1.10
Introduction
to Word Problems
Part II
from Basic Algebra: One Step at
a Time © 2002
P.
84-100
Dr. Robert J. Rapalje
Seminole Community College
ANSWERS TO ALL EXERCISES ARE INCLUDED AT THE
END OF THIS PAGE
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IV. COIN PROBLEMS
When
solving coin problems (and mixture problems of various types later on!), it is
helpful to form a three-column chart. This
chart is simply a convenient way to organize the information you need to write
the equation. While the number of
rows in the chart varies from one problem to another, there will always be three
columns. For coin problems, the
first column is the number of coins (of each type!).
The second column is the value of each type of coin (in dollars or in
cents--cents is recommended!). The
third column is the “values” column. The
third column is obtained by taking the product of the quantities in the first
two columns. The equation will
always be in the third column. The
examples that follow illustrate this method.
EXAMPLE
12. A
certain number of quarters and four times as many pennies are worth $1.45.
How many of each coin are there?
Solution:
Let x = number of quarters
4x = number of
pennies
Remember
to fill in the first column of the chart, then the second column of the chart.
Finally, multiply the quantities in the first two columns to obtain the
third column.
No.
Coins ×
Each (¢)
= Values
|
Q
|
x
|
25
|
25(x)
|
|
P
|
4x
|
1
|
1(4x)
|
|
|
|
|
145¢
|
Write the equation:
(The equation will always be found in the third column.)
25(x) + 1(4x) =
145
Solve the equation: 25x
+ 4x = 145
29x = 145
x
= 5
Answer
the question:
x = 5 Quarters
4x = 20 Pennies
Check:
5 Quarters = $ 1.25
20
Pennies = .20
$ 1.45
EXERCISES:
50.
A certain number of dimes and four times as many pennies are worth $.98.
How many of each are there?
Solution: Let x
= number of __________
_____ = number of
__________
No.
Coins
×
Each (¢)
= Values
51.
A certain number of quarters and three more dimes than quarters are worth
$7.30. How many of each are there?
Solution: Let x
= number of __________
_____ = number of
__________
No. Coins ×
Each (¢)
= Values
52.
A certain number of dimes and three less pennies than dimes are worth
$7.67. How many of each are there?
Solution:
Let x = number of __________
_____ = number of
__________
No. Coins ×
Each (¢)
= Values
53.
A certain number of nickels and some dimes are worth $7.20.
The number of dimes is three less than twice the number of nickels. How many of each are there?
Solution:
Let x = number of __________
_____ = number of
__________
No. Coins ×
Each (¢)
= Values
EXAMPLE
13. A
box contains 30 coins, in nickels and dimes, worth $2.40.
How many of each coin are there?
Solution: Let x
= number of nickels
30 – x = number of
dimes
No. Coins
× Each (¢)
=
Values
|
N
|
x
|
5
|
5(x)
|
|
D
|
30
– x
|
10
|
10(30
- x)
|
|
|
|
|
240
|
Write
the equation:
5(x) + 10(30 - x) = 240
Solve the equation:
5(x) + 10(30 - x) = 240
5x
+ 300 - 10x = 240
-5x
+ 300 = 240
-5x
= -60
x
= 12 Nickels
Answer the question: x
= 12 Nickels
Check: 12(.05)
= $ .60
30-x = 18
Dimes
18(.10) = 1.80
$2.40
[Note:
You may also let x=number of dimes, and 30-x= number of nickels.
See Example 14.]
EXAMPLE
14. A
box contains 30 coins, in nickels and dimes, worth $2.40.
How many of each coin are there? [Note: This time, let x=no. dimes,
30-x=no. nickels.]
Solution: Let
x =
number of dimes and
30 – x = number of nickels
No. Coins
× Each (¢)
=
Values
|
D
|
x
|
10
|
10(x)
|
|
N
|
30
– x
|
5
|
5(30
- x)
|
|
|
|
|
240
|
Write
the equation:
10(x) + 5(30 - x) = 240
Solve the equation:
10(x) + 5(30 - x) = 240
10x
+ 150 - 5x = 240
5x + 150
= 240
5x
= 90
x
= 18 Dimes
Answer the question:
x
= 18 Dimes
Check:
18(.10) = $ 1.80
30-x = 12 Nickels
12(.05) = .60
$2.40
EXERCISES.
