourly rate is 60/12 = 5 doghouses per hour. Therefore, each robot produces 5
 2.5 = 12.5 doghouses in 2 1/2 hours. (We could also simply divide the 150 total
minutes by 12 to get the same result.)
However, since we are interested only in completed doghouses, and the robots are
working independently, we need to drop the decimal. Each robot completes 12
doghouses, for a total of 12  10 = 120 doghouses.

Problem 26 Solution: B

The question gives us the initial value and the rate of doubling, so we can set up
our solution in a straightforward manner. The hardest part is cutting through all the
language to find what we need. The initial value is stated outright: 5 million
transistors/square mm. To find the number of doubling periods, we just need to convert
from months to years:

(30 years  12 months per year) / 18 months per doubling period = 360/18 = 20 doubling
periods

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So we have an initial value of 5 million, doubled 20 times. Thankfully, we don’t have to

calculate that number! Since the answer choices are in exponential form, let’s set things
up that way. Also, note that the question asks for the density in millions of
transistors/square mm, so we can use 5, rather than 5,000,000, in our calculation.
(Answer choice C is waiting if we don’t make that conversion):

5 doubled 20 times = 5  220

Problem 27 Solution: E

To begin with, we do not need to find the total number of liters produced in a day.
That would be, in the case of Soda Q, 500 liters/sec  60 sec/min  60 min/hour  24
hours/day = 43,200,000 liters per day! We don’t need that information, so there’s no
need to incur the headache (or the toothache). What we need to know is the ratio of Q’s
rate to V’s rate, and that is simply 500/300 = 5/3. Now let’s suppose that an equal
number of bottles of Q and V were filled each day. In that case, the ratio of volumes
would need to be 5/3 to accommodate Q’s greater output. Since twice as many bottles of
V are produced, the bottles will need to be half as large, which leads to a ratio of 5/(3/2)
= 10/3. The answer is E.
The most comfortable way to arrive at this figure under time pressure might be to
use Smart Numbers. Let’s simply compare the output of the two sodas for one second.
Since twice as many bottles of Soda V are produced, let’s say the output in one second is
100 bottles. Therefore, 50 bottles of Soda Q are produced in the same time. The volume
of the Q bottles is 500 liters / 50 bottles = 10 liters/bottle. The volume of the V bottles is
300 liters / 100 bottles = 3 liters/bottle. 10 liters / 3 liters = 10/3.

Problem 28 Solution: A

We can set up a table to show the exponential growth of the phone tree. Since
each family contacts 5 others, the number of families contacted at each 15-minute
interval increases by a factor of 5.

Time # of families contacted

0 minutes 5
2
15 minutes 5 =25
30 minutes 53=125
45 minutes 54=625
60 minutes 55 =3125

The trick comes when we have to interpret this data. After 1 hour, how many total
families have been contacted? It isn’t simply 3125. That’s how many families were

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contacted in the last round of the phone tree. The total number of families contacted is
actually the sum of the contacts made in each round: 5 + 25 + 125 + 625 + 3125 = 3905.
The answer is A.
If you’re wondering why we have to add in this case, when we don’t have to do so
in, say, a bacteria doubling problem, it’s because the setup is slightly different. The total
number of families isn’t simply getting multiplied by 5 each round. Rather, in each
round, the number of contacts made at one time is multiplied by 5. The total number of
families contacted is the sum of the contacts made in all of the rounds combined.

Problem 29 Solution: A

In order to determine whether the train was on the bridge at 2:00 p.m., we need to
know something about the distance between the train and the bridge, and how fast the
train was traveling. Statement 1 gives us the total distance for a particular length of time,
which allows us to find the train’s rate. The train’s total distance covered during this
period can be expressed as follows:

Total Distance = Distance north of bridge + bridge + Distance south of bridge

Total Distance = 200 + 10 + 140 = 350

The train traveled 350 miles in the 7-hour period described, so we can determine that its
rate was 350/7 = 50 miles/hour. Since we now know the train’s rate and the distance it
needed to travel to reach the bridge, we should be able to determine the train’s distance
relative to the bridge at any point in time between 9:00 and 4:00. Statement 1 is
sufficient.
However, if we want to be industrious and find the yes/no answer, we will need to
plug our numbers back in. We know the train was 200 miles north of the bridge at 9:00
a.m., and that the bridge is 10 miles long. This means that, in order to be on the bridge at
2:00, the train must have traveled between 200 and 210 miles. In the 5 hours between
9:00 and 2:00, it must have traveled 50 miles/hour  5 hours = 250 miles. Therefore, it
had already crossed over the bridge. The answer to the question is no. (Remember only
to do this work if you are not certain a solution is possible. As soon as you see that the
problem can be solved, stop working the equations!)
Statement 2 gives us what might appear to be important details—the train’s start
time, and a limit on its rate. However, without information about the distance between
the train and the bridge at any point in time, there is no way we can determine whether
the train was on the bridge at 2:00 p.m. Statement 2 is insufficient. The answer is A.

