Algebra 1 BMP #5 Practice Problems

Algebra 1 BMP #5 Practice Problems

ABOUT THESE PRACTICE PROBLEMS These practice problems are provided to help you and your students prepare for the upcoming benchmark exam. This set of problems is not designed to mimic the benchmark exam. These are not simply the exam questions with different numerical values. They are, in fact, problems that highlight the concepts that could appear on the exam. There are more practice problems than test questions because we could not possibly test several weeks of teaching in one 45-minute exam. A concept that has multiple applications might be tested only once in the benchmark exam. We, however, have tried to provide practice problems that cover a variety of the concept’s applications in order to better prepare your students for anything they might encounter on the benchmark exam or the CST. Further, in every benchmark exam and in every set of practice problems, we have included material from previous benchmark periods. This spiraling is designed to force students to continuously review everything they’ve learned thus far this year. Some of these practice problems may require the use of a calculator, and some may be short-answer or essay questions rather than multiple-choice. While every question on every benchmark exam is multiple-choice and does not require a calculator, we recognize this is not always the best way of assessing mathematical skills. The open-ended and calculator-active questions are designed to help you assess your students’ work when they do not have the aid of multiple-choice, integer solutions. It may also help your students to know that we are most likely to provide open-ended practice problems on particularly critical skills. Finally, there is no one right way to use these practice problems. Some teachers give them as a practice test. Some teachers assign them as homework problems. Some use them as class work done individually or in groups. Many teachers, however, break these problems into smaller sets and do some combination of the above. A few teachers ignore them entirely, preferring to review with other materials or not to review at all. It is entirely up to you. — Algebra and Geometry Benchmark Teams

Algebra 1 BMP #5 Practice Problems 2

Multiple Choice or Find the Answer as indicated 1. Find the number that must divide each term in the equation so that the equation can be solved by the method of completing the square:

2x2

- 9x = 20 [A]

x [B] 20 [C] –9 [D] 2 2. Solve by completing the square:

2x2

- 4x - 4 = 0 [A]

1 ± 3 [B]

-1 ± 3 [C]

-1 ± 2 3 [D]

1 ± 2 3 3. Name the most efficient method to solve the quadratic equation:

(x + 6)2 = 7.

[A]

Solve for x 2 and use the square root method. [B] The square root method [C] The quadratic formula [D] Factoring and the zero-product principle 4. Solve:

81x2 - 16 = 0

[A]

-81

16,

81

16 [B]

-16

81,

16

81 [C]

-4

9,

4

9 [D]

-9

4,

9

4 5. Solve by the quadratic formula:

x2 =

-5x + 4

[A]

-5 + 41

2,

5 - 41

2 [B]

-5 + 41, - 5 - 41

[C]

5 + 41, 5 - 41 [D]

-5 + 41

2,

-5 - 41

2

6. Evaluate:

25

144

[A]

5

12 [B]

5

144 [C]

5

72 [D]

1

2 Solve: 7.

16x2 - 49 = 0

[A]

-16

49,

16

49 [B]

-49

16,

49

16 [C]

-4

7,

4

7 [D]

-7

4,

7

4


8.

x2

+ 6x + 7 = 0 [A]

3 + 2, 3 - 2 [B]

6 + 2 2, 6 - 2 2 [C]

-6 + 2 2, - 6 - 2 2 [D]

-3 + 2, - 3 - 2 9. The number of new cars purchased in a city can be modeled by the equation

C = 18t2

+ 149t + 6227, where C is the number of new cars and t = 0 corresponds to the number of new cars purchased in 1966. In what year will the number of new cars reach 13,000? [A] 1982 [B] 2032 [C] 1985 [D] 1997 10. Factor:

x2

-10x + 24 [A]

(x + 6)(x - 4) [B]

(x - 6)(x - 4) [C]

(x - 6)(x + 4) [D]

(x + 6)(x + 4) 11. Solve the equation

4x2

+ 7x - 2 = 0.

[A]

1,-1

2 [B]

-2,1

4 [C]

-1,1

2 [D]

2,-2 12. Solve:

6x - 7 = x - 4

[A]

-3

5 [B]

5

3 [C]

-1

6 [D]

3

5 13. State the x- and y-intercepts of

y = - 7x + 5.

[A] x-intercept: 5; y-intercept: –7 [B] x-intercept: 5; y-intercept:

5

7

[C] x-intercept: –7; y-intercept: 5 [D] x-intercept:

5

7; y-intercept: 5

14. The cost of a school banquet is $85 plus $13 for each person attending. Determine the linear equation that models this problem. What is the cost for 57 people? [A] y = 85x – 13; $4832 [B] y = 85x + 13; $4858 [C] y = 13x – 85; $656 [D] y = 13x + 85; $826


15. Write an equation, in point-slope form, of the line that passes through the point (5, 7) and has the slope

3

4. Then rewrite the equation in slope-intercept form.

[A]

y - 7 = 3

4(x - 5); y =

3

4x +

13

4 [B]

y - 5 = 3

4(x - 7); y =

3

4x -

1

4

[C]

y + 7 = 3

4(x + 5); y =

3

4x -

13

4 [D]

y + 5 = 3

4(x + 7); y =

3

4x +

1

4 16. Lev earns $5.35 per hour working after school. He needs at least $290 for a stereo system. Write an inequality that describes how many hours he must work to reach his goal. [A]

290 ! x " 5.35 [B]

5.35 ! x " 290 [C]

