1.05
Applications
from
Intermediate
Algebra: One Step at a Time
© 1998
P.
51-76
Dr.
Robert J. Rapalje
Seminole State College of Florida
After
solving equations and inequalities, the question "What good is
this?" inevitably arises! Answers
to questions such as "What good is math?" usually involve
applications--that is, "word problems!"
Word problems range in difficulty from simple to absurd. While it is helpful to arrange them in "categories"
for study, it will be even more helpful to see that the
"different" categories or types of problems are actually more
similar than they are different. It
may also be helpful to identify five steps in setting up and solving
word problems:
STEP 1:
IDENTIFY THE VARIABLE. State exactly what it is
that the variable represents.
For example, "Let x = the number of
dimes" or "Let x = rate of the first
plane" or "Let x = amount of pure
alcohol to be added."
Then express all other quantities to be used in the
problem in terms of x.
This is the most important, often the most difficult,
and usually the most overlooked step of the problem.
STEP 2:
WRITE THE EQUATION. Having completed step 1,
use this step in writing the equation.
This is often no more than translating a sentence of
the problem into an equation.
Read the problem carefully.
STEP 3:
SOLVE THE EQUATION. This is usually the easy
part!
STEP 4:
ANSWER THE QUESTION. After solving for x,
there may be other quantities to be determined. Be sure you have answered the question before going on to
the next exercise.
STEP 5:
CHECK.
Check the answers in
the worded problem itself and make sure the solution actually
works. Reject any
extraneous or "inappropriate" answers.
|
In
this section, the following "categories" of applications will
be considered:
I. Number problems, consecutive number problems
II. Perimeter
problems
III. Coin
problems
IV. Interest
problems
V. Mixture
problems
VI. Distance,
rate, time problems
VII. Digit
problems (From the Florida CLAST!)
I. NUMBER
PROBLEMS, CONSECUTIVE NUMBER PROBLEMS.
When
finding consecutive numbers (integers), as 7, 8, 9
or 29, 30, 31, it is
usually convenient to let x represent the first number.
Then since the difference between the numbers is 1, it follows
that x+1 represents the second number, x+2
represents the third number, etc.
When
finding consecutive odd numbers (integers), as 7, 9, 11 or 29, 31, 33, notice that the difference between the numbers
is 2. For this reason, after letting x represent the
first number, the second number will be 2 more than the first (that is, x+2),
the third number will be 2 more than the second (that is, x+4),
and so on. Likewise for
consecutive even numbers, such as 8, 10, 12
or 22, 24, 26, the difference between consecutive numbers is 2.
Therefore, if x represents the first, then x+2
and x+4 represent the second and third numbers
respectively.
In summary, for
consecutive numbers:
Let
x = first
number
x+1 =
second number
x+2 = third
number.
For
consecutive even numbers or consecutive odd numbers:
Let
x = first
number
x+2 =
second number
x+4 = third
number.
The
numbers represented by x, x+2, and x+4
will all be even, or they will all be odd, depending upon whether the
value of x
turns out to be even or odd.
EXAMPLE 1:
Three
numbers are such that the second is six less than the first, and the third
is ten more than the sum of the first two numbers.
The sum of the numbers is 38.
Find the numbers.
SOLUTION: Let
x =
first number
x – 6 =
second number
x + x
– 6 + 10 = third number
or
2 x + 4
Equation:
x + (x – 6) + (2 x +
4) = 38
4 x – 2
= 38
4
x = 40
Answer
the question: x = 10
first number
x –
6 = 4 second number
2 x + 4
= 24 third number
Check:
10 + 4 + 24 = 38
EXAMPLE 2:
Find
three consecutive numbers whose sum is 189.
SOLUTION: Let
x = first number
x +1 = second number
x +2 = third number
Equation:
x + x
+1 + x +2
= 189
3 x + 3 =
189
3 x =
186
x =
62
Answer
the question: x
= 62 first number
x + 1 = 63 second
number
x + 2 = 64
third number
Check:
62 + 63 + 64 = 189
EXAMPLE 3:
Find
three consecutive odd numbers whose sum is 189.
SOLUTION: Let
x = first number
x + 2 = second number
x + 4 = third number
Equation:
x + x
+2 + x +4
= 189
3 x + 6 =
189
3 x =
183
x =
61
Answer
the question: x
= 61 first number
x + 2 = 63
second number
x + 4 = 65
third number
Check:
61 + 63 + 65 = 189
EXAMPLE 4:
Find
three consecutive even numbers such that the first plus twice the
second plus three times the third is equal to 100.