Problem 30 Solution: B

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To determine how long Ferris’ breaks were, we will need to know the difference
between the amount of work the two should have completed in two hours, and the
amount they actually did complete (1 job).

Audrey and Ferris are working together, so we should find each worker’s individual rate,
and then add them together to find the combined rate.

Rate  Time = Work

Audrey 1/4 job/hour  4 hours = 1 job
Ferris 1/3 job/hour  3 hours = 1 job
Audrey & 1/4 + 1/3 = 3/12 + 4/12  2 hours = 7/6 job
Ferris = 7/12 job/hour

Combining the two workers’ rates, we find that they can complete 7/12 job/hour, and
should therefore complete 14/12 = 7/6 job in one hour. Therefore Ferris’ breaks cost
them 7/6 – 1 = 1/6 job worth of productivity.
So how do we find how long Ferris was out? We need to look at his work by
itself. Ferris produced 1/6 job less work than he would have if he had not taken any
break. Let’s fit that into the RTW equation with his rate to find the total time he spent
not working.

Ferris’ breaks 1/3 job/hour  1/2 hour = 1/6 job

At the rate of 1/3 job/hour, Ferris must have spent 1/2 hour on break to miss 1/6 job.
Therefore each of his 3 breaks was 30 minutes ÷ 3 = 10 minutes long. The answer is B.
We can make this problem easier with two different approaches. One is to use
smart numbers to eliminate the fractions. Instead of using a common denominator of 12,
we can define the job as, say, making 12 toys. Now we can say that Audrey makes 3
toys/hour and Ferris makes 4 toys/hour. The work is all the same as above, but when we
get to the point where we are figuring out Ferris’ missed time, the work becomes much
easier. Together, the two should produce 14 toys, and they produce only 12. Therefore
Ferris’ slacking costs them 2 toys. Since his rate is 4 toys/hour, he must have missed 1/2
hour of work.
For those who like theoretical approaches, let’s look at one more method. We
might notice that Audrey works 3/4 as fast as Ferris. However, 2 hours is how long the
job should take if both workers performed at Audrey’s rate of 1 job per 4 hours. This
would involve Ferris working at 3/4 of his normal rate. To do this, he would need to be
on break for 1/4 of the 2 hours, or 30 minutes.

Problem 31 Solution: E

To find the turtle’s average speed for the trip, we need to fill in the following
formula:

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Total distance / Total time = Average rate

Since the distance has not been defined, we can call each equal leg of the trip D.
Therefore, the turtle’s total distance is 3D. (Note that since we have assigned a variable
that will not exist in the answer choices, we need to make sure that D cancels out before
we are done.)
We can find the turtle’s total time by calculating the time for each leg of the
journey. In each case, the time is equal to D/rate.

Rate  Time = Distance

Up 4 mi/hr  D/4 hr = D miles
Across x mi/hr  D/x hr = D miles
Down x2 mi/hr  D/x2 hr = D miles

We can now add up the separate legs to find the total time. We just need a common
denominator:

D D D
  
4 x x2
Dx 2 4 Dx 4 D
  
4 x2 4 x2 4 x2
Dx 2  4 Dx  4 D

4x2
D  x2  4x  4
4 x2

All that’s left to do is plug this total time, along with the total distance of 3D, into our
solution formula:

Average rate = Total distance / Total time =

3D

D  x2  4 x  4
4x2
3D  4 x 2

D  x2  4x  4
12 x 2
x2  4x  4

Notice that the D’s cancel out, as predicted. To match this to answer choice E, we simply
need to recognize the denominator as a special product: (x+2)2.
Alternately, we can solve this problem with Smart Numbers. Let’s make x = 5.
2
Since x = 25, we want a distance for each leg that is divisible by 4, 5, and 25. 100 will
work nicely. Now let’s find the time for each leg of the trip.