5.35 ! x " 290 [D]

x + 5.35 ! 290 x

! 54 hours x

! 55 hours x

! 56 hours x > 55 hours Solve: 17. –8

£ 2x + 10

£ 4 [A] 14

£ x

£ 2 [B] –9

£ x

£ –3 [C] –3

£ x

£ –9 [D] 2

£ x

£ 14 18.

x - 6 = 4 [A] {2} [B] {10, 2} [C] {–2, –10} [D] {–10} 19. Graph: –y <

x - 4

[A]

x

y

-10 10

-10

10

[B]

x

y

-10 10

-10

10

[C]

x

y

-10 10

-10

10

[D]

x

y

-10 10

-10

10

20. Solve by substitution:

3x - 2y = -15

y = -x - 5

[A]

0, 15

2

Ê Ë

ˆ ¯ [B] (–5, 0) [C] (–4, –1) [D] no solution


21. Solve by linear combinations:

2x - 2y = 16

x + 2y = -1

[A] (26, –3) [B] no solution [C]

0, -8( ) [D] (5, –3) 22. Marc sold 423 tickets for the school play. Student tickets cost $2 and adult tickets cost $6. Marc’s sales totaled $1686. How many adult tickets and how many student tickets did Marc sell? [A] 213 adult, 210 student [B] 205 adult, 218 student [C] 218 adult, 205 student [D] 210 adult, 213 student 23. Simplify. Leave your answer in exponential form.

54

¥

510

[A]

2514 [B]

56 [C]

514 [D]

540

24. Simplify the product:

(2 fg3)2( fg)

6 [A]

4 f8g12 [B]

2 f3g12

[C]

2 f8g12

[D]

4 f8g9

25. Simplify:

24x7y2

-6x6y4

[A]

- 4x

y2

[B]

2x13

y6

[C]

- 2x

13

y6

[D]

4x

y2

26. While standing on the edge of a 50 foot cliff, Joe threw a rock up and over the edge of the cliff . The initial velocity of the rock was 80 feet per second and acceleration due to gravity is always a constant of 32

ft/sec2. The height of an object t seconds after launch is given by the equation

h = -1

2gt

2+ v

0t + h

0. Solve

the equation to find how long it took the rock to hit the ground at the bottom of the cliff after the rock was thrown. Estimate your answer to the nearest tenth of a second. [A] 4.7 seconds [B] -.6 seconds [C] 3.2 seconds [D] 5.6 seconds 27. The distance h traveled in t seconds by an object dropped from a height is

h = 16t2. If an object is

dropped from a height of 147 feet, how long will it take before the object hits the ground? Estimate your answer to the nearest whole second. [A] 1 seconds [B] 4 seconds [C] 3. seconds [D] 2 seconds


Graph: 28.

y = - x2

- 2x + 5

[A]

x

y

-10 10

-10

10

[B]

x

y

-10 10

-10

10

[C]

x

y

-10 10

-10

10

[D]

x

y

-10 10

-10

10

Graph: 29.

y = x2

+ 2

[A]

x

y

–10 10

–10

10

[B]

x

y

–10 10

–10

10

[C]

x

y

–10 10

–10

10

[D]

x

y

–10 10

–10

10

30. Solve:

x2

- 3x - 54 = 0 [A] 6, -9 [B] 3, -18 [C] 2, -27 [D] 9, -6 31. Solve the equation

4x2

+ 7x - 2 = 0.

[A] –2, 1 [B]

-2,1

4 [C]

-6,1

4 [D] 4, –2

32. Which of the following values must be added to both sides of the equation x2 + 12x = 15 to complete the square? [A] 6 [B] 36 [C] –15 [D] 2 33. Solve: 2(5 – 3a) = –4(a + 4) 34.Determine the solutions of x2 + 4x – 5 = 0.


35. Determine the roots of 7x2 + 4x = 5. 36. How many roots does the function f(x) = x2 + 12x + 36 have? 37. How many zeros does the function y = 2x2 + x – 6 have? 38. Multiply: (3x – 5)(x – 4)

39. How many pounds of dried pineapple selling at 35¢ a pound should be mixed with dried apricots selling at 70¢ a pound to make a mixture of 100 pounds to sell at 42¢ per pound?


Answers – Algebra 1 BMP#5 Practice Problems

[1] [D] CA 14.0 [2] [A] CA 14.0 [3] [B] CA 14.0 [4] [C] CA 14.0 [5] [D] CA 20.0 [6] [A] [7] [D] CA 14.0 [8] [D] CA 20.0 [9] [A] CA 21.0 [10] [B] 11.0 [11] [B] CA 11.0, CA 14.0 [12] [D] CA 4.0, CA 5.0 [13] [D] CA 6.0 [14] [D] [15] [A] CA 7.0 [16] [B] CA 5.0 [17] [B] CA 5.0 [18] [B] CA 3.0 [19] [D] CA 6.0 [20] [B] CA 9.0 [21] [D] CA 9.0 [22] [D] CA 9.0 [23] [C] CA 2.0 [24] [A] CA 2.0 [25] [A] CA 2.0, CA 13.0 [26] [D] [27] [C] [28] [A] CA 21.0 [29] [A] CA 21.0 [30] [D] CA 11.0, CA 14.0 [31] [B] CA 11.0, CA 14.0 [32] [B] CA 14.0 [33] 13

[34] -1 ± 21

2

[35] -2 ± 39

7 [36] one [37] two [38] 3x

2-17x + 20

[39] 20 pounds of dried pineapple, 80 pounds of dried apricots

Documents

Algebra 1 BMP #5 Practice Problems