SOLUTION: Let
x = first number
x + 2 = second number
x + 4 = third number
Equation:
x + 2(x +2) + 3(x +4)
= 100
x + 2 x + 4 + 3 x + 12
= 100
6 x + 16 = 100
6 x = 84
x =
14
Answer
the question: x
= 14 first number
x +2 = 16
second number
x + 4 = 18
third number
Check:
14 + 2(16) + 3(18) = 100
14 + 32
+ 54 = 100
EXERCISES:
1. Three numbers are such that the second number is 5 more than the first,
and the third number is 4 more than three times the second.
The sum of the three numbers is 134.
Find the numbers.
2. Three numbers are such that the second number is 3 less than
the first, and the third is twice the second.
The sum of the numbers is 91.
Find the numbers.
3. Three numbers are such that the second number is 4 more than three times
the first, and the third number is 12 less than the sum of the first two
numbers. The sum of the three
numbers is 44. Find the
numbers.
4. Three numbers are such that the first number is 10 less than twice the
second, and the second number is 4 more than three times the third.
Twice the second number is equal to the sum of the first and third
numbers. Find the numbers.
5. Find three consecutive integers whose sum is 432.
6. Find three consecutive even integers whose sum is 432.
7. Find two consecutive odd integers such that twice the second plus
the first is 121.
8. Find three consecutive integers such that the first plus twice the
second plus three times the third is equal to 200.
9. Find three consecutive even integers such that three times the first
plus six times the second is equal to ten times the third.
10. Find three consecutive integers such that twice the first minus the
third is 3 less than the second.
II. PERIMETER
PROBLEMS.
The
perimeter of a geometric figure (shape) is the total distance around the
outside of that figure. For a rectangle, the perimeter consists of two
widths and two lengths. For a
triangle, the perimeter is just the sum of the three sides.
For a circle, the perimeter (for circles it is called the circumference) is π times
the diameter of the circle.
Rectangle: P = 2W + 2L
Triangle: P = a + b+ c
Circle: C
= π d (also C=2 π r)
|
EXAMPLE 5:
The length of a rectangle is 8 more than twice
the width. The perimeter of
the rectangle is 106 centimeters. Find
the dimensions of the rectangle.
SOLUTION: Let
x = width of
rectangle
2 x +8 =
length of rectangle
Equation:
2W +
2L =
Perimeter
2(x) + 2(2 x +8) = 106
2x +
4x + 16 = 106
6 x = 90
x = 15
Answer
question: x =
15 cm width of rectangle
2 x + 8 = 38 cm length of rectangle
Check:
2W +
2L = P
2(15) + 2(38) = 106
30 +
76 = 106
EXERCISES:
11. The length of a rectangle is 5 meters longer than the width.
The perimeter is 50 meters. Find
the dimension of the rectangle.
SOLUTION: Let
x = width of the rectangle
_____ = length of the rectangle
Equation:
2 W
+ 2 L
= Perimeter
2(
) +
2( ) =
______
Check:
12.
The length of a rectangle is 5 meters more than twice
the width. The perimeter is
130 meters. Find the
dimensions.
13.
The length of a rectangle is 5 meters less than twice
the width. The perimeter is
50 meters. Find the length
and width of the rectangle.
14.
The length of a rectangle is 50 feet less than three
times the width. If the
perimeter is 500 feet, find the length and width of the rectangle.
15.
The length of a rectangle is 50 meters less than
twice the width. The
perimeter is 1100 meters. Find
the dimensions.
16. The width of a rectangle is 50
feet less than the length. The
perimeter is 400 feet. Find
the dimensions of the rectangle.
17.
The length of a rectangle is 3 less than 5 times the
width. The perimeter is 10
times the width. Find the
dimensions and the perimeter of the rectangle.
18. EXTRA
CHALLENGE:
The
perimeter of a rectangle is 46 meters.
Twice the length is 4 more than 5 times the width.
Find the length and width of the rectangle.
SOLUTION:
Let x = width
_____ = two lengths
Equation: Two widths +
two lengths = Perimeter
__________ + __________ = 46
III. COIN
PROBLEMS.
EXAMPLE 6:
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:
19.
A certain number of quarters and four times as many
pennies are worth $1.45. How
many of each coin are there?
SOLUTION:
No. Coins
Each (¢)
Values
20.
A certain number of quarters and three times as many
dimes are worth $1.65. How
many of each coin are there?
21.
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?
22.
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, then
30 - x = number of dimes.
No. Coins
Each (¢)
Values
23.
A box contains 20 coins in quarters and dimes worth
$2.90. How many of each coin
are there?
24.
A box contains 20 coins in quarters and dimes worth
$3.80. How many of each coin
are there?
25.
