Math 115 Word Problem Solutions
3.2.41: Twelve added to a certain number is 21. What is the number?
a. Choose a variable and indicate what it represents in the problem.
Let x = a certain number.
b. Set up an equation that represents the situation described.
12  x  21
c. Solve the equation.
12  x  21
12  x  12  21  12
x9
3.2.43: Nine subtracted from a certain number is 13. Find the number.
a. Choose a variable and indicate what it represents in the problem.
Let x = a certain number.
b. Set up an equation that represents the situation described.
x  9  13
c. Solve the equation.
x  9  13
x  9  9  13  9
x  22
3.2.55: Dress socks cost $2.50 a pair more than athletic socks. Randall purchased one pair
of dress socks and six pairs of athletic socks for $21.75. Fine the price of a pair of
dress socks.
a. Choose a variable and indicate what it represents in the problem.
Cost of Dress socks = x + 2.50
Cost of Athletic Socks = x
b. Set up an equation that represents the situation described.
6 x  1( x  2.50)  21.75
c. Solve the equation.
6 x  1( x  2.50)  21.75
6 x  x  2.5  21.75
7 x  2.5  21.75
7 x  19.25
x  2.75
 Cost of Athletic Socks = $2.75
Cost of Dress Socks = 2.75 + 2.50 = $5.25
3.3.35: Find two consecutive odd numbers whose sum is 72.
Set up and solve an algebraic equation.
First odd number = x
Second odd number = x + 2
x  ( x  2)  72
2 x  2  72
2 x  70
x  35
 First odd number = 35
Second odd number = 35 + 2 = 37
Math 115 Word Problem Solutions
3.3.37: Find three consecutive even numbers whose sum is 114.
Set up and solve an algebraic equation.
First even number = x
Second even number = x + 2
Third even number = x + 4
x  ( x  2)  ( x  4)  114
x  x  x  2  4  114
3x  6  114
3 x  108
x  36
 First even number = 36
Second even number = 36 + 2 = 38
Third even number = 36 + 4 = 40
3.3.45: If two angles are supplementary and the larger angle is 20° less than three times the
smaller angle, find the measure of each angle.
Set up and solve an algebraic expression.
First angle = x
Second angle = 3x – 20
NOTE: Two angles for which the sum of their measure is 180° are called supplementary
angles.
x  (3x  20)  180
4 x  20  180
4 x  200
x°
(3x-20)°
x  50
 First angle = 50°
Second angle = 3(50) – 20 = 130°
3.3.53: At a university-sponsored concert, there were three times as many women as men.
A total of 600 people attended the concert. How many men and how man women
attended?
Set up and solve an algebraic equation.
Number of men = m
Number of women = 3m
m  3m  600
4m  600
m  150
 Number of men = 150
Number of women = 3(150) = 450
Math 115 Word Problem Solutions
3.3.57: At a local restaurant, $275 in tips is to be shared between the server, bartender, and
busboy. The server gets $25 more than three times the amount the busboy receives.
The bartender gets $50 more than the amount the busboy receives. How much will
the server receive?
Set up and solve an algebraic equation.
Busboy’s share = b
Bartender’s share = 50 + b
Server’s share = 3b + 25
b  (50  b)  (3b  25)  275
b  b  3b  50  25  275
5b  75  275
5b  200
b  40
 Server’s share = 3(40) + 25 = $145
3.4.63: Find three consecutive whole numbers such that twice the sum of the two smallest
numbers is 10 more than three times the largest number.
Set up and solve an algebraic equation.
First number = x
Second number = x + 1
Third number = x + 2
2[ x  ( x  1)]  3( x  2)  10
2(2 x  1)  3 x  6  10
4 x  2  3x  16
x  2  16
x  14
 First number = 14
Second number = 14+1 = 15
Third number = 14+2 = 16
3.4.67: Find a number such that 20 more than one-third of the number equals three-fourths
of the number.
Set up and solve an algebraic expression.
Let x = a certain number.
1
3
20  x  x
3
4
1 

3 
12  20  x   12  x 
3 

4 
240  4 x  9 x
240  5 x
x  48
Math 115 Word Problem Solutions
3.4.75: Max has a collection of 210 coins consisting of nickels, dimes, and quarters. He has
twice as many dimes as nickels, and 10 more quarters than dimes. How many coins
of each kind does he have?
Set up and solve an algebraic expression.
Amount of dimes = d
d
Amount of nickels =
2
Amount of quarters = d  10
d
 d       d  10   210
2
d
2d   10  210
2
d
2d   200
2
d

