Download Alg2 - CH3 Practice Test −5x + 4y = 6 3x − y = 2 П М У ФФФФФ ФФФФ

Alg2 - CH3 Practice Test
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
ÏÔ −5x + 4y = 6
Ô
1. Use a graph to solve the system ÔÔÌ
. Check your answer.
ÔÓ 3x − y = 2
a.
c.
The solution to the system is (2, 4).
b.
____
The solution to the system is (–2, 4).
d.
The solution to the system is (–2, –4).
The solution to the system is (2, –4).
2. Two snow resorts offers private lessons to their customers. Big Time Ski Mountain charges $5 per
hour plus $50 insurance. Powder Hills charges $30 per hour plus $10 insurance. For what number
of hours is the cost of lessons the same for each resort?
a. 3 hours
c. 5 hours
b. 4 hours
d. 6 hours
____
____
____
____
3. Jake fills a tank that can hold 200 gallons of water. The tank already has 50 gallons of water in it
when Jake starts filling it at the rate of 10 gallons per minute. Karla fills a tank that can hold 300
gallons of water. That tank already has 100 gallons of water in it when Karla starts filling it at the
rate of 5 gallons per minute. Jake and Karla start filling the tanks at the same time. How long after
they start filling the tanks do the tanks have the same volume of water? What is that volume of
water?
a. 5 minutes; 150 gallons
c. 10 minutes; 150 gallons
b. 5 minutes; 250 gallons
d. 10 minutes; 250 gallons
ÔÏÔ
Ô 3x − 3y = 9
4. Use elimination to solve the system ÌÔ
.
ÔÔ x + 3y = 7
Ó
a. (1, 4)
c. (3, 0)
b. (0, –3)
d. (4, 1)
5. A zookeeper needs to mix a solution for baby penguins so it has the right amount of medicine.
Solution A has 20% medicine. Solution B has 4% medicine. How many ounces of each solution is
needed to obtain 10 ounces of 8% medicine?
a. 10 ounces of A and 20 ounces of B
c. 2 ounces of A and 10 ounces of B
3
3
b. 4 ounces of A and 6 ounces of B
d. 2.5 ounces of A and 7.5 ounces of B
6. Mina’s Catering Service is organizing a formal dinner for 280 people. The hall has two kinds of
tables, one that seats 4 people and one that seats 10 people. The hall can contain up to a total of 52
tables. Write and graph a system of inequalities that can be used to determine the possible
combinations of tables that can be used for the event so there are enough seats for all the people.
a. ÏÔÔ x + y ≤ 52
c. ÏÔÔ x + y ≤ 52
Ô
ÔÌÔ
Ó 4x + 10y ≥ 280
b. ÏÔÔ x + y ≤ 52
Ô
ÌÔ
ÔÓ 4x + 10y ≤ 280
Ô
ÔÌÔ
Ó 4x + 10y = 280
d. ÏÔÔ x + y ≥ 52
Ô
ÌÔ
ÔÓ 4x + 10y ≤ 280
____
7. Graph the system of inequalities, and classify the figure created by the solution region.
ÏÔ y ≤ 4x + 4
ÔÔ
ÔÔ
ÔÔ y ≤ −0.25x + 4
ÌÔ
ÔÔ y ≥ 4x − 1
ÔÔ
ÔÔ y ≥ −0.25x − 1
Ó
a.
The shaded region is a plane minus a rectangle.
b.
There is no region common to all four inequalities.
c.
The shaded region is a rectangle.
d.
____
The region is the entire plane.
8. Which point gives the minimum value of P = 3x − 2y in the feasible region shown?
____
a. S(4, 1)
c. T(2, 0)
b. R(1, 4)
d. U(0, 1)
9. Graph (–2, 3, 1) in three-dimensional space.
a.
c.
b.
____
d.
10. A teacher prepares 3 different tests. The teacher uses 3 types of questions which are each worth a
certain number of points. The table shows the number of questions of each type on each of the
three tests. Find the number of points each type of question is worth.
Question Question Question
Total
Test 1
Test 2
Test 3
Type A
36
3
0
Type B
3
45
1
Type C
0
1
39
Points
150
103
41
a. Question type A is worth 4 points, type B is worth 2 points, and type C is worth 1
point.
b. Question type A is worth 1 point, type B is worth 2 points, and type C is worth 4
points.
c. Question type A is worth 4 points, type B is worth 48 points, and type C is worth 1
point.
d. There is no solution to this problem.