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?
26.
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?
27.
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?
28.
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?
29.
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?
30.
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?
IV. INTEREST
PROBLEMS.
EXAMPLE 7:
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:
Principle
× Rate
= Interest
|
6%
|
x
|
0.06
|
0.06(x)
|
|
9%
|
x + 3000
|
0.09
|
0.09(x+3000)
|
|
|
|
|
$4170
|
Equation:
0.06x +
0.09(x + 3000) =
$4170
0.06
x + 0.09 x
+ 270
= $4170
0.15
x
+ 270
= $4170
0.15 x = $3900
x =
$3900/0.15
x
= $26000 @
6%
Answer
the question:
x
= $26000 @ 6%
x
+ 3000 = $29000 @
9%
Check:
Interest equals 26000(0.06)
= $ 1560.00
29000(0.09) = 2610.00
$ 4170.00 Total
EXERCISES:
31. 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
32.
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?
33.
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?
34.
A total of $2500 was invested, part at 12% simple
interest, and the rest at 10%. The
total interest earned for the year was $260.
How much was invested at each rate?
35.
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?
36.
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?
V. MIXTURE
PROBLEMS.
EXAMPLE 8:
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:
Amt Solution ×
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 =
14
x
= 14 liters of
10% solution
Answer
the question:
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
EXAMPLE 9:
How much liquid must be drained from a 20 liter radiator at 20%
antifreeze and replaced with pure antifreeze to bring the strength up to
50%?
SOLUTION:
Amt Solution ×
Strength =
Pure Stuff
|
Begin
with
|
20
|
0.20
|
0.20(20)
|
|
Drain
|
(–) x
|
0.20
|
– 0.20(x)
|
|
Add
|
(+) x
|
1.00
|
+1.00(20)
|
|
End up
with
|
20
|
0.50
|
=0.50(20)
|
Equation:
0.20(20) -
0.20(x) +
1.00(x) =
0.50(20)
4
+ 0.80x
= 10
0.80x
= 6
x
= 6/(0.80)
x = 7.5
liters of antifreeze
EXAMPLE 10 :
How much water must be added to 60 liters of 20% acid
solution in order to dilute the solution to 8%?
SOLUTION:
Amt Solution × Strength
= Pure
Stuff
|
20%
|
60
|
0.20
|
0.20(60)
|
|
Water
|
x
|
0.00
|
0.00(x)
|
|
8%
|
x
+ 60
|
0.08
|
0.08(x + 60)
|
Equation:
.20(60)
+ .00(x)
= 0.08(x
+ 60)
12 +
0 =
0.08x + 4.8
7.2
= 0.08x
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
EXAMPLE 11:
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.
If there are twice as many pecans as cashews, and the total value
of the nuts is $56, how many of each are there?
[HINT: Let x
= kg cashews;
2x = kg pecans
3x = kg cashews and pecans combined
20 - 3x = kg peanuts
]
SOLUTION:
No. Kg.
×
Each Kg.
=
Total Value
|
Cashews
|
x
|
6.00
|
6.00(x)
|
|
Pecans
|
2x
|
2.50
|
2.50(2x)
|
|
Peanuts
|
20 - 3x
|
1.50
|
1.50(20
- 3x)
|
|
Total
|
20
|
|
56.00
|
Equation:
6.00(x)
+ 2.50(2x)
+ 1.50(20 - 3x)
= 56.00
6.00x
+ 5.00x
+ 30.00
- 4.50x
= 56.00
6.50x + 30.00
= 56.00
6.50
x
= 26.00
x =
26.00/(6.50)
x
= 4 kg cashews
2x =
8 kg pecans
20 - 3x =
8 kg peanuts
EXERCISES:
37.
How much 10% alcohol solution must be added to 20
liters of 50% solution to make a 20% solution?
SOLUTION:
Amt Solution × Strength
= Pure
Stuff
|
10%
|
x
|
0.10
|
|
|
50%
|
20
|
0.50
|
|
|
20%
|
x + 20
|
0.20
|
|
Equation:
38.
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?
39.
How much water must be added to 20 liters of 50%
alcohol solution to dilute it to 10%?
SOLUTION:
Amt Solution
× Strength
= Pure
Stuff
|
Water
0%
|
x
|
0.00
|
|
|
50%
|
20
|
0.50
|
|
|
10%
|
x + 20
|
0.10
|
|
Equation:
40.
How much water must be added to 50% alcohol solution
to obtain 100 liters of 10% solution?
41.
How much pure alcohol must be added to 20 liters of
10% alcohol solution to create a 50% solution?