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Rate  Time = Distance

Up 4 mi/hr  25 hr = 100 miles
Across x = 5 mi/hr  20 hr = 100 miles
Down x2 = 25 mi/hr  4 hr = 100 miles
Entire Trip 49 hr = 300 miles

The turtle’s average rate is 300/49 mi/hr. There is no need to simplify, as we will just
need to plug in for x in each answer choice and see if the result matches. When we plug x
= 5 into the answer choices, only E produces 300/49.

Problem 32 Solution: B

The first thing to recognize about this problem is that since the vat is being
simultaneously filled and emptied, we need to subtract the rates. Also, since the volume
of chocolate is decreasing with time, we should expect to find a negative combined rate.
Since this is a VIC problem, we can solve either by direct algebra or by picking smart
numbers. Let’s start with the algebra:
If we know the fill tube’s rate is z, we can determine that the dispenser tube’s rate
is 2z. However, we need to allow for the stipulation that this tube does not run on
weekends. Since the tubes otherwise run continuously, we can multiply the dispenser’s
rate by 5/7 to reflect its 2 “days off” per week. (Note that this is an average rate, which
we can only use as a daily rate because the tubes are running for exactly 3 complete
weeks.) To find the combined rate, we can subtract the output from the input. Let’s get
this down in a chart:

Rate  Time = Work

Fill tube z gal/day  21 days = 21z gallons
Dispenser tube 2z(5/7) gal/day  21 days = 42z(5/7) gallons
Both tubes z – 2z(5/7) gal/day  21 days = 21z – 42z(5/7) gallons

Now we need to look at the vat’s capacity. Let’s call it C. The change in volume of
chocolate in 3 weeks is equal to:

New volume – old volume =

(1/4)C – (2/3)C =
-5C/12

The volume of chocolate has gone down by 5/12 of total capacity. Let’s fill this
in as total work. Since we are expecting a negative rate, we should keep the negative
sign. Now we have two ways of expressing the total work: as the product of the
combined rate and time, and as the change in capacity. Any time we have two values for
the same box, we have ourselves an equation:

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21z – 42z(5/7) gallons = -5C/12

We can now solve for capacity (C) in terms of z:

21z – 42z(5/7) = -5C/12

21z – 210z/7 = -5C/12
21z(1 – 10/7) = -5C/12
21z(-3/7) = -5C/12
-9z = -5C/12
-9z(-12/5) = C
108z/5 = C

We can make this process go a little easier by picking smart numbers. Since we
have fractions with 3 and 4 in the denominator, let’s make z = 12. Now that we are using
actual numbers, there’s no point in using an average rate for the dispenser tube. Let’s just
have it pump 2z = 24 gallons/day, and adjust its time to reflect the weekends off. Since
the machines are not always running at the same time, we won’t subtract their rates
directly. We will just calculate the total reduction in volume. The vat loses 108 gallons
in 3 weeks.

Rate  Time = Work

Fill tube 12 gal/day  21 days = 252 gallons
Dispenser tube 24 gal/day  15 days = 360 gallons
Both tubes 21 days = 252 – 360= -108 gallons

Now we’ll work with the capacity. We can’t pick a smart number, because the difference
between 2/3 full and 1/4 full has to be exactly 108 gallons. So let’s find the capacity:

New volume – old volume =

(1/4)C – (2/3)C = -5C/12 = -108 gallons
C = -108(-12/5) gallons
C = 108(12)/5 gallons

At this point, we might recognize that we conveniently have z in the numerator, so C =

108z/5, and the answer is B. If we don’t see this, we still don’t need to actually solve for
108(12)/5. We just need to plug z = 12 into each answer choice to see if it comes close.

Problem 33 Solution: D

This is a difficult problem, but we can get through it with direct algebra if we
keep ourselves very organized. To begin with, no variables are named. How many do
we need? We are asked for the ratio of the kangaroo’s descending rate to the dingo’s
constant rate, so we will need variables for those. We also have an unspecified distance
from Bush A to Bush B, so we’ll need a variable for that.