54.
A box contains 20 coins in quarters and dimes worth $2.90. How many
of each coin are there?
Solution: Let
x = number of
__________
_____
= number of __________
No. Coins ×
Each (¢)
= Values
55.
A box contains 20 coins in quarters and dimes worth $3.80.
How many of each coin are there?
56. A box contains 35 coins in quarters and
nickels worth $3.15. How many of
each coin are there?
EXAMPLE
15. A
box contains nickels, dimes, and quarters worth a total of $2.10.
There are twice as many dimes as quarters, and the number of nickels is
two less than the number of dimes. How
many of each coin are there?
Solution
No. Coins
Each (¢)
Values
|
Q
|
x
|
25
|
25x
|
|
D
|
2x
|
10
|
10(2x)
|
|
N
|
2x
- 2
|
5
|
5(2x
– 2)
|
|
|
|
|
210¢
|
Equation:
25x +
20
x + 10 x - 10 =
210
55
x - 10 = 210
55 x =
220
x
= 4
Quarters
Answer
the question:
2
x = 8 Dimes
2
x - 2 =
6 Nickels
Check:
4(.25) = $1.00 Quarters
8(.10)
= .80 Dimes
6(.05) =
.30
Nickels
$2.10 Total
EXERCISES:
57.
A certain number of quarters, four times as many pennies as quarters, and
6 more dimes than pennies are worth $3.36.
How many of each coin are there?
Solution:
Let
_______ = no quarters
_______
= no pennies
_______ = no dimes
No. Coins
Each (¢)
Values
58.
A certain number of dimes, twice as many pennies as dimes, and 6 less
quarters than pennies are worth $6.56. How
many of each coin are there?
Solution:
Let
_______ = no __________
_______ = no __________
_______ = no __________
No. Coins ×
Each (¢)
= Values
59.
A box contains $6.60 in nickels, dimes, and quarters.
There are three times as many nickels as quarters, and the number of
dimes is 4 less than the number of nickels.
How many of each coin are there?
60.
A box contains $8.00 in nickels, dimes, and quarters.
There are three times as many nickels as quarters, and the number of
dimes is 4 less than the number of nickels.
How many of each coin are there?
61.
A certain number of pennies, four times as many dimes as pennies, and a
number of quarters which is 16 less than twice the number of dimes, are worth
$24.92. How many of each coin are
there?
62.
A sum of money consists of nickels, dimes, and quarters amounting to
$1.90. If there are twice as many
nickels as quarters and three less dimes than nickels, how many of each coin are
there?
63.
A box contains nickels, dimes, and quarters worth $12.60.
The number of dimes is 2 less than three times the number of nickels, and
the number of quarters is 4 less than twice the number of dimes. How many of each coin are there?
64.
A box contains nickels, dimes, and quarters worth $69.50.
The number of nickels is 10 more than twice the number of dimes. There
are as many quarters as nickels and dimes combined.
How many of each coin are there?
V.
MIXTURE and INTEREST PROBLEMS
EXAMPLE
16.
Twenty kilograms of nuts consisting of cashews worth $6.00 per kg, pecans
worth $2.50 per kg, and peanuts worth $1.50 per kg are mixed.
There are twice as many pecans as cashews, and there are as many
kilograms of peanuts as pecans. The
total value of the nuts is $56. How
many of each are there?
Solution:
Let x
= kilograms cashews
2x = kilograms pecans
2x = kilograms
peanuts
Type
Nut No. Kilograms
× Each
($) =
Values
|
Cashews
|
x
|
6.00
|
6.00(x)
|
|
Pecans
|
2x
|
2.50
|
2.50(2x)
|
|
Peanuts
|
2x
|
1.50
|
1.50(2x)
|
|
Total
|
20
|
|
$56.00
|
Write
the equation:
6.00(x) + 2.50(2x) + 1.50(2x) = 56.00
Solve
the equation:
6.00(x) + 2.50(2x) + 1.50(2x)
= 56.00
6.00x
+ 5.00x
+ 3.00x =
56.00
14x
= 56.00
x
= 4 kg cashews
Answer
the question:
x =
4 kg cashews
Check: 4($6.00)
= $24.00
2x
= 8 kg pecans
8($2.50)
= 20.00
2x = 8
kg peanuts
8($1.50) = 12.00
$56.00
EXERCISES.
