© Scarborough
MATH 141 Week In Review 2 KEY
Spring 2013
1
1. A nutritionist wants to create a lunch for a client containing 880 calories and 186g of carbohydrates.
The meal is to be created using two foods: Food I and Food II. Determine the number of units of each
type of food needed to meet the nutritional requirements of this lunch. The nutritional information for
each food is listed in the table below.
Food Calories per unit Grams of carbohydrate per unit
I
35
4
II
20
6
a. Clearly define your variables. Set up a system of equations that can be used to solve the problem.
Let x = number of units of Food I
Let y = number of units of Food II
35 x  20 y  880
num ber of calories
4 x  6 y  186
gram s carbohydrates
b. Solve the system of equations to determine the number of units of each type of food needed to
meet the nutritional requirements of this lunch. Use at least 2 different methods to solve the
system of equations.
 35
rref 
4
20
6
880 
1
 
186 
0
0
1
12 

23 
 35
OR rref 
4
20 880 
1
 
6 186 
0
Therefore 12 units of Food I and 23 units of Food II are needed.
OR
35 x  20 y  880 m ult by 3
4 x  6 y  186
m ult by  10
105 x  60 y  2640
40 x  60 y  1860
65 x
 780
x  12
4  12  6 y  186
6 y  138
y  23
OR
0 12 

1 23 
© Scarborough
MATH 141 Week In Review 2 KEY
1
 35

4
880 

186 
20
6
R1  R 2

row Sw ap ([ A ],1, 2)
4

 35
6
20
186 

880 
4
1

* row  , A ns ,1) 
4


* row  (  35, A ns ,1, 2)
1

0
 1 .5 R 2  R1  R1

* ro w    1 .5, A n s , 2,1 
1

0
1.5
 32.5
0
1
46.5 

 747.5 
32.5
2
R1  R1

1
 35 R1  R 2  R 2
Spring 2013
1

 35
1.5
20
46.5 

880 
R2  R2

 1

* row 
, A ns , 2 
 32.5

1

0
1.5
1
46.5 

23 
12 

23
FORESHADOWING (will be covered week 3):
 35
X  A B 
4
1
20 

6 
1
 880   12 

 
186   23 
2. Which augmented matrices are in row-reduced form, where a, b, and c are any real numbers?
a.
1

0
0

0
0
0
1
1
0
3 a

4 b
5 c 
No, since in any two successive, nonzero rows, the leading 1 in the lower row should lie to the
right of the leading 1 in the upper row.
b.
1

0
0

0
0
1
0
0
1
0
0
1
0
0
2
3 a

4 b
5 c 
Yes.
c.
1

0
0

3 a

4 b
5 c 
No, since the first nonzero entry in each nonzero row should be a 1.
© Scarborough
d.
1

0
0

0
0
1
1
0
1
MATH 141 Week In Review 2 KEY
Spring 2013
3
3 a

4 b
5 c 
No, since if a column in the coefficient matrix contains a leading 1, then the other entries in that
column should be zeros.
e.
1

0
0

0
0
1
0
0
0
0
0
0
0
1
0
3 a

4 b
0 0 
Yes.
f.
1

0
0

3 a

0 0
5 c 
No, since each row consisting entirely of zeros should lie below all rows having nonzero entries.
3. A clown sells balloons and toy ducks. One balloon sells for $0.75 and one toy duck sells for $1.25. If
three times as many balloons were sold as toy ducks with a total daily revenue of $84, how many of
each were sold.
a. Clearly define your variables. Set up a system of equations that can be used to solve the problem.
Let x = number of balloons
y = number of toy ducks
$ price
ratio
0.75x + 1.25y = 84
x = 3y
or
x – 3y = 0
b. Solve the system to determine the number of balloons and toy ducks sold. Use at least 2 different
methods to solve the system of equations.
 0.75
rref 
 1
1.25
3
84 
1
 
0 
0
0
1
72 

24 
 0.75
OR rref 
 1
Therefore 72 balloons and 24 toy ducks were sold.
OR
0.75x + 1.25y = 84
x = 3y
0.75 (3y) + 1.25y = 84
3.5y = 84
y = 24
x = 3(24)
x = 72
1.25 84 
1
 
3 0 
0
0 72 

1 24 
© Scarborough
MATH 141 Week In Review 2 KEY
Spring 2013
4
OR
 0.75

 1
84 

0 
1.25
3
R1  R 2

row Sw ap ([ A ],1, 2)
1
1

0
3
3.5
0 

84 
3.5
3
 1

 0.75
1.25
R2  R2

 1

* row 
, A ns , 2) 
 3.5

1

0
3
1
0 

84 
0 

24 
 0.75 R1  R 2  R 2

* row  (  0.75, A ns ,1, 2)
3 R 2  R1  R1

* row  (3, A ns , 2,1)
1

0
0
1
72 

24 
FORESHADOWING (will be covered week 3):
 0.75
X  A B 
 1
1
1.25 

3 
1
 84   72 
  
 0   24 
4. Solve the system of equations. If the system has an infinite number of solutions, write the solution set in
parametric form. If the system of equations has an infinite number of solutions, give two particular
solutions.
5 x  25  12.5 y
0.2 x  0.5 y  1  0
5 x  12 . 5 y  25
0 .2 x  0 .5 y  1
 5
rref 
 0.2
 12.5
 0.5
25 
1
 