2  2d    2  200 
2

4d  d  400
5d  400
d  80
 Amount of dimes = 80
80
Amount of nickels =
= 40
2
Amount of quarters = 80 + 10 = 90
Math 115 Word Problem Solutions
3.4.83: In triangle ABC, the measure of angle A is 2° less than one-fifth of the measure of
angle C. The measure of angle B is 5° less than one-half of the measure of angle C.
Find the measure of the three angles of the triangle.
Set up and solve an algebraic expression.
Measure of angle C = c
1

Measure of angle A =  c  2 
5

1

Measure of angle B =  c  5 
2

NOTE: The sum of the measures of the three angles of a triangle is 180°.
1
 1

c   c  2    c  5   180
5
 2

A
1

 c  5
1

1
1
2

 c  2
c  c  c  7  180
5


5
2
1
1
c  c  c  187
5
2
c°
1
1 

10  c  c  c   10 187 
C
5
2 

10c  2c  5c  1870
17c  1870
c  110
 Measure of angle C = 110°
1

Measure of angle A =  110   2    22  2  20
5

1

Measure of angle B =  110   5  55  5  50
2

3.4.85: The supplement of an angle is 10° smaller than three times its complement. Find
the size of the angle.
Set up and solve an algebraic expression.
Let x = an angle.
Supplement of the angle = 180  x
Complement of an angle = 90  x
180  x  3  90  x   10
180  x  270  3 x  10
180  x  260  3 x
180  2 x  260
2 x  80
x  40
 The measure of angle x = 40°.
B
Math 115 Word Problem Solutions
3.6.69: Suppose that the perimeter of a rectangle is to be no greater than 70 inches, and the
length of the rectangle must be 20 inches. Find the largest possible value for the
width of the rectangle.
Set up and solve an appropriate inequality.
l
Let P = perimeter
Let l = length.
w
Let w = width
P  2l  2 w
P  70
2l  2 w  70
2  20   2 w  70
40  2 w  70
2 w  30
w  15
 The largest possible width, w, of the rectangle is 15 inches.
3.T.22: Suppose that a triangular plot of ground is enclosed with 20 meters of fencing. The
longest side of the lot is two times the length of the shortest side, and the third side
is 10 meters longer than the shortest side. Find the length of each side of the plot.
Set up and solve an algebraic expression.
Let s = the shortest side.
Longest side = 2s
Third side = s + 10
 s    2s    s  10   70
2s
4 s  10  70
s  10
4 s  60
s  15
 Shortest side = 15 meters
Longest side = 2 15 = 30 meters
Third side = 15  10  25 meters
4.1.47: 15% of what number is 6.3?
Set up and solve an appropriate equation.
Let n = a number.
(15%)n  6.3
.15n  6.3
n  42
4.1.49: 76 is what percent of 95?
Set up and solve an appropriate equation.
Let p = the percentage.
76  p(95)
76
p
95
p  0.8
p  80%
s
Math 115 Word Problem Solutions
4.1.51: What is 120% of 50?
Set up and solve an appropriate equation.
Let n = a number.
n  120%  (50)
n  (1.20)(50)
n  60
4.1.55: 160% of what number is 144?
Set up and solve an appropriate equation.
Let n = a number.
(160%)n  144
1.60n  144
n  90
4.1.59: Suppose that a car can travel 264 miles using 12 gallons of gasoline. How far will it
go on 15 gallons?
Solve using a proportion.
Let m = miles traveled.
Gallons
12 15
:

Miles Traveled 264 m
12m  (15)(264)
12m  3960
m  330 miles
4.1.67: A preelection poll indicated that three out of every seven eligible voters were going
to vote in an upcoming election. At this rate, how many people are expected to vote
in a city of 210,000?
Solve using a proportion.
Let a = the number of actual voters.
3
a
Actual Voters

:
7 210, 000
Eligible Voters
 7  a    3 210, 000 
7 a  630, 000
a  90, 000
4.1.73: An inheritance of $180,000 is to be divided between a child and the local cancer
fund in the ratio of 5 to 1. How much money will the child receive?
Solve using a proportion.
Let i = the child’s inheritance.
Child's Share
:
Cancer Fund's Share
5
i

