Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
538
23.36 A collection of 44 coins containing pennies, dimes, and quarters is worth
$5.90. There are twice as many dimes as quarters. How many pennies are in
the collection?
Let q represent the number of quarters, d = 2q represent the number of dimes,
and p = 44 – 3q represent the number of pennies. Multiply the total number of
each coin by its value and set the sum equal to 5.90.
Solve for q.
There are q = 13 quarters, 2(13) = 26 dimes, and 44 – 3(13) = 44 – 39 = 5
pennies.
Work
How much time does it save to work together?
23.37 Two computer programmers are instructed to write code for an application.
Working by herself, Donna could complete the task in five days, but when
working with Chris, she is able to complete the program in three days. How
long would it take Chris to complete the task by himself?
To set up a work problem, create one fraction for each of the individuals
involved. The numerator of each fraction is the length of time the individual
worked, and the denominator is the length of time it would take that individual
to complete the task working alone. Add the fractions and set the sum equal to
one.
Chris and Donna worked together for the same length of time, three days.
Donna can complete the task alone in five days. Let x represent the length of
time it would take Chris to complete the task alone.
There are
q quarters and
2q dimes. The
combined number of
quarters and dimes
is q + 2q = 3q. The
total number of coins
is 44, so subtract the
combined number of
quarters and dimes
from 44 to calculate
the number of
pennies.
This is the
expression for
pennies dened in
the beginning of the
problem: p = 44 – 3q.
The sum
of the fractions
always equals 1.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
539
Multiply the entire equation by 5x, the least common denominator, to eliminate
fractions.
Solve for x.
It would take Chris days to complete the task by himself.
Note: Problems 23.38–23.39 refer to a pair of carpenters named Lisa and Karen. Lisa can
build a bookshelf in 30 minutes, and Karen can build the same bookshelf in 45 minutes.
23.38 How long will it take Lisa and Karen to build the bookshelf together?
Let x represent the length of time Lisa and Karen work together to complete
the bookshelf. Create a rational expression for each carpenter, dividing x (the
length of time worked by both) by the length of time it would take each to
complete the bookshelf alone. Add the fractions and set the sum equal to one.
Multiply the entire equation by 90, the least common denominator, to eliminate
fractions.
Working together, Lisa and Karen complete the bookshelf in 18 minutes.
“1” represents “1
complete bookshelf.”
The fractions on
the left side of the
equation represent
the portions of that
complete job that each
person is responsible for.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
540
Note: Problems 23.38–23.39 refer to a pair of carpenters named Lisa and Karen. Lisa can
build a bookshelf in 30 minutes.
23.39 Karen is injured and can no longer complete the bookshelf alone in 45
minutes. It now takes Lisa and Karen 28 minutes 48 seconds to complete the
bookshelf working together. How long would it take Karen to complete the
bookshelf working alone?
Divide 48 seconds by 60 to convert into minutes: . It takes the
pair of carpenters 28.8 minutes to build the bookshelf together, and Lisa can
complete the bookshelf in 30 minutes by herself. Let x represent the length of
time it would take Karen to build the bookshelf.
Multiply the entire equation by 30x, the least common denominator, to
eliminate fractions.
Solve for x.
It would take Karen 720 minutes, or hours, to complete the task alone.
Note: Problems 23.40–23.41 refer to Rob and Matt, who go into business together mowing
lawns during the summer. Rob works twice as fast as Matt.
23.40 If it takes Rob and Matt 90 minutes to mow one lawn together, how long would
it take each of them to mow it separately?
Let x be the length of time it takes Rob to mow the lawn alone, and let 2x be
the time Matt needs to complete the same task. They spend 90 minutes working
together.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
541
Multiply the entire equation by 2x, the least common denominator, to eliminate
fractions.
Rob could mow the lawn alone in 135 minutes, and it would take Matt
2(135) = 270 minutes.
Note: Problems 23.40–23.41 refer to Rob and Matt, who go into business together mowing
lawns during the summer. Rob works twice as fast as Matt.
23.41 Rob and Matt are mowing the lawn described in Problem 23.40, a job they can
complete in 90 minutes working together. Twenty minutes after they begin,
Rob has to leave and does not return. How long will it take Matt to complete
the remainder of the job alone?
Rob and Matt begin work together, but Rob works for only 20 minutes. If x
represents the amount of time it will take Matt to complete the job after Rob
leaves, Matt works a total of x + 20 minutes. According to Problem 23.40, Rob
can mow the lawn by himself in 135 minutes and it takes Matt 270 minutes.
Multiply the entire equation by 270, the least common denominator, to
eliminate fractions.
It will take Matt 210 minutes, or hours, to finish mowing the lawn.
He works 20
minutes with Rob
and then x minutes
after Rob leaves.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
542
Note: Problems 23.42–23.43 refer to an aquarium that contains a plecostomus, an algae-
eating fish. Under normal circumstances, this plecostomus can completely clean the algae
from the sides of the aquarium in seven days.
23.42 Due to poor maintenance, algae are blooming in the fish tank at an
accelerated rate. In fact, if the plecostomus were absent, the algae would cover
the sides of the aquarium in four days. How long will it take algae to cover the
sides of the tank despite the efforts of the fish?
Let x represent the time it will take algae to cover the sides of the aquarium.
Both the fish and the algae “work” for the entire time period.
Multiply the entire fraction by 28, the least common denominator, to eliminate
fractions.
In days, the sides of the tank will be covered with algae.
Note: Problems 23.42–23.43 refer to an aquarium that contains a plecostomus, an algae-
eating fish. Under normal circumstances, this plecostomus can completely clean the algae
from the sides of the aquarium in seven days.
23.43 The owner of the aquarium described in Problem 22.42 cleans the tank
and thereby restores the ecosystem. However, two months later, the algae
begin to bloom again. To combat the algae problem this time, the aquarium
owner adds a second plecostomus to the tank. Working alone under normal
circumstances, the new fish could clean the sides of the tank in five days.
How long will it take both fish, working together, to clean the sides of the
aquarium as they combat an algae bloom that (if left unchecked) would once
again cover the sides of the tank in four days?
Let x represent the time it takes both fish working together to clean the sides of
the tank. The pair of fish and the algae “work” for that entire period of time.
Note that the algae is working against the progress of the fish, so its fraction
should be subtracted from, rather than added to, the fractions representing the
work of the fish.
Instead
of adding
the fractions,
you subtract the
sh fraction from
the algae fraction.
The sh is actively
working against the
algae, so all of its
work is slowing
down the algae
growth.
The algae
could take
over the tank
in 4 days. Neither
sh can work that
fast—one takes 5
days to clean the
aquarium and one
takes 7—but working
together, they can
eliminate the
algae.
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.