R6 Math 208 Learning Team Problems Chapters 4, 5.1, 5.5) (Rockswold & Krieger
3rd Ed)
Section 4.1: P. 256: 68, 72
Section 4.2: P. 264: 58, 66
Section 4.3: P. 275: 62, 64
Section 4.4: P. 286: 74, 76
Section 5.1: P. 316: 96
Section 5.5: P. 354: 94
4.1 #68: Renting a Car A rental car costs $25 plus $0.25 per mile.
(a) Write an equation that gives the cost of driving the car x miles.
(b) Use the intersection-of-graphs method to determine
the number of miles that the car is driven if the rental cost is $100.
(c) Solve part (b) numerically with a table of values.
(a) Let x number of miles driven, then:
Cost = 25 + .25*(Miles) or C = 25 + .25x
(b) 100 = 25 + .25x; so .25x = 75 and x = 300
(c)
x
C
0
25
100
50
200
75
300
100
4.1 #72: Dimensions of a Triangle An isosceles triangle has
a perimeter of 17 inches with its two shorter sides
equal in length. The longest side measures 2 inches
more than either of the shorter sides.
(a) Write a system of two equations in two variables
that describes this information. Be sure to specify
what each variable means.
(b) Solve your system graphically. Explain what your
results mean.
Let x be the length of the shorter side; y be the longer side. As in:
/\
/ \
x /
\ x
/
\
/
\
/--------------\
y
Then x + x + y = 17 (perimeter)
And y = x + 2
(2 more than shorter side)
(a) Then 2x + y = 17
-x + y = 2
(b) Solving by R1 + 2*R2 -- >
2x + y = 17
-2x + 2y = 4
----------------3y = 21
y=7
so in 2nd equation, x = y – 2
or
x=5
(c) Graph these lines for intersection (5, 7)
4.2 #58: Ticket Prices Two hundred tickets were sold for a
baseball game, which amounted to $840. Student
tickets cost $3, and adult tickets cost $5.
(a) Let x be the number of student tickets sold and y
be the number of adult tickets sold. Write a system
of linear equations whose solution gives the
number of each type of ticket sold.
(b) Use the method of substitution to solve the system.
(a) X = student tickets Y = adult tickets
The number of tickets sold is 200, so x + y = 200
The amount of revenue from the cost of the tickets is $840
So, 3x + 5y = 840
Together, we have
x + y = 200
3x + 5y = 840
(b) y = 200 – x so substitute y in the 2nd equation:
3x + 5(200 – x) = 840
3x + 1000 – 5x = 840
-2x + 1000 = 840
-2x = 840 – 1000
-2x = -160
x = 80, so y = 200 – 80 = 120
4.2 #66: Mixture Problem A mechanic needs a radiator to
have a 40% antifreeze solution. The radiator currently
is filled with 4 gallons of a 25% antifreeze solution.
How much of the antifreeze mixture should be
drained from the car if the mechanic replaces it with
pure antifreeze?
% Antifreeze x Amount Liquid = Amount Antifreeze
25%
(4 – y) Gallons = 1 - .25y Gallons
100%
y Gallons
= y Gallons
---------------------------------------------------------------------40%
4 Gallons
=
1.6 Gallons
So 1 - .25y + y = 1.6 # combine like y terms
1 + .75y = 1.6 # combine number terms
.75y = 0.6 # solve for y
(3/4)y = 3/5 # Multiply both sides by (4/3)
y = (4*3)/(3*5) = 4/5 = 0.8 Gallons
4.3 #62: Airplane Speed An airplane travels 3000 miles
with the wind in 5 hours and takes 6 hours for the
return trip into the wind. What is the speed of the
wind and the speed of the airplane without any wind?
X
X
6 hrs
X
5 hrs
Wind --- >
X
X
3000 miles
X
Let r = airplane speed; let w = wind speed
So D = r1*t1 = r2*t2 where r1 = r + w, r2 = r – w, t1 = 5, t2 = 6
And (r+w)*5 = (r – w)*6 = 3000
r + w = 600
r – w = 500
--------------2r = 1100
r = 550 mph
w = 50 mph
4.3 #64: Mixing Antifreeze A car radiator holds 2 gallons of
fluid and initially is empty. If a mixture of water and
antifreeze contains 70% antifreeze and another mixture
contains 15% antifreeze, how much of each
should be combined to fill the radiator with a 50%
antifreeze mixture?
% * Amount of liquid = Amount of A/F
.7
x
.7x
.15
2–x
.15(2 – x)
.50
2
1
So .7x + .15(2 – x) = 1
.55x + .3 = 1
.55x = .7
x = .7/.55 = 14/11 70% A/F and 2 – x = 8/11 15% A/F
4.4 #72: Working on Two Projects An employee is required
to spend more time on project X than on project Y.
The employee can work at most 40 hours on these
two projects. Shade the region in the xy-plane that
represents the number of hours that the employee can
spend on each project.
Let X = time on project X; Y = time on project Y
Then X > Y
X + Y ≤ 40
Intercepts are (0, 40), (40, 0) and shaded area is to the NWSE of
X = Y in the first quadrant within the region bounded by the X
And Y axes and the line X + Y = 40 in the first quadrant.
4.4 #74: Target Heart Rate (Refer to the preceding exercise.)
A target heart rate T that is half a person’s maximum
heart rate is given by T = 110 – (1/2)A, where A
is a person’s age. The formula for maximum heart rate is:
R = 220 – A (R units is beats per minute)
(a) What is T for a person 30 years old? 50 years old?
(b) Sketch a graph of the system of inequalities.
T ≥ 110 – (1/2)A
T ≤ 220 - A
Assume that A is between 20 and 60.
(c) Interpret your graph.
(a) Let T = 110 – (1/2)(30) = 95 bpm
T = 110 – (1/2)(50) = 85 bpm
(b) both slopes are negative. Intercepts of 1st inequality (A, T) are: (0, 110), (220, 0)
for 2nd inequality: (A, T) intercepts are: (0, 220), (220, 0)
(c) If 20 < A < 60, then look at the projection of lines sectioned at A = 20, and A = 60
The beats per minute for a person age 20 to 60 is in the range (80 - 100) with
maxima being in the range of (160 - 200)
5.1 #96: Compound Interest If P dollars are deposited in an
account that pays 9% annual interest, then the amount
of money in the account after 4 years is P(1 + 0.09)4
Find the amount when P = $500.
Amount = (500)(1.09)4 = 500(1.411582) = 705.7908 = $705.79
5.5 #94: Average Family Net Worth A family refers to a group
of two or more people related by birth, marriage, or
adoption who reside together. In 2000, the average family
net worth was $280,000, and there were about 7.2 x 107
families. Calculate the total family net worth
in the United States in 2000. (Source: U.S. Census Bureau.)
(a) Total family net worth = (average net worth)*(number of families)
or 280,000*(7.2 x 107 ) = ( 2.8 x 105 ) * ( 7.2 x 107 ) = 20.16 x 1012 =
= 2.016 x 101 x 1012 = 2.016 x 1013