1 180, 000  i
1 i    5180, 000  i 
i  900, 000  5i
6i  900, 000
i  $150, 000
Math 115 Word Problem Solutions
4.2.23: Tom bought an electric drill at a 30% discount sale for $35. What was the original
price of the drill?
Original Selling Price  Discount  Discount Selling Price
Let p = original price.
100%  p    30%  ( p)  $35
1.00 p  .30 p  35
.70 p  35
p  $50
4.2.25: Find the cost of a $4800 wide-screen plasma television that is on sale for 25% off.
Original Selling Price  Discount  Discount Selling Price
Let d = the discounted selling price.
$4800   $4800  25%   d
4800   4800 .25   d
4800  1200  d
d  $3600
4.2.29: Pierre bought a coat for $126 that was listed for $180. What rate of discount did he
receive?
Original Selling Price  Discount  Discount Selling Price
Let r = the rate of discount.
$180  ($180)( r )  $126
180  180r  126
180  126  180r
54  180r
54
r
180
r  .30
r  30%
4.2.31: A retailer has some toe rings that cost him $5 each. He wants to sell them at a profit
of 70% of the cost. What should be the selling price of the toe rings?
Selling Price  Cost  Profit
Let s = selling price.
s  $5  (70%)($5)
s  5  (.70)(5)
s  5  3.5
s  $8.50
Math 115 Word Problem Solutions
4.2.35: Jewelry has a very high markup rate. If a ring costs a jeweler $400, at what price
should it be sold to gain a profit of 60% of the selling price.
Selling Price  Cost  Profit
Let s = selling price
s  $400  (60%)( s )
s  400  .60s
.40s  400
s  $1000
4.2.37: If the cost of a pair of shoes for a retailer is $32 and he sells them for $44.80, what is
his rate of profit based on the cost?
Selling Price  Cost  Profit
Let r = rate of profit.
$44.80  $32   $32  p
44.80  32  32 p
12.8  32 p
p  .40
p  40%
4.2.39: Find the annual interest rate if $560 in interst is earned when $3500 is invested for 2
years.
i  Prt
$560   $3500  r  2 
560   3500  2  r
560  7000r
r  .08
r  8%
4.2.43: What will be the interest earned on a $5000 certificate of deposit invested at 3.8%
annual interest for 10 years?
i  Prt
i   $5000  3.8% 10 
i   5000 .03810 
i  $1900
4.2.45: How much is a month’s interest on a mortgage balance of $145,000 at a 6.5%
annual interest rate?
i  Prt
1
i   $145000  6.5%   
 12 
1
i  145000 .065   
 12 
i  $785.42
Math 115 Word Problem Solutions
4.3.15: A dirt path 4 feet wide surrounds a rectangular garden that is 38 feet long and 17
feet wide. Find the area of the dirt path.
A  lw
Area of garden  17  38   646
Area of garden with dirt path   46  25   1150
Area of Dirt Path  Whole area  Area of Garden
 1150  646
 504 sq. ft.
17  4  4  25 ft
38  4  4  46 ft
4 ft
17 ft
4 ft
38 ft
4.3.19: Find the length of an altitude of a trapezoid with bases of 8 inches and 20 inches and
an area of 98 square inches.
1
A  h  b1  b 2 
2
8 in
1
98  h  8  20 
2
1
h
98  h  28
2
98  14h
20 in
h  7 in
4.3.31: If the total surface area of a right circular cylinder is 104π square meters, and a
radius of the base is 4 meters long, find the height of the cylinder.
S   r 2   rs
65    5     5  s
2
65  25  5 s
40  5 s
s  8 ft
s
5 ft
Math 115 Word Problem Solutions
4.4.21: The width of a rectangle is 3 inches less than one-half of its length. If the perimeter
of the rectangle is 42 inches, find the area of the rectangle.
P  2l  2 w
l
1


42  2l  2  l  3 
2

1
42  2l  l  6
w  l 3
2
48  3l
l  16
1
16   3
2
w  83
w5
w
A  lw
A  16  5   8 sq in
4.4.25: The second side of a triangle is 1 centimeter longer than three times the first side.
The third side is 2 centimeters longer than the second side. If the perimeter is 46
centimeters, find the length of each side of the triangle.
First Side = F
Second Side = 3F+1
Third Side = (3F+1)+2
Perimeter = First Side + Second Side + Third Side
F
46   F    3F  1   3F  1  2 
3F+1
46  3F  3F  F  2  1  1
46  7 F  4
42  7 F
F 6
 First Side = 6 cm
Second Side = 3  6  1  18  1  19 cm
Third Side = 3  6   1  2  18  1  2  19  2  21 cm
(3F+1)+2
Math 115 Word Problem Solutions
4.4.33: The distance between Jacksonville and Miami is 325 miles. A freight train leaves
Jacksonville and travels toward Miami at 40 miles per hour. At the same time, a
passenger train leaves Miami and travels toward Jacksonville at 90 miles per hour.
How long will it take the two trains to meet?
D  rt
Freight Train Leaving Jacksonville: D  40t
Passenger Train Leaving Miami: D  90t
Freight Train’s Distance + Passenger Train’s Distance = Total Distance
40t  90t  325
130t  325
325
t
130
1
t  2 hours
2
4.4.35: A car leaves a town traveling at 40 miles per hour. Two hours later a second car
leaves the town traveling the same route and overtakes the first care in 5 hours and
20 minutes. How fast was the second car traveling?
20
1
or 5 hours
5 hours and 20 minutes = 5
60
3
D
r
t
1
 1