____
ÏÔ x − y + z = 6
ÔÔ
Ô
11. Classify the system ÔÌÔ 7x − 2y − 2z = 1 as consistent or inconsistent, and determine the number of
ÔÔ
ÔÔ 5x + 6y + 3z = 11
Ó
____
solutions.
a. Consistent. One solution.
b. Inconsistent. Infinitely many solutions.
c. Inconsistent. No solutions.
d. Consistent. Infinitely many solutions.
12. A sprinter is running up a hill. Using parametric equations, the horizontal distance traveled by the
sprinter after t seconds can be modeled by x = 6t feet, and the vertical distance traveled by the
sprinter can be modeled by y = 2t feet. Use the parametric equations to write an equation for y in
terms of x.
a. y = 6x
c. y = 1 x
b. y = 3x
d. y =
6
1
3
x
Numeric Response
13. Manny works 40 hours per week. He must work for his parents where he earns $8 per hour. He
also works for a computer company where he earns $20 per hour. What is the minimum number of
hours Manny can work for the computer company to earn a total of $464 per week from both jobs?
14. Find the z-intercept of the graph of −2x − 12y + 5z = 3. Express your answer as a decimal.
Matching
Match each vocabulary term with its definition.
a. linear system
b. system of equations
c. consistent system
d. dependent system
e.
f.
g.
h.
i.
____
____
____
____
____
____
15.
16.
17.
18.
19.
20.
inconsistent system
independent system
system of linear inequalities
complex system
standard system
a system of equations containing only linear equations
a system of equations or inequalities that has no solution
a set of two or more equations containing two or more variables
a system of inequalities in two or more variables in which all of the inequalities are linear
a system of equations that has exactly one solution
a system of equations or inequalities that has at least one solution
Match each vocabulary term with its definition.
a. objective function
b. ordered triple
c. parameter
d. parametric equations
e. z-coordinate
f. z-plane
g. three-dimensional coordinate system
h. z-axis
____
____
____
____
____
____
21.
22.
23.
24.
25.
26.
the third axis in a three-dimensional coordinate system
a space that is divided into eight regions by an x-axis, a y-axis, and a z-axis
a set of three numbers that can be used to locate a point in a three-dimensional coordinate system
one of the constants in a function or equation that may be changed
a pair of equations that define the x- and y-coordinates of a point in terms of a third variable
the function to be maximized or minimized in a linear programming problem
Alg2 - CH3 Practice Test
Answer Section
MULTIPLE CHOICE
1. ANS: A
Solve each equation for y.
ÔÏÔ y = 5 x + 3
4
2
ÔÌ
ÔÔ
ÔÓ y = 3x − 2
Then graph each equation.The lines appear to intersect at the point (2, 4). Check by substituting the
x- and y-values into each equation.
Feedback
A
B
C
D
Correct!
Solve each equation for y. Then graph the two lines.
Solve each equation for y. Are the slopes positive or negative?
Solve each equation for y. Then graph the two lines.
PTS: 1
DIF: Average
REF: Page 183
OBJ: 3-1.2 Solving Linear Systems by Using Graphs and Tables
NAT: 12.5.4.g
TOP: 3-1 Using Graphs and Tables to Solve Linear Systems
2. ANS: B
Write an equation for the cost of lessons for each company.
Let x equal the number of hours of the lesson and y equal the cost of insurance.
Big Time Ski
Mountain:
Powder Hills:
y = 5x + 50
y = 30x + 10
y = 5x + 50
x
1
2
3
4
5
y = 30x + 10
y
55
60
65
70
75
x
1
2
3
4
5
y
40
50
60
70
80
When x = 4, the y-values are both 70.
The cost of a private lesson for 4 hours is $70 at either resort.
So the cost is the same for a 4-hour lesson at each resort.
Feedback
A
Set up a table to solve the problem. At this price Powder Hills is cheaper.
B
C
D
Correct!
Set up a table to solve the problem. At this price, Big Time Ski Mountain is
cheaper.
Set up a table to solve the problem. At this price, Big Time Ski Mountain is
cheaper.
PTS: 1
DIF: Average
REF: Page 185
OBJ: 3-1.4 Application
NAT: 12.5.4.g
TOP: 3-1 Using Graphs and Tables to Solve Linear Systems
3. ANS: C
Step 1: Write two equations.