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Dingo’s rate = r
Kangaroo’s descending rate = k
Distance from Bush A to Bush B = d

Our solution equation looks like this: k/r = ___. It’s helpful to set this up right away,
before we forget that we have been asked for a ratio.
Note that the dingo’s rate is also the kangaroo’s ascending rate. Also note that we
have not assigned a variable for time. Since the dingo and kangaroo arrive at Bush B at
the same time, we can set their times equal to each other. Therefore our initial skeleton
equation looks like this:

Dingo’s distance / dingo’s rate = Kangaroo’s distance / kangaroo’s rate

The left side of that equation is nice and straightforward: d/r. The right side is trickier,
because the kangaroo travels in an arc, and does not move at a constant rate.
Let’s start by finding the kangaroo’s total distance. Since the arc is a semicircle
extending from Bush A to Bush B, we know that d is the diameter. If we had a whole
circle, its circumference would be πd. Since we are dealing with a semicircle, the
distance is πd/2. Let’s break this into two halves and fit it into a chart:

Rate  Time = Distance

First half r  πd/4r = πd/4
Second half k  πd/4k = πd/4
Total πd/4r + πd/4k = πd/2

Each half of the arc is a quarter-circle, or πd/4. We can divide by the rates for each half
to find the time. Therefore the kangaroo’s total time = πd/4r + πd/4k. We can set this
equal to the dingo’s time to solve for k.

d d d
 
r 4r 4k
d  dk  dr
 
r 4rk 4rk
d  dk   dr

r 4rk
 dk   dr
d
4k
4kd   dk   dr
4kd   d (k  r )
4k = π(k + r)
4k = πk + πr
4k – πk = πr
k(4 – π) = πr

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r
k
4 

Now we can plug into our solution equation:

k r 
 r 
r 4  4 

Problem 34 Solution: C

Since we want to know what Y’s total distance would have been, let’s create a
variable for Y’s rate. Our solution will look like this:

Distance = Y’s rate  Y’s time

To solve, we will need to figure out Y’s rate, and then figure out how long it would have
spent flying.
The initial scenario lends itself to a nice straightforward RTD chart. The flocks
are flying toward each other, so we can add their rates. We’ll create a variable for Y’s
rate.

Rate  Time = Distance

Flock X .8y  40/3 hr =
Flock Y y  40/3 hr =
X&Y 1.8y  40/3 hr = 1,200 miles

Now we can solve for y:

1,200 ÷ (40/3) = 1.8y

(1,200  3)/40 = 1.8y
90 = 1.8y
90 = (9/5)y
90(5/9) = y
50 = y

To find Y’s total time for the changed scenario, we will need to separate out the two legs
of the journey:

Leg 1 = Distance that Flock Y flies before Flock X starts

Leg 2 = Distance that Flock Y flies to meet up with Flock X

Leg 1 is already solvable. 50 mi/hr  6 hrs = 300 miles.

We know the distance for Leg 2 is 1,200 – 300 = 900 miles. To find the time for
this leg, we will need to find the combined rate for X and Y. We can solve for X’s rate

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and add, or simply plug right into the combined rate formula from our chart. X and Y
together travel at a rate of 1.8y = 1.8(50 mi/hr) = 90 mi/hr. Now we have all we need to
find the time.

Rate  Time = Distance

Leg 1 (Y only) 50 mi/hr  6 hr = 300 miles
Leg 2 (X & Y together) 90 mi/hr  10 hrs = 900 miles
Total 16 hrs = 1,200 miles

To solve, we just have to plug in Flock Y’s rate and total time. 50 mi/hr  16 hrs = 800
miles.

Problem 35 Solution: E

To determine if either brother covered 100 or more square feet, we would need to
know an individual rate and time for at least one of the two. Note that to get a definitive
answer of yes, we only need to establish that one of the two painted at least 100 square
feet. However, to get a definitive answer of no, we would need to be able to prove that
neither of the brothers painted 100 square feet or more.
From statement 1, we can infer that Nesuhi’s rate was 60 sq. ft/hr. However, we
do not know how much time either Ahmet or Nesuhi spent painting, so we don’t know if
either brother painted more than 100 square feet. Statement 2 tells us the relationship of
the two brothers’ painting times, but nothing about their rates.
Now let’s combine the statements. If we define Ahmet’s time as t, we can
conclude that Ahmet painted 40t square feet, and Nesuhi painted 60(t+1)= 60t + 60
square feet. Therefore, if Ahmet painted for an hour, then he painted 40 square feet and
Nesuhi painted 120 square feet. However, if Ahmet only painted for 15 minutes, then he
painted 10 square feet and Nesuhi painted 75 square feet. Together the statements are
insufficient. The answer is E.