40  5  2 
5 +2
First Car
40
3
 3

1
 1
r 5 
5
Second Car
r
3
 3
 1

 1
40  5  2   r  5 
 3

 3
 16 6 
 16 
40     r  
 3 3
 3
 22  16
40    r
 3  3
880 16
 r
3
3
r  55 miles per hour
Math 115 Word Problem Solutions
4.4.37: Two trains leave at the same time, one traveling east and the other traveling west.
1
At the end of 9 hours they are 1292 miles apart. If the rate of the train traveling
2
east is 8 miles per hour faster than the rate of the other train, find their rates.
D
Train Traveling East
Train Traveling West
9
1
 r  8
2
1
9 r
2
r
r 8
r
t
1
9
2
1
9
2
1
1
9  9  r  8   1292
2
2
9.5r  9.5r  76  1292
19r  1216
r  64
 Rate of Eastbound Train = 64+8 = 72 miles per hour
Rate of Westbound Train = 64 miles per hour.
4.4.39: Jeff leaves home and rides his bicycle out into the country for 3 hours. On his
return trip along the same route, it takes him three-quarters of an hour longer. If
his rate on the return trip was 2 miles per hour slower than on the trip out into the
country, find the total roundtrip distance.
Leaving Home
Returning Home
D
3r
3

 3    r  2
4

r
r
r2
t
3
3
3
4
3

3r   3    r  2 
4

3r  3.75  r  2 
3r  3.75r  7.5
.75r  7.5
r  10
 The distance he traveled away from home = 3(10) 30 miles. Thus, the roundtrip
distance is 2(30) = 60 miles.
Math 115 Word Problem Solutions
4.5.13: How many milliliters of pure acid must be added to 100 milliliters of a 10% acid
solution to obtain a 20% solution?
Let l = milliliters of pure acid.
100%  l   10% 100    20% 100  l 
1.00l  .10 100   .20 100  l 
l  10  20  .2l
.8l  10  20
.8l  10
l  12.5 millilters
4.5.15: How many centiliters of distilled water must be added to 10 centiliters of a 50% acid
solution to obtain a 20% acid solution?
Let c = centiliters of distilled water
A solution that is 50% acid is 50% water.
A solution that is 20% acid is 80% water.
100%  c    50% 10   80% 10  c 
c  .5 10   .8 10  c 
c  5  8  .8c
.2c  5  8
.2c  3
c  15 centiliters
4.5.17: Suppose that we want to mix some 30% alcohol solution with some 50% alcohol
solution to obtain 10 quarts of a 35% solution. How many quarts of each kind
should we use?
If we let q = quarts of 30% alcohol solution
Then, 10 – q = quarts of 50% alcohol solution.
 30%  q   50% 10  q    35% 10 
.3q  .5 10  q   .35 10 
.3q  5  .5q  3.5
5  .2q  3.5
.2q  1.5
q  7.5
 Amount of 30% alcohol solution needed = 7.5 quarts
Amount of 50% alcohol solution needed = 10  7.5  2.5 quarts
Math 115 Word Problem Solutions
4.5.19: How much water needs to be removed from 30 liters of a 20% salt solution to
change it to a 50% salt solution.
Let g = gallons of water.
A 30% salt solution has 70% water.
A 40% salt solution has 60% water.
 70% 12   100%  g    60%  20  g 
.7  20   g  .6  20  g 
14  g  12  .6 g
14  12  .4 g
2  .4 g
g  5 gallons
4.5.21: Suppose that a 12-quart radiator contains a 20% solution of antifreeze. How much
solution needs to be drained out and replaced with pure antifreeze to obtain a 50%
solution?
Let q = quarts of antifreeze to be drained out and replaced.
 20% 12    20%  q   100%  q    40% 12 
.2 12   .2q  q   4 12 
2.4  .8q  4.8
.8q  2.4
q  3 quarts
4.5.25: Thirty ounces of punch that contains 10% grapefruit juice is added to 50 ounces of
a punch that contains 20% grapefruit juice. Find the percent of grapefruit juice in
the resulting mixture.
Let p = percent of grapefruit juice.
10%  30    20%  50    p 80 
.1 30   .2  50   80 p
3  10  80 p
13  80 p
13
p
80
p  .1625
p  16.25%
Math 115 Word Problem Solutions
4.5.27: Suppose that the perimeter of a square equals the perimeter of a rectangle. The
width of the rectangle is 9 inches less than twice the side of the square, and the
length of the rectangle is 3 inches less than twice the side of the square. Find the
dimensions of the square and the rectangle.
s
w  2s  9
s
l  2s  3
P  2l  2w
2l  2 w  4 s
2  2 s  3  2  2 s  9   4 s
4 s  6  4 s  18  4 s
P  4s
 The dimensions of the square are 6 in by 6 in.
The length of the rectangle is l  2  6  3  12  3  9 in
The width of the rectangle is w  2  6  9  3 in
8s  24  4 s
24  4 s
s6
4.5.31: Pam is half as old as her brother Bill. Six years ago Bill was four times older than
Pam. How old is each sibling now?
Bill’s age now = b
b
Pam’s age now =
2
Bill’s age 6 years ago = b  6
b
Pam’s age 6 years ago =  6
2
b

