Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
529
Discard the negative solution. The width of the prism is in.
Substitute w into the length formula defined above.
Use common denominators to simplify the expression.
The dimensions of the prism are h = 5 in, in, and
in.
Speed and Distance
Distance equals rate times time
23.23 Calculate the average speed (in km/h) of a turtle that travels 2.5 km in four
hours.
The formula for distance traveled at an average rate r for a length of time t is
d = rt. Substitute d = 2.5 and t = 4 into the formula and calculate r.
The turtle traveled at a rate of r = 0.625 km/h.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
530
23.24 How far will a jogger moving at an average speed of 6 mph travel in 75
minutes?
The speed is reported in miles per hour, so express time in terms of hours,
rather than minutes: hours. Substitute r = 6 and into the
distance formula to calculate d.
The jogger travels miles.
23.25 If it takes Bus A 3.25 hours to complete a fixed route at an average speed of 35
mph, how long will it take Bus B (in hours) to complete the same route at an
average speed of 30 mph? Round the answer to the hundredths place.
Let d
A
be the distance Bus A travels (in miles) at rate r
A
mph for time t
A
hours.
Similarly, let d
B
be the distance Bus B travels at rate r
B
mph for time t
B
hours.
The distance formula for Bus A is d
A
= r
A
t
A
, and the distance formula for Bus B
is d
B
= r
B
t
B
.
The buses travel the same distance, so d
A
= d
B
. Substitute d
A
= r
A
t
A
and d
B
= r
B
t
B
into that equation.
Substitute r
A
= 35, t
A
= 3.25, and r
B
= 30 into the equation and solve for t
B
.
Bus B will complete the route in approximately 3.79 hours.
The units
need to match
up. The speed is
written in terms of
miles per HOUR so
time needs to be
measured in hours.
Either that or convert
6 mph into
miles per minute and
keep t = 75. You’ll
get the same
value for d.
Multiply 0.79
(the decimal part
of the answer) by 60
to approximate minutes:
(0.79)(60) ≈ 47.4. Bus B
took just over 3 hours
and 47 minutes to
complete the route.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
531
23.26 If Don travels 320 km at a speed 5 km/h faster than Mike, who travels 300 km
during the same period of time, what was the average speed of both drivers?
Let d
D
represent the distance traveled (in kilometers) by Don at a rate of
r
D
km/h for t
D
hours. Let d
M
represent the distance traveled by Mike at a rate
of r
M
km/h for t
M
hours. The distance formula for Don’s trip is d
D
= r
D
t
D
, an the
distance formula for Mike’s trip is d
M
= r
M
t
M
.
Solve each of the distance formulas for t.
Both drivers travel for the same period of time, so t
D
= t
M
. Substitute the above
rational expressions into the equation.
Don’s speed is 5 km/h faster than Mike’s speed, so r
D
= r
M
+ 5. Substitute
d
D
= 320, d
M
= 300, and r
D
= r
M
+ 5 into the equation.
Cross multiply and solve for r
M
.
Mike drove at an average speed of 75 km/h, and Don drove at an average speed
of 80 km/h.
In Problem
23.25, the
distances traveled
were equal, and you
substituted into the
equation d
A
= d
B
. In
this problem, the time
traveled is equal.
You’re solving for t so
that you can plug into
the equation
t
D
= t
M
.
Don’s speed is
r
D
= r
M
+ 5 = 75 + 5 =
80 km/h.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
532
23.27 Jack and Jill are exercising on a 400-meter-long track. Jack walks at an average
speed of 2 m/sec. After Jack has completed one full lap, Jill starts from the
same position Jack did, jogging along the same path at an average speed of
5 m/sec. How long does Jill have to run to catch up to and pass Jack?
Let d
JA
represent the distance (in meters) Jack travels at an average rate of r
JA
m/sec for t
JA
seconds. Let d
JI
represent the distance Jill travels at an average rate
of r
JI
m/sec for t
JI
seconds. The formula for Jack’s distance walked is d
JA
= r
JA
t
JA
,
and the formula for Jill’s distance jogged is d
JI
= r
JI
t
JI
.
After Jill has caught up to Jack, she will have traveled the same distance he has.
At that moment, d
JA
= d
JI
. Substitute the distance formulas into the equation.
Jill begins jogging after Jack has completed a lap, a distance of 400 m. It
will take Jack 400 ÷ 2 = 200 seconds to travel that distance at 2 m/sec.
Therefore, Jack’s travel time is 200 seconds longer than Jill’s: t
JA
= t
JI
+ 200.
Substitute r
JA
= 2, r
JI
= 5, and t
JA
= t
JI
+ 200 into the equation.
Jill will catch up to Jack in seconds.
23.28 It takes Phil 10 minutes to swim from the shore to a buoy, at an average speed
of 3.5 ft/sec. How long will it take Phil to swim back to shore along the same
route if his speed is slowed to 2 ft/sec by the currents in the water? Express the
answer in minutes and seconds.
Let d
1
= r
1
t
1
represent the distance from the shore to the buoy and d
2
= r
2
t
2
represent the distance from the buoy to the shore. According to the problem,
Phil swims the same route each way, so the distances traveled each way are
equal.
d = rt = (2)(200) = 400
Which
is
minutes?
Because d
1
= r
1
t
1
, you can
substitute r
1
t
1
for d
1
in the equation d
1
= d
2
.
Same thing for d
2
= r
2
t
2
.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
533
Time is expressed in minutes but speed is expressed in ft/sec. Convert time
into seconds, thereby making the units compatible: 10 minutes = 10(60) = 600
seconds. Substitute r
1
= 3.5, t
1
= 600, and r
2
= 2 into the equation and solve for t
2
.
It takes Phil 1,050 seconds to swim from the buoy to the shore. Divide 1,050 by
60 to calculate the number of minutes Phil swam.
1,050 ÷ 60 = 17.5 minutes
Phil’s swim from the buoy to the shore lasted 17 minutes and 30 seconds.
23.29 Two cars are placed at opposite ends of a straight, 0.75-mile-long test track
to determine the effectiveness of driver’s side air bags in a head-on collision.
The first car begins driving toward the second at an average speed of 30 mph,
and 10 seconds later, the second car begins driving toward the first car at an
average speed of 40 mph. How long does the first car drive before colliding
head-on with the second car? Report the answer in hours.
Let d
1
= r
1
t
1
represent the distance traveled by the first car and d
2
= r
2
t
2
represent
the distance traveled by the second car. The cars begin at opposite ends of the
0.75-mile-long track and will collide when the sum of the distances traveled by
both cars is 0.75 miles.
The second car starts driving 10 seconds after the first car, so at the time of
collision, the first car will have traveled for 10 seconds longer. Notice that the
speeds of the cars are expressed as miles per hour, so convert 10 seconds into
hours: . Therefore, t
1
= t
2
+ 10 seconds, and hours.
Substitute r
1
= 30, r
2
= 40, and into the equation.
Multiply by
60 to convert the
decimal part of the
answer back to seconds:
(60)(0.5) = 30.
To convert
from seconds
to minutes, you
divide by 60. To
convert from minutes
to hours, divide by
60 again. If you want
to convert straight
from seconds to
hours, divide
by (60)(60) =
3,600.
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