1 
0
 2.5
0
5

0
 5
OR rref 
 0.2
 12.5 25 
1
 
 0.5 1 
0
 2.5 5 

0 0
Now we have x  2 .5 y  5 so x  2 .5 y  5
The solution is  2.5 t  5, t  , where t is any real number. A couple of particular solutions are
(5, 0) and (7.5, 1).
FORESHADOWING (will be covered week 3): Using a matrix equation, X  A  1 B to solve this
 5
system of equations will not work since 
 0.2
matrices and thus have no inverses.
 12.5 

 0.5 
, and consequently A  1 B , are singular
© Scarborough
MATH 141 Week In Review 2 KEY
Spring 2013
5
5. Grape juice is sold for $4 per gallon and pomegranate juice is sold for $20 per gallon. If a fourth as
many gallons of grape juice are needed as gallons of pomegranate juice, how many gallons of each juice
should be sold to have $3066 in revenue?
a. Clearly define your variables. Set up a system of equations that can be used to solve the problem.
Let x = the number of gallons of grape juice
Let y = the number of gallons of pomegranate juice
$ revenue
4x + 20y =3066
ratio
4x = y
(or x 
1
y
4
or 4x – y = 0)
b. Solve the system of equations to determine the number of gallons sold of each type of juice. Use
at least 2 different methods to solve the system of equations.
4 x  20  4 x   3066
84 x  3066
x  36.5
y  4  36.5 
y  146
Therefore 36.5 gallons of grape juice and 146 gallons of pomegranate juice should be sold.
OR
4 x  20 y  3066
4x – y  0
m ult by  1
4 x  20 y  3066
4x  y  0
21 y  3066
y  146
x  36.5
OR
4
rref 
4
OR
20
1
3066 
1
 
0 
0
0
1
36.5 

146 
4
OR rref 
4
20 3066 
1
 
1 0 
0
0 36.5 

1 146 
© Scarborough
MATH 141 Week In Review 2 KEY
1
4

4
20
1
R R
3066  4 1 1  1



0  * row  1 ,[ A ],1   4
4

1
R2
1

 1

* row 
, A ns , 2   0
21
21


5
1

766.5   4 R 1  R 2  R 2  1



0  * row  (  4 , A ns ,1, 2 )  0
766.5   5 R 2  R1  R1  1



146  * row  (  5 , A ns , 2 ,1)  0
5
1

0
1
5
 21
Spring 2013
6
766.5 

 3066 
36.5 

146 
FORESHADOWING (will be covered week 3):
4
X  A B 
4
1
1

6. The matrix  0
0

0
2
0
20 

 1
1
 3066   36.5 



 0   146 
0 1

4 4  represents a system of equations.
1 0 
a. What is the next Gauss-Jordan row operation needed? Give the matrix that would result from
performing this Gauss-Jordan row operation.
1

0
0

0
2
0
1
0 1 1
 R2  R2 
4 4 2
0

0
1 0 

0
1
0
0 1

2 2
1 0 
b. After performing the Gauss-Jordan row operation needed in part a, what is the next and final
Gauss-Jordan row operation?
 2 R3  R 2  R 2
c. Give the solution, if it exists, as a point which represents the solution to this system of equations.
If the solution does not exist, write “no solution.”
1

0
0

0
2
0
1
0 1 1
 R2  R2 
4 4 2
0

0
1 0 

The solution is (1, 2, 0).
0
1
0
1
0 1
  2 R3  R 2  R 2 
2 2
0

0
1 0 

0
1
0
0 1

0 2
1 0 
© Scarborough
7. The matrix
MATH 141 Week In Review 2 KEY
2
1

0

 0
3 

1 4

0 0 
Spring 2013
7
0
0
0
represents a system of equations. Give four particular solutions to this
system of equations.
Since x  2 y  3 and z   4 , the solutions are of the form  2 t  3, t ,  4  where t is any real number.
Some particular solutions are  3, 0,  4  ,  5,1,  4  ,  7, 2,  4  , and 1,  1,  4  .
8. An investor will invest all of $56,000 in stocks. The investor estimated that the high-risk stock will have
a rate of return of 18% per year; the medium-risk stocks will have a rate of return of 10% per year, and
the low-risk stock will have a rate of return of 4% per year. The investment in the medium-risk stock is
to be three times the sum of the investments in the other two stock categories. If the investment goal is
to have a return of $5880 per year on the total investment, determine how much the investor should
invest in each type of stock. How much did the investor invest in high- and medium-risk stocks? How
much return did the investor get on the low- and medium risk stocks?
Let x = $ invested in high-risk stock
Let y = $ invested in medium-risk stock
Let z = $ invested in low-risk stock
x  y  z  56000
0.18 x  0.10 y  0.04 z  5880
3( x  z )  y
$ of money invested
$ of return on money
ratio stipulation
x  y  z  56000
0.18 x  0.10 y  0.04 z  5880
3x  y  3z  0
 1

rref  0.18
 3

1
0.10
1
56000 
1


0.04 5880    0
0
3
0 

1
0
1
0
0 8000 

0 42000 
1 6000 
Therefore the investor should invest $8000 in the high-risk stock, $42,000 in the medium-risk
stock, and $6000 in the low-risk stock. The investor invested $8000 + $42,000 in high- and
medium-risk stocks. The investor’s return on low- and medium stock was
0.04  6000   0.10  42000   $4440.