b  6  4  6
2

b  6  2b  24
6  b  24
b  18
 Bill’s age now = 18 years.
18
 9 years.
Pam’s age now =
2
Math 115 Word Problem Solutions
4.5.33: Nina received an inheritance of $12,000 from her grandmother. She invested part
of it at 6% interest, and she invested the remainder at 8% interest. If the total
yearly interest from both investments was $860, how much did she invest at each
rate?
If we let x = the amount invested at 6%
Then, 12, 000  x  the amount invested at 8%.
 6%  x   8% 12000  x   860
.06 x  .08 12000  x   860
.06 x  960  .08 x  860
960  .02 x  860
960  .02 x  860
100  .02 x
x  5000
The amount invested at 6% = $5,000
The amount invested at 8% = 12,000-5000 = $7,000.
4.5.35: Sally invested a certain sum of money at 9%, twice that sum at 10%, and three
times that sum at 11%. Her total yearly interest from all three investments was
$310. How much did she invest at each rate?
If we let x = the sum of money invested at 9%
Then, 2x = the sum of money invested at 10%
And, 3x = the sum of money invested at 11%.
 9%  x   10%  2 x   11%  3x   310
.09 x  .1 2 x   .11 3 x   310
.09 x  .2 x  .33 x  310
.62 x  310
x  500
 The sum of money invested at 9% = $500
The sum of money invested at 10% = 2  500   $1000
The sum of money invested at 11% = 3  500  $1,500.
4.5.37: Fawn invested a certain amount of money at 3% interest and she invested $1250
more than that amount at 5%. Her total yearly interest was $134.50. How much
did she invest at each rate?
If we let x = the sum of money invested at 3%
Then, x + 1250 = the sum of money invested at 5%.
 3%  x    5%  x  1250   134.50
.03 x  .05  x  1250   134.50
.03x  .05 x  62.5  134.5
.08 x  72
x  900
 The sum of money invested at 3% = $900
The sum of money invested at 5% = 900  1250  $2,150.
Math 115 Word Problem Solutions
4.R.14: The sum of two numbers is 96, and their ratio is 5 to 7. Find the numbers.
Let x = a number.
Let y = another number.
x  y  9

x 5
y  7

x
5
y
7
5
y  y  96
7
5
7
y  y  96
7
7
12
y  96
7
y  56
5
 56   40
7
The numbers are 40 and 56.
4.R.20: Seventy-eight yards of fencing were purchased to enclose a rectangular garden.
The length of the garden is 1 yard shorter than three times it width. Find the length
and width of the garden.
P  2l  2 w
x
78  2  3w  1  2 w
w
78  6 w  2  2 w
78  8w  2
80  8w
l  3w 1
w  10
 The width is w = 10 yards
The length is l  3 10 1  30 1  29 yards.
4.T.14: Find the area of a circular region if the circumference is 16π centimeters. Express
the answer in terms of π.
C  2 r
16  2 r
16
r
2
r  8 cm
r 8
A   r2
A   8
2
A  64 sq. cm.
Math 115 Word Problem Solutions
4.T.22: The election results in a certain precinct indicated that the ratio of female voters to
male voters was 7 to 5. If a total of 1500 people voted how many women voted?
If we let f = the number of female voters
Then, 1500  f  the number of male voters.
7
f
Female Voters