Problem 36 Solution: C

The first thing to note is that we cannot simply average the runners’ rates. Since
they are running equal distances, the runners’ times will all be different, and this will
weight the average rate toward the slowest runner. We need to use the formula for
average rate: Average Rate = Total Distance / Total Time.
Since the distance has not been defined, let’s call each stage of the race D. Now
we can set up our chart:

Rate  Time = Distance

Runner 1 10 ft/sec  D/10 sec = D feet
Runner 2 15 ft/sec  D/15 sec = D feet

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Runner 3 z feet/sec  D/z sec = D feet

Total ?  D/10 + D/15 + D/z = 3D feet

We can now fill in our solution formula:

3D
Average Rate =
D D D
 
10 15 z
3D

3zD 2 zD 30 D
 
30 z 30 z 30 z
3D

3zD  2 zD  30 D
30 z
3D

D  3 z  2 z  30 
30 z
3D

D  5 z  30 
30 z
90 zD

D  5 z  30 
90 z

5 z  30
18 z

z6

Alternately, we can plug in smart numbers. Let’s make z = 20. We will want to
choose a distance that divides evenly by all 3 rates. 60 will work nicely. Now let’s try
that chart again:

Rate  Time = Distance

Runner 1 10 ft/sec  6 seconds = 60 feet
Runner 2 15 ft/sec  4 seconds = 60 feet
Runner 3 20 ft/sec  3 seconds = 60 feet
Total ?  13 seconds = 180 feet

The Target Value equals the average rate, which is 180/13 ft/sec. Plugging 20 into the
answer choices, C yields 18(20)/(20+6)=360/26=180/13. While this approach has its
merits, the time saved up-front may well be lost crunching the unpleasant fractions in the
answer choices.

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Problem 37 Solution: C

We are being asked for the time it would take Grace to walk d miles. This can be
expressed as d / Grace’s rate. We have been given the combined distance and time. Now
we need to use this information, along with the relationship between the two rates, to find
Grace’s individual rate.
Let’s call Grace’s and Sergei’s rates g and s. From the question, we can derive
the following equations:

d = (g + s)t (The total distance is equal to the combined rate  time.)

s = g + 1 (Sergei’s rate is 1 mile/hr faster than Grace’s.)

Substituting g + 1 for s, we get

d = (g + g + 1)t
d = (2g + 1)t
d/t = 2g + 1
d/t – 1 = 2g
d t
= 2g
t
d t
=g
2t

We can now plug this rate into our solution formula, d/g:

d 2dt
Grace’s time to walk d miles =  .
d t d t
2t

The answer is C.
We can also solve with smart numbers. To make our numbers work out cleanly,
we will want to choose a distance that is divisible not only by the two walkers’ rates, but
by their combined rate as well. Let’s try walking rates of 2 and 3. Since these add to 5,
we’ll make the d equal to 2 3 5 = 30 miles. t will be 30 miles / 5 miles/hr = 6 hours.
The target value is Grace’s time of 30 miles / 2 miles/hr = 15 hours. Plugging d =
30 and t = 6 into the answer choices, we find that only C yields an answer of 15 hours.

Rate  Time = Distance

Grace 2 miles/hr  15 hours = 30 miles
Sergei 3 miles/hr  =
Grace & Sergei 5 miles/hr  6 hours = 30 miles

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Problem 38 Solution: C

In this problem, the two trains are working together, so we can combine their
rates. We have been asked for time, so we can rephrase the question as follows:

Time = Distance / Combined rate = ?

Since statement 2 is simpler, let’s start there. If we call the trains’ rates a and b, this
statement gives us the equation b = a/2. Since this tells us only the relative rates of the
two trains, and provides no information on distance or time, it is clearly insufficient. We
can scratch B and D.
Statement 1 has a bit more weight to it. It provides hypothetical adjustments to
the two rates, which we can express as follows:

Distance / Adjusted rates = 1 hour

Distance / (2a + 3b) = 1 hour
Distance = 2a+ 3b

Let’s try plugging this into our solution equation:

2a  3b
Time = =?
ab

Unfortunately, we can’t simplify any further, so Statement 1 is insufficient. We can

scratch A.
Now let’s try combining the two statements. If we substitute the value for b that
we derived from Statement 2 into the equation above, we get the following:

a
2a  3
Time = 2 =?
a
a
2

At this point, since we’re dealing with Data Sufficiency, we might acknowledge that we
have sufficient data to solve. The answer is C. However, to round things out, here are
the final steps:

a
2a  3
Time = 2
a
a
2
7a / 2

3a / 2
= 7/3 hours

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Problem 39 Solution: B

To solve this problem, we need to fill in the following equation:

Average Rate = Total Work / Total Time

We know that the total time is 8 hours, so we just need to find the brothers’ total work.
The first brother works for the full 8 hours at a rate of x, so we can express his work as
8x. The second brother starts t hours later, so he works for 8-t hours. Since his rate is 2x,
his total work is 2x(8-t). How long does the third brother work? He is t hours later than
the second brother, so he is at work for (8-2t) hours. His rate is 4x, so his total work is
4x(8-2t).