Let V1 represent the volume of water in the tank Jake fills, and let V2 represent the volume of water
in the tank Karla fills. Let t represent the time after they start filling the tanks.
V1 = 50 + 10t
V2 = 100 + 5t
Step2: Solve for the value of t when V1 = V2.
50 + 10t = 100 + 5t
5t = 50
t = 10
Step3: Find the volume of water in both tanks when t = 10.
50 + 10t = 100 + 5t
50 + 10(10) = 100 + 5(10) = 150
Feedback
A
B
C
D
In 5 minutes, Jake's tank holds 250 gallons of water and Karla's tank holds 125
gallons of water.
In 5 minutes, Jake's tank holds 250 gallons of water and Karla's tank holds 125
gallons of water.
Correct!
In 10 minutes, both tanks hold 150 gallons of water.
PTS:
TOP:
4. ANS:
Step 1
1
DIF: Advanced
NAT: 12.5.4.g
3-1 Using Graphs and Tables to Solve Linear Systems
KEY: multi-step
D
3x
– 3y = 9
x
+ 3y = 7
The y-terms have opposite coefficients.
4x
= 16
Add the equations to eliminate the y terms.
x
= 4
Step 2 3(4) – 3y =
12 – 3y = 9
9
Substitute for x in one of the original equations.
Simplify and solve for y.
– 3y = –3
y =1
(4, 1)
Write the solution as an ordered pair.
Feedback
A
B
C
D
You switched the x- and y-coordinates.
Add the equations to eliminate the variable, not subtract.
This is a solution of the first equation, but it is not a solution of the second
equation. Use elimination to find a solution of both equations.
Correct!
PTS: 1
DIF: Basic
REF: Page 191
OBJ: 3-2.2 Solving Linear Systems by Elimination
NAT: 12.5.4.g
TOP: 3-2 Using Algebraic Methods to Solve Linear Systems
KEY: linear equations | system of equations | solving | elimination
5. ANS: D
Begin by writing an equation for the volume needed and an equation for the amount of medicine
needed.
Let a = volume of solution A.
Let b = volume of solution B.
The equation for the volume of solution: a + b = 10 or a = 10 − b
The equation for the amount of medicine: 0.20a + 0.04b = 0.08(10)
Solve the system by substituting the first equation into the second.
0.20(10 − b) + 0.04b = 0.08(10) Substitute (10 − b) for a.
Distribute.
2 − 0.20b + 0.04b = 0.8
Simplify.
−0.16b = −1.2
b = 7.5
a = 10 − 7.5
a = 2.5
Substitute b in the original equation.
The mixture will contain 2.5 ounces of solution A and 7.5 ounces of solution B.
Feedback
A
B
C
D
Write two equations: one for volume and one for solution concentration.
The total percent of medicine should be 8%.
The total volume is 10 ounces.
Correct!
PTS: 1
NAT: 12.5.4.g
6. ANS: A
DIF: Average
REF: Page 193
OBJ: 3-2.4 Application
TOP: 3-2 Using Algebraic Methods to Solve Linear Systems
Let x represent the number of tables of 4 and let y represent the number of tables of 10.
The total number of tables can be modeled by the inequality x + y ≤ 52. The number of seats
available for the invitees can be modeled by the inequality 4x + 10y ≥ 280.
ÔÏÔ x + y ≤ 52
The system of inequalities is ÔÌÔ
.
ÔÓ 4x + 10y ≥ 280
Graph the solid boundary line x + y = 52 and shade below it. Graph the solid boundary line
4x + 10y = 280 and shade above it. The solution region is the overlapping region.
Feedback
A
B
C
D
Correct!
Can the number of seats available be less than 280?
Can't the number of seats available be greater than 280?
Can the number of tables be greater than 52? Can the number of seats available
be less than 280?
PTS: 1
DIF: Average
REF: Page 200
OBJ: 3-3.2 Application
NAT: 12.5.4.g
TOP: 3-3 Solving Systems of Linear Inequalities
7. ANS: C
Graph the solid boundary lines y = −0.25x + 4 and y = 4x + 4, and shade below them.
Graph the solid boundary lines y = −0.25x − 1 and y = 4x − 1, and shade above them.
The solution region is a four-sided figure, or quadrilateral.
The boundary lines y = 4x + 4 and y = 4x − 1 have the same slope and are parallel.