:
5 1500  f
Male Voters
5 f  7 1500  f 
5 f  10500  7 f
12 f  10500
f  875
 The number of female voters is f = 875
4.T.23: A car leaves a city traveling at 50 miles per hour. One hour later a second car
leaves the same city traveling the same route at 55 miles per hour. How long will it
take the second car to overtake the first car?
D
r
t
50  t  1 50
t
First Car
Second Car
55 t  1
55t
50  t  1  55t
50t  50  55t
50  5t
t  10 hours
6.1.85: The square of a number equals nine times that number. Find the number
Let x = the number.
x2  9x
x2  9x  0
x  x  9  0
x  0 or x  9  0
x9
 x = 0 or 9.
Math 115 Word Problem Solutions
6.1.87: The area of a square is numerically equal to five times its perimeter. Find the
length of a side of the square.
Let x = the length of a side of a square.
x
Area of a Square  x 2
x
Perimter of a Square  4 x
x2  5  4x 
x 2  20 x
x 2  20 x  0
x  x  20   0
x  0 or x  20  0
x  20
Length cannot equal 0.  x  20 .
6.1.89: Suppose that the area of a circle is numerically equal to the perimeter of a square,
and that the length of a radius of the circle is equal to the length of a side of the square.
Find the length of a side of the square. Express your answer in terms of π.
Let r = the length of a radius of the circle.
Let s = the length of a side of the square.
rs
 r 2  4r
r
 r 2  4r  0
r  r  4   0
s
r  0 or  r  4  0
A   r2
r  4
4
r

Length cannot equal 0.  s 
s
4
P  4s
.

6.2.65: Forty-nine less than the square of a number equals zero. Find the number.
Let x = the number.
x 2  49  0
 x  7  x  7   0
x7  0
x7
or
x7 0
x  7
Math 115 Word Problem Solutions
6.2.69: The sum of the areas of two squares is 234 square inches. Each side of the larger
square is five times the length of a side of the smaller square. Find the length of a
side of each square.
Let s = the length of a side of the smaller square.
Let 5s = the length of a side of the larger square.
x 2  25 x 2  234
x
26 x 2  234
A  x2
2
5x
x 9
x3
 The length of a side of the smaller square is s  3 inches
The length of a side of the larger square is s  5  3  15 inches
6.3.51: Find two consecutive integers whose product is 56.
First number = x
Second number = x + 1
x  x  1  56
A   5 x   25 x 2
x 2  x  56
x 2  x  56  0
x7  0
or
x8  0
x7
x  8
 The two consecutive integers are 7 and 8 or -7 and -8.
6.3.53: Find two consecutive even whole numbers whose product is 168.
First number = x
Second number = x + 2
x  x  2   168
x 2  2 x  168
x 2  2 x  168  0
 x  12  x  14   0
x  12  0
or
x  14  0
x  12
x  14
 The first number is x = 12
The second number is x  2  2  12  14
2
Math 115 Word Problem Solutions
6.3.55: Find four consecutive integers such that the product of the two larger integers is 22
less than twice the product of the two smaller integers.
First number = x
Second number = x + 1
Third number = x + 2
Fourth number = x + 3
2  x  x  1   22   x  2  x  3
2  x 2  x   22  x 2  3 x  2 x  6
2 x 2  2 x  22  x 2  5 x  6
x 2  3 x  28  0
 x  7  x  4   0
x  7  0 or x  4  0
x7
x  4
 The four consecutive integers are 7, 8, 9, 10 or -4, -3, -2, -1.
6.5.71: Find two numbers whose product is 15 such that one of the numbers is seven more
than four times the other number.
Let n = the first number.
Then, 4n  7  the other number.
n  4n  7   15
4n 2  7 n  15  0
 4n  5 n  3  0
4n  5  0 or n  3  0
4n  5
n  3
5
n
4
5
5
 If n  , then the other number is 4    7  5  7  12 .
4
4
If n  3 , then the other number is 4  3  7  12  7  5 .
Math 115 Word Problem Solutions
6.5.73: Find two numbers whose product is -1. One of the numbers is three more than
twice the other number.
Let a = the first number.
Then, 2a + 3 = the other number.
a  2a  3  1
2a 2  3a  1
2a 2  3a  1  0
 2a  1 a  1  0
2a  1  0
2a  1
a
a 1  0
a  1
or
1
2
1
 1
 If a   , then the other number is 2     3  1  3  2
2
 2
If a  1 , then the other number is 2  1  3  2  3  1
6.5.75: A number is one more than twice another number. The sum of the squares of the
two numbers is 97. Find the number.
Let n = the number.
 n    2n  1
2
2
 97
n 2  4n 2  4n  1  97
5n 2  4n  96  0
 5n  24  n  4   0
n40
5n  24  0
5n  24
n4
24
n
5
 If n  4 , then the other number is 2  4  1  8  1  9.
If n  
or
24
48 5
43
 24 
, then the other numbers is 2     1      .
5
5 5
5
 5 
Math 115 Word Problem Solutions
6.T.25: If numerically the volume of a cube equals twice the total surface area, find the
length of an edge of the cube.
Let s = the length of an edge of the cube.
Volume of the Cube = s 3
Surface Area of the Cube = 6s 2
s 3  2  6s 2 
s 3  12s 2
s 3  12s 2  0
s 2  s  12   0
s2  0
s0
or
s  12  0
s  12
 Since length cannot equal 0, the length of an edge of the cube is s  12.
7.4.61: If by jogging at a constant rate Joan can complete a race in 40 minutes, how much
of the course has she completed at the end of m minutes?
D  rt
Let D = how much of the course Joan has completed.
Time t = m minutes.
1
Rate r 
40
m
D
.
40
7.4.63: If Sandy drove k kilometers at a rate of r kilometers per hour, how long did it take
her to make the trip?
D  rt
D
t
r
Let t = the time it took Sandy to drive the distance.
Distance D  k kilometers.
Rate r  r kilometers per hour.
k
t .
r
7.4.69: If the area of a rectangle is 47 square inches, and the length is l inches, what is the
width of the rectangle?
A  lw
47  lw
47
w
inches.
l
l
w
Math 115 Word Problem Solutions
7.5.41: The numerator of a fraction is 8 less than the denominator. The fraction in its
5
simplest form is . Find the fraction.
6
If we let d = the denominator
Then, d  8 = the numerator.
d 8 5