65.
A merchant mixes some candy worth $3.50 per pound with cheap stuff worth
$1.00 per pound. There are 10 more pounds of cheap stuff than the more
expensive candy. If the total value
of the mixture is $28, how many pounds of each are there?
No. Pounds
× Each
= Values in $$
66.
A man is making a nut mixture of almonds worth $4 per pound and cashews
worth $6 per pound. If there are 3
pounds less cashews than almonds, and the total mixture is worth $122, how many
pounds of each are there?
67.
A merchant mixes a total of 50 pounds of candy, some worth $2 per pound,
the rest worth $4 per pound. If the
total value of the mixture is $160, how many pounds of each are there?
68.
A nibble-mix recipe calls for mixing of cheerios, corn chex, and peanuts
together. There are to be two more
pounds of cheerios than nuts and twice as much corn chex as cheerios.
The peanuts are $5 per pound, and both types of cereal cost $3 per pound.
How much of each should be mixed in order to make a total mixture worth $60.
EXAMPLE
17. A woman invests a
sum of money at 6% and $3000 more than this at 9%.
If the total interest earned in one year is $4170, how much was invested
at each rate?
Solution:
Use the familiar formula from business:
Principle
× Rate
× Time =
Interest
In particular,
for these problems, since time = 1 year, Principle
× Rate
= Interest
Identify
the variable:
Let x
= Principle invested at 6% (0.06)
x + 3000 = Principle invested at 9% (0.09)
Principle
× Rate =
Interest
|
6%
|
x
|
0.06
|
0.06(x)
|
|
9%
|
x
+ $3000
|
0.09
|
0.09(x+3000)
|
|
|
|
|
$4170.00
|
Write
the equation: 0.06x
+ 0.09(x + 3000) =
$4170
Solve the equation:
0.06x +
0.09(x + 3000) = $4170
0.06x +
0.09x +
270 = $4170
0.15x
+ 270
= $4170
0.15x
= $3900
x
= $3900/0.15
x = $26000
@ 6%
Answer the question:
x + 3000 = $29000
@ 9%
Check:
26000(0.06) =
$1560.00
29000(0.09) =
2610.00
$
4170.00 Total
EXERCISES:
69.
A sum of money was invested at 8% simple interest, and three times as
much at 10%. The total interest
earned for the year was $190. How
much was invested at each rate?
Solution:
Principle ×
Rate =
Interest
70.
A sum of money was invested at 12% simple interest, and $1000 less than
this at 10%. The total interest
earned for the year was $1000. How
much was invested at each rate?
71.
A sum of money was invested at 5% annual interest, and $500 less than
twice this amount was invested at 12%. If
the total interest earned for the year was $375, how much was invested at each
rate?
72.
A total of $1,000 was invested, some at 8% and the rest at 6% simple
interest. The total interest earned
for the year was $76. How much was
invested at each rate?
73.
A total of $10,000 was invested, some at 12% and the rest at 10% simple
interest. The total interest earned
for the year was $1060. How much
was invested at each rate?
74.
A man has $10,000 to invest, some in a relatively safe account earning 5%
interest per year, and the rest in more speculative investments earning 12% per
year. If the total interest earned
for the year was $955, how much was invested at each rate?
EXAMPLE
18.
How much water must be added to 60 liters of 20% acid solution in
order to dilute the solution to 8%?
Solution:
Use the formula for mixtures: Amt of Solution × Strength = Amt Pure Stuff
Let x = number
of liters of water added
Amt. Sol.
× Strength
= Pure
Stuff
|
20%
|
60
|
0.20
|
0.20(60)
|
|
Water (0%)
|
x
|
0.00
|
0.00(x)
|
|
8%
|
x
+ 60
|
0.08
|
0.08(x +
60)
|
Write/Solve
the Equation: .20(60) +
.00(x) =
0.08(x + 60)
12.0
+ 0
= 0.08x + 4.8
- 4.8
- 4.8
7.2
= 0.08x
0.08x =
7.2
x
= 7.2/(0.08)
x =
90
liters of water
Check:
60(0.20)
= 12 liters acid
+ 90 Water
= No acid
150(0.08)
= 12 liters acid
EXERCISES.