The boundary lines y = −0.25x − 1 and y = −0.25x + 4 have the same slope and are parallel.
The slope of y = 4x + 4 and y = 4x − 1 and the slope of y = −0.25x − 1 and y = −0.25x + 4 are
opposite reciprocals. Thus the two sets of boundary lines are perpendicular.
The solution region is a rectangle.
Feedback
A
B
C
D
Shade below the solid boundary lines y = -0.25x + 4 and y = 4x + 4. Shade above
the solid boundary lines y = -0.25x – 1 and y = 4x – 1.
Shade below the solid boundary lines y = -0.25x + 4 and y = 4x + 4. Shade above
the solid boundary lines y = -0.25x – 1 and y = 4x – 1.
Correct!
The solution region is the set of points that satisfy all four inequalities.
PTS: 1
DIF: Average
REF: Page 201
OBJ: 3-3.3 Application
NAT: 12.5.4.g
TOP: 3-3 Solving Systems of Linear Inequalities
8. ANS: B
The points representing the intersections of the lines are R, S, T, and U. Substitute the coordinates
of these points into the function P = 3x − 2y.
P(R(1, 4)) → P = 3(1) − 2(4) = −5
P(S(4, 1)) → P = 3(4) − 2(1) = 10
P(T(2, 0)) → P = 3(2) − 2(0) = 6
P(U(0, 1)) → P = 3(0) − 2(1) = −2
The point R(1, 4) gives the minimum value of P.
Feedback
A
B
C
D
To minimize P, find the point whose coordinates give the minimum value of P.
Correct!
To minimize P, find the point whose coordinates give the minimum value of P.
To minimize P, find the point whose coordinates give the minimum value of P.
PTS: 1
DIF: Advanced
TOP: 3-4 Linear Programming
9. ANS: A
Start from the origin. Move 2 units back along the x-axis, 3 units right, and 1 unit up.
Feedback
A
B
C
D
Correct!
The x-value of the point is negative. Move 2 units back along the x-axis.
This graph shows the point (–3, 3, 0). Move forward 1 unit and up 1 unit.
This graph shows the point (3, –2, 1). The x- and y-values are reversed.
PTS: 1
DIF: Basic
REF: Page 214
OBJ: 3-5.1 Graphing Points in Three Dimensions
TOP: 3-5 Linear Equations in Three Dimensions
10. ANS: A
Step 1 Let x represent the number of points for a question type A, y for a question type B, and z for
a question type C.
ÏÔ 36x + 3y + 0z = 150
ÔÔ
ÔÔ
ÔÌÔ 3x + 45y + z = 103
ÔÔ
ÔÓ 0x + y + 39z = 41
(1)
Test 1 points
(2)
Test 2 points
(3)
Test 3 points
Step 2 Use substitution. Solve for y in equation (1).
36x + 3y = 150
(1)
Solve for y.
y = −12x + 50
Step 3 Substitute for y in equation (2) and (3).
ÔÏÔ 3x + 45( −12x + 50 ) + z = 103
(2)
ÔÌ
Substitute −12x + 50.
ÔÔ
(3)
−12x + 50 + 39z = 41
Ó
ÔÏÔ −537x + z = −2147
ÔÌ
ÔÔ
Ó −12x + 39z = −9
(4)
Simplify to find a 2-by-2 system.
(5)
Step 4 Solve equation (4) for z.
(4)
−537x + z = −2147
z = 537x − 2147
Solve for z.
Step 5 Substitute for z in equation (5).
−12x + 39( 537x − 2147 ) = −9 Substitute 537x − 2147 for z.
(5)
20, 931x = 83, 724
Solve for x.
x=4
Step 6 Substitute for x to solve for z and then for y.
(4)
(3) y + 39z = 41
z = 537x − 2147
z = 537( 4 ) − 2147
y + 39( 1 ) = 41
y=2
z=1
The solution to the system is (4, 2, 1). So, type A question is worth 4 points, type B is worth 2
points, and type C is worth 1 point.
Feedback
A
B
C
D
Correct!
You reversed the points values for type A question and type C question.
You substituted a wrong value in one of the equations to solve for the points for
question type B. Check your answer.
Use substitution to solve a 3-by-3 system.
PTS: 1
DIF: Average
REF: Page 222
TOP: 3-6 Solving Linear Systems in Three Variables
11. ANS: A
Step 1 Choose a variable to eliminate.