Rate  Time = Work

First brother x sq ft/hr  8 hours = 8x sq ft
Second brother 2x sq ft/hr  8-t hours = 2x(8-t) sq ft
Third brother 4x sq ft/hr  8-2t hours = 4x(8-2t) sq ft

The total work is 8x + 2x(8-t) + 4x(8-2t). Let’s plug this into our solution equation and
try to simplify.

Average Rate = Total Work / Total Time

8 x  2 x(8  t )  4 x(8  2t )

8
4 x  x(8  t )  2 x(8  2t )

4
4 x  8 x  xt  16 x  4 xt

4
28 x  5 xt

4
x(28  5t )

4

The answer is B.
We might also solve this problem with smart numbers. Since we know we will be
dividing by 8 at the end, let’s use x = 8. We also need to make t quite small, or the third
brother will never show up. We can use t = 3. Now let’s set up that chart again.

Rate  Time = Work

First brother 8 sq ft/hr  8 hours = 64 sq ft
Second brother 16 sq ft/hr  8–3 = 5 hours = 80 sq ft
Third brother 32 sq ft/hr  5–3 =2 hours = 64 sq ft
Total = 208 sq ft

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The brothers’ total output is 208 square feet, so their average rate over the 8 hours is
208/8 = 26 sq ft. Our target value is 26. We can dispatch some of the answers without
performing full calculations.

A yields 8(13)/12 = 8 13/12. This is barely more than 8.

B yields 8(13)/4 = 2(13) = 26. B is a keeper.
C yields 8(19)/4. Since 8(19) is larger than 8(13), this will clearly be larger than 26.
D yields (56-15)/3 = 41/3. This is less than 26, and is not even an integer. To get 26,
we’d need 78/3.
E yields 56/3. Still a non-integer, and still less than 78/3.

Problem 40 Solution: A

This is clearly a chase problem, so we can subtract the two spies’ rates. However,
there are two complications: the change in Spy X’s rate, and the conversion from miles
per hour to miles per minute. Let’s do the conversion first. Spy Y drives at 80 miles/hr /
60 min/hr = 4/3 mile/min. Since Spy X’s rate changes, let’s break the problem into two
distinct legs. In Leg 1, Spy X is driving at half speed. We can therefore call her rate X/2.

Rate  Time = Distance

Spy X X/2 mile/min  5 min =
Spy Y 4/3 mile/min  5 min =
Spy X – Spy Y (X/2 – 4/3) miles/min  5 min = 5(X/2 – 4/3) miles

In Leg 2, Spy X returns to maximum speed. We can now call her rate X.

Rate  Time = Distance

Spy X X mile/min  4 min =
Spy Y 4/3 mile/min  4 min =
Spy X – Spy Y (X - 4/3) miles/min  4 min = 4(X – 4/3) miles

Now we have the two legs worked out, but what exactly do these figures represent?
Since we subtracted Y’s rate from X’s, the combined rates represent X’s rate relative to
Y. The distances represent X’s gain (or loss) in miles for the two legs of the chase.
So what do we do with those distances? We know that X catches up with Y at the
end of the 2nd leg. So we can add the distances and set them equal to 1, which is the
initial distance between the two spies. (If you chose to subtract X’s rate from Y’s, you
would use a distance of -1, showing that Y lost a mile to X.)

5(X/2 – 4/3) + 4(X – 4/3) = 1

5X/2 – 20/3 + 4X – 16/3 = 1
13X/2 – 36/3 = 1
13X/2 -12 = 1

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13X/2 = 13
X/2 = 1
X=2

Spy X’s maximum rate is 2 miles per minute. There is no conversion trap here (2 is not
in the answer choices), but we do need to remember to convert back to miles per hour. 2
miles/min  60 min/hr = 120 miles/hr. The answer is A.

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