75.
How much 10% alcohol solution must be added to 20 liters of 50% solution
to make a 20% solution?
Solution: Let x = amount of
10% alcohol solution
Amt.
Sol. ×
Strength
= Pure
Stuff
|
10%
|
x
|
0.10
|
0.10(x)
|
|
50%
|
20
|
0.50
|
|
|
20%
|
x
+ 20
|
0.20
|
|
76.
How much water must be added to 20 liters of 50% alcohol solution to
dilute it to 10%?
Solution: Let x = amount of
water
Amt.
Sol. ×
Strength
= Pure
Stuff
|
Water (0%)
|
x
|
0
|
|
|
50%
|
20
|
0.50
|
|
|
10%
|
x + 20
|
0.10
|
|
77.
How much pure alcohol must be added to 20 liters of 10% alcohol solution
to create a 50% solution?
Solution: Let x =
amount of pure 100% alcohol
Amt.
Sol. ×
Strength
= Pure
Stuff
|
100%
|
x
|
1.00
|
|
|
10%
|
20
|
0.10
|
|
|
50%
|
x
+ 20
|
0.50
|
|
78.
How much pure alcohol must be added to 100 liters of 10% alcohol solution
to create an 80% solution?
EXAMPLE
19.
Some 10% alcohol solution is to be mixed with some 30% alcohol
solution to make 20 liters of 16% solution.
How much of each must be used?
Solution:
Let x = amount of 10% solution.
Amt. Sol.
× Strength
= Pure
Stuff
|
10%
|
x
|
0.10
|
0.10(x)
|
|
30%
|
20 - x
|
0.30
|
0.30(20 - x)
|
|
16%
|
20
|
0.16
|
0.16(20)
|
Equation:
0.10(x) + 0.30( 20 - x )
= 0.16(20)
0.10x +
6 - 0.30x =
3.20
-0.20x +
6 =
3.20
-0.20x
= -2.80
x
= -2.80/(-0.20)
x
= 14 liters of 10%
solution
Answer
the question:
x =
14 liters of 10% solution
20 - x
= 6 liters of
30% solution
Check:
14(0.10) = 1.4
liters alcohol
6(0.30)
= 1.8 liters alcohol
20(0.16) =
3.2 liters Total
79.
Some 80% acid solution is to be mixed with some 35% acid solution to make
300 liters of 50% solution. How
much of each acid should be used?
80.
How much 25% acid solution should be mixed with some 85% solution in
order to obtain 300 liters of 65% solution?
81.
How much water must be added to 50% alcohol solution to obtain 100 liters
of 10% solution?
82.
How much pure alcohol must be mixed with some 30% alcohol solution to
obtain 700 liters of 75% solution?
ANSWERS 1.10
p.
65 - 100:
50.
7D, 28P; 51.
20Q, 23D; 52. 70D, 67P; 53.
30N, 57D; 54. 6Q, 14D;
55. 12Q, 8D;
56. 7Q, 28N; 57.
4Q, 16P, 22D; 58. 13D, 26P, 20Q;
59. 10Q, 26D, 30N; 60. 12Q, 36N, 32D; 61.
12P, 48D, 80Q; 62.
4Q, 8N, 5D; 63. 8N, 22D, 40Q;
64. 70D, 150N, 220Q;
65. 4 lb. @ $3.50, 14 lb. @ $1.00;
66. 14 lb. almonds, 11 lb.
cashews; 67. 20 lb. @ $2,
30 lb. @ $4; 68. 3 lb. nuts, 5 lb. cheerios, 10 lb. corn chex; 69. $ 500
@ 8%,
$1500 @ 10%; 70.
$5000 @ 12%, $4000 @ 10%; 71. $1500 @ 5%,
$2500 @ 12%; 72. $ 800 @ 8%,
$ 200 @ 6%;
73. $3000 @ 12%,
$7000 @ 10%;
74.
$3500 @ 5%, $6500
@ 12%;
75. 60 L;
76. 80 L;
77. 16 L; 78. 350 L;
79. 100 L;
80. 100 L; 81.
80 L; 82. 450 L.
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