SOLUTION:
Amt Solution × Strength
= Pure
Stuff
|
100%
|
x
|
1.00
|
|
|
10%
|
20
|
0.10
|
|
|
50%
|
x + 20
|
0.50
|
|
Equation:
42.
How much pure alcohol must be added to 100 liters of
10% alcohol solution to create an 80% solution?
43.
How much liquid must be drained from an 18 liter
radiator that is 10% antifreeze and replaced with pure antifreeze in order
to bring the strength up to 50%?
SOLUTION:
Amt
Solution ×
Strength = Pure
Stuff
|
Begin
with
|
18
|
|
|
|
Drain
|
(–) x
|
|
|
|
Add
|
(+) x
|
|
|
|
End
up with
|
18
|
|
|
Equation:
44.
How much liquid must be drained from a 24 liter
radiator that is 25% antifreeze and replaced with pure antifreeze in order
to bring the strength up to 50%?
45.
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?
46.
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?
47.
EXTRA CHALLENGE
Fifty
tickets were sold to a chicken barbeque for a total of $219.
Children's tickets sold for $2.50, youth tickets sold for $3.50,
and adult's tickets sold for $5.00. There
were 10 more youth tickets than children's tickets.
How many of each ticket were sold?
48.
EXTRA CHALLENGE
A
total of 180 tickets are sold, some at $3, some at $5, and some at $10
each. The total value of the tickets was $1100.
The number of $5 tickets was 20 more than the number of $3 tickets.
How many of each ticket were sold?
VI. DISTANCE,
RATE, TIME PROBLEMS.
EXAMPLE 12:
Two bicycles start at the same point traveling in the
opposite direction. The speed
of the second bike in miles per hour is 12 less than three times the
first. At the end of 6 hours,
the bicycles are 144 miles apart. Find
the speed of each bicycle.
SOLUTION:
Rate ×
Time
= Distance
|
1st
|
x
|
6
|
6(x)
|
|
2nd
|
3x - 12
|
6
|
6(3x - 12)
|
|
Total
|
|
|
144
|
Equation:
6x +
6(3x - 12) =
144
6x
+ 18x
- 72 =
144
24x
= 216
x
= 9
Answer
the question:
x
= 9 mph
1st bicycle
3x - 12 = 15
mph 2nd bicycle
EXAMPLE 13:
A train
leaves the
Sanford terminal
averaging 45 mph.
A second train leaves the same terminal two hours later averaging
60 mph. How long will it take
the second train to catch the first train?
[HINT:
Let x = time of second train to catch first.
The distances are equal.]
SOLUTION:
Rate
×
Time
= Distance
|
1st
|
45
|
x
+ 2
|
45
(x
+ 2
)
|
|
2nd
|
60
|
x
|
60(x)
|
Equation:
1st Distance
= 2nd
Distance
45(x + 2)
= 60(x)
45x + 90
= 60x
90 =
15x
x
= 6 hours for
second train
EXERCISES:
49.
Two boys are riding bicycles in the opposite
direction. One travels 15 mph
faster than the other. At the
end of 3 hours they are 102 miles apart.
How fast is each boy riding?
50.
Two cars are driving in opposite directions, one at
55 mph and the other at 65 mph (on the interstate!). How long will it take before the cars are 300 miles apart?
VII. DIGIT
PROBLEMS (This skill is on the Florida CLAST!!)
The
value of a two digit number is 10 times the ten’s digit (t), plus the unit’s digit (u).
VALUE
of 2 digit number = 10t + u
EXAMPLE 14:
The ten’s digit of a number is 5 more than the unit’s
digit. The value of the
number is 8 times the sum of its digits.
Find the digits and value of the number.
SOLUTION:
Let
u = unit’s digit
u + 5 = ten’s digit (t)
Value
of the number = 10t
+ u = 8(t + u)
Equation:
10(u
+ 5) + u = 8(u + u+5 )
10u
+ 50 + u = 8u + 8u + 40
11u
+ 50 = 16u + 40
-5u
= -10
u
= 2
Unit’s digit
u
+ 5 = 7 Ten’s
digit
The number is 72.
51.
The ten’s digit of a two digit number is 2 less
than the unit’s digit. The
value of the number is 7 times the unit’s digit.
Find the digits and the value of the number.
52.
The ten’s digit of a two digit number is 2 more
than the unit’s digit. If
the number itself is 16 times the unit’s digit, find the digits and the
value of the number.
ANSWERS
1.05
p.55-76:
1. 22,27,85;
2. 25,22,44;
3. 6,22,16;
4. 58,34,10;
5. 143,144,145;
6.
142,144,146;
7. 39,41;
8.
32,33,34;
9. -28,-26,-24; |