2ÊÁË x − y + z = 6 ˆ˜¯
7x − 2y − 2z = 1
−3ÊÁË x − y + z = 6 ˆ˜¯
5x + 6y + 3z = 11
OBJ: 3-6.2 Application
2x − 2y + 2z = 12 Eliminate z.
Multiply equation 1 by 2.
7x − 2y − 2z = 1 Add equation 1 to equation 2.
9x − 4y = 13
−3x + 3y − 3z = −18 Multiply equation 1 by –3.
5x + 6y + 3z = 11
Add to equation 3.
2x + 9y = −7
ÔÏÔ 9x − 4y = 13
Ô
ÌÔ
ÔÓ 2x + 9y = −7
Write the results of each part to
get a 2-by-2 system.
Step 2 Eliminate x using multiplication and addition. Then solve for y.
2ÊÁË 9x − 4y = 13 ˆ˜¯
18x − 8y = 26
−9ÊÁË 2x + 9y = −7 ˆ˜¯ 
→
−18x − 81y = 63
−89y = 89
y = −1
Step 3 Solve for x using the value for y.
9x − 4y = 13 Use the first equation in the 2-by-2 system to solve for x.
9x − 4( −1 ) = 13 Substitute –1 for y.
x = 1 Solve for x.
Step 4: Substitute for x and y in equation 1, and solve for z.
x − y + z = 6 Substitute 1 for x and 1 for y.
( 1 ) − − 1 + z = 6 Solve for z.
z=4
The solution is (1, –1, 4).
Feedback
A
B
C
D
Correct!
Obtain a 2-by-2 system using elimination. Solve the 2-by-2 system and
substitute the values into one of the original equations to solve for the third
variable.
Obtain a 2-by-2 system using elimination. Solve the 2-by-2 system and
substitute the values into one of the original equations to solve for the third
variable.
Obtain a 2-by-2 system using elimination. Solve the 2-by-2 system and
substitute the values into one of the original equations to solve for the third
variable.
PTS: 1
DIF: Average
REF: Page 223
OBJ: 3-6.3 Classifying Systems with Infinitely Many Solutions or No Solution
TOP: 3-6 Solving Linear Systems in Three Variables
12. ANS: D
Solve for t in terms of x.,
x = 6t
1
6
x=t
Substitute for t in the other equation to get y in terms of x.
y = 2t
y = 2ÊÁË 16 x ˆ˜¯
y = 13 x
Feedback
A
B
C
D
Solve one of the parametric equations for t, then substitute into the other
equation.
Solve one of the parametric equations for t, then substitute into the other
equation.
Solve one of the parametric equations for t, then substitute into the other
equation.
Correct!
PTS: 1
DIF: Average
REF: Page 231
OBJ: 3-Ext.2 Writing Functions Based on Parametric Equations
TOP: 3-Ext Parametric Equations
NUMERIC RESPONSE
13. ANS: 12
PTS: 1
DIF: Advanced
NAT: 12.5.4.g
TOP: 3-3 Solving Systems of Linear Inequalities
14. ANS: 0.6
PTS:
1
DIF:
Average
TOP: 3-5 Linear Equations in Three Dimensions
MATCHING
15. ANS:
TOP:
16. ANS:
TOP:
17. ANS:
TOP:
18. ANS:
TOP:
19. ANS:
TOP:
20. ANS:
TOP:
A
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
E
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
B
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
G
PTS: 1
DIF: Basic
3-3 Solving Systems of Linear Inequalities
F
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
C
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
REF: Page 182
21. ANS:
TOP:
22. ANS:
TOP:
23. ANS:
TOP:
H
PTS: 1
DIF: Basic
3-5 Linear Equations in Three Dimensions
G
PTS: 1
DIF: Basic
3-5 Linear Equations in Three Dimensions
B
PTS: 1
DIF: Basic
3-5 Linear Equations in Three Dimensions
REF: Page 214
REF: Page 18
REF: Page 182
REF: Page 199
REF: Page 184
REF: Page 183
REF: Page 214
REF: Page 214
24. ANS:
TOP:
25. ANS:
TOP:
26. ANS:
TOP:
C
PTS: 1
3-Ext Parametric Equations
D
PTS: 1
3-Ext Parametric Equations
A
PTS: 1
3-4 Linear Programming
DIF:
Basic
REF: Page 230
DIF:
Basic
REF: Page 230
DIF:
Basic
REF: Page 206