d
6
6  d  8   5d
6d  48  5d
6d  5d  48
d  48
48  8 40
 .
48
48
7.6.37: Suppose that Celia rides her bicycle 60 miles in 2 hours les time than it takes Tom to
ride his bicycle 85 miles. If Celia rides 3 miles per hour faster than Tom, find their
respective rates.
D
r
t
60
Celia 60
t 2
t2
85
Tom
85
t
t
60 85
 3
t 2 t
60t  85  t  2   3t  t  2 
 The fraction is
60t  85t  170  3t 2  6t
3t 2  19t  170  0
 3t  34  t  5   0
34
or t  5
3
60
60
 Celia's Rate 

 20 mph
52 3
85
Tom's Rate 
 17 mph
5
t
Math 115 Word Problem Solutions
7.6.39: One day, Jeff rides his bicycle out into the country 40 miles. On the way back, he
takes a different route that is 2 miles longer and it takes him an hour longer to
return. If his rate on the way out into the country is 4 miles per hour faster than his
rates back, find both rates.
D
Into the Country
40
Way Back
42
r
40
t
42
t 1
t
t
t 1
40 42

4
t
t 1
40  t  1  42t  4t  t  1
40t  40  42t  4t 2  4t
4t 2  6t  40  0
2  2t 2  3t  20   0
2  t  4  2t  5   0
t  4
or
t
5
2
40 80

 16 mph
5
5
2
42
42 84
Rate back from country 


 12 mph
5 1 7
7
2
2
7.6.43: It takes Barry twice as long to deliver papers as it does Mike. How long would it
take each if they can deliver the papers together in 40 minutes?
Let x = how long it takes Mike to deliver the papers.
1 1
1


x 2 x 40
1 1 
 1 
40 x     40 x  
 x 2x 
 40 
40  20  x
x  60
 Mike takes 60 minutes
Barry takes 2  60  120 minutes.
 Rate into the country =
Math 115 Word Problem Solutions
7.6.45: Mark can overhaul an engine in 20 hours, and Phil can do the same job by himself
in 30 hours. If they both work together for a time, and then Mark finishes the job
by himself in 5 hours, how long did they work together?
Time to Finish
Entire Job
Mark
20
Phil
30
Working
Rate
1
20
1
30
Time
Working
t 5
t
Fractional Part
Of Job Done
t 5
20
t
30
t 5 t

1
20 30
t 5 t 
60 
   60 1
 20 30 
3  t  5   2  t   60
3t  15  2t  60
5t  45
t  9 hours
7.6.47: A copy center has two copiers. Copier A can produce copies at a rate of 40 pages
per minute, and copier B does 30 pages per minute. How long will copier B need to
run if copier A has been copying by itself for 6 minutes, and then both copier A and
B are used until 520 copies are made?
40  x  6   30 x  520
40 x  240  30 x  520
70 x  280
x  4 minutes
7.R.31: The sum of a number and twice its reciprocal is 3. Find the number.
Let x = the number.
1
x  2   3
x
2

x  x    x  3
x

x 2  2  3x
x 2  3x  2  0
 x  1 x  2   0
x 1
or
x2
Math 115 Word Problem Solutions
7.R.35: An inlet pipe can fill a tank in 10 minutes. A drain can empty the tank in 12
minutes. If the tank is empty and both the inlet pipe and drain are open, how long
will it be before the tank overflows?
x
Inlet Pipe 
10
x
Drain 
12
x x
 1
10 12
 x x
60     60 1
 10 12 
6 x  5 x  60
x  60 minutes
2
7.T.23: The sum of a number and twice its reciprocal is 3 . Find the number.
3
Let x = the number.
2
1
x  2   3
3
x
2 11
x 
x 3
2

 11 
3x  x    3x  
x

3
2
3 x  6  11x
3 x 2  11x  6  0
 x  3 3x  2   0
x3
or
x
2
3
Math 115 Word Problem Solutions
7.T.24: Wendy can ride her bicycle 42 miles in the same time that it takes Betty to ride her
bicycle 36 miles. Wendy rides 2 miles per hour faster than Betty. Find Wendy’s
rate.
D
r
t
42
t
Wendy 42
t
36
2
t
Betty
36
t
42 36

2
t
t
 42   36

t    t   2
 t   t

42  36  2t
6  2t
t 3
36
 Wendy's Rate 
 2  12  2  14 miles per hour
3
7.T.25: Garth can mow a lawn in 20 minutes, and Alex can mow the same lawn in 30
minutes. How long would it take the two of them working together to mow the
lawn?
x
x

1
20 30
x 
 x
60     60 1
 20 30 
3 x  2 x  60
5 x  60
x  12 minutes
8.6.25: The sum of two numbers is 30, and their difference is 12. Find the numbers.
8.6.29: One number is twice another number. The sum of three times the smaller and five
times the larger is 78. Find the numbers.
8.6.31: Three lemons and 2 apples cost $2.35. Two lemons and 3 apples cost $2.40. Find
the cost of 1 lemon and the cost of 1 apple.
8.6.33: Larry has $1.45 in change in his pocket, consisting of dimes and quarters. If he has
a total of ten coins, how many of each kind does he have?
8.6.37: A 10% salt solution is to be mixed with a 15% salt solution to produce 10 gallons of
a 13% salt solution. How many gallons of the 10% solution and how many gallons
of the 15% solution will be needed?
8.6.39: Sidney invested $13,000, part of it at 5% and the rest at 6%. If his total yearly
interest was $730, how much did he invest at each rate?
8.7.47: Suppose that the sum of two numbers is 46, and the difference of the numbers is 22.
Find the numbers.
8.7.49: Suppose that a youth hostel rents double rooms at $35 per day and single rooms at
$25 per day. If a total of 50 rooms was rented one day for $1600, how many of each
kind were rented?
Math 115 Word Problem Solutions
8.7.51: In a class of 50 students, the number of women is 2 more than five times the number
of men. How many women are there in the class?
8.7.55: A deposit slip listed $700 in cash to be deposited. There were 100 bills, some of
them five-dollar bills and the remainder ten-dollar bills. How many bills of each
denomination were deposited?
8.7.57: Suppose that Sue has three times as many nickels as pennies in her collection.
Together her pennies and nickels have a value of $4.80. How many pennies and
how many nickels does she have?
8.7.59: One solution contains 40% alcohol, and a second solution contains 70% alcohol.
How many liters of each solution should be mixed to make 30 liters that contain
50% alcohol?
10.R.25: Each of three consecutive odd whole numbers is squared. The three results are
added and the sum is 251. Find the numbers.
10.R.29: A company has a rectangular parking lot 40 meters wide and 60 meters long.
They plan to increase the area of the lot by 1100 square meters by adding a strip
of equal width to one side and one end. Find the width of the strip to be added.
10.T.22: Abu rode his bicycle 56 miles in 2 hours less time than it took Stan to ride his
bicycle 72 miles. If Abu’s rate was 2 miles per hour faster than Stan’s rate, find
Abu’s rate.
10.T.23: Find two consecutive odd whole numbers whose product is 255.