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This is a sample test for the elementary algebra diagnostic test at the Harbor College. The solution and a link to the review of the topics are provided.

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Presentation on theme: "This is a sample test for the elementary algebra diagnostic test at the Harbor College. The solution and a link to the review of the topics are provided."— Presentation transcript:

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2 This is a sample test for the elementary algebra diagnostic test at the Harbor College. The solution and a link to the review of the topics are provided for each question. The review–links of all the topics may be accessed at herehere This is an interactive PowerPoint file. For the best viewing, download it and run it in the viewing–mode. http://www.lahc.edu/math/algebra/ele-alg- place-review.html

3 A B D Question 1. Answer for Question 1: A bag contained 48 pieces of candies. Joe took 2/3 of the bag, Mary took 3/4 of what was left, and Chuck got the rest. How many pieces of candies did Chuck get? 8 6 4 3 C

4 Solution to Q1 The correct answer is 4. The statement “(fraction or %) of an amount” is translated to the multiplication operation.

5 Solution to Q1 1 4 The correct answer is 4. The statement “(fraction or %) of an amount” is translated to the multiplication operation. Joe took ¾ of the bag so ¼ of the bag was left, i.e.(48) = 12 pieces were left. 12

6 Link to review Next Question Solution to Q1 1 4 The correct answer is 4. The statement “(fraction or %) of an amount” is translated to the multiplication operation. Joe took ¾ of the bag so ¼ of the bag was left, i.e.(48) = 12 pieces were left. 12 Mary took 2/3 so Chuck got 1/3 of 12, 1 3 i.e.(12) = 4 pieces. 4

7 Correct! Next Question A B Question 1. 8 6 C D 4 3 A bag contained 48 pieces of candies. Joe took 2/3 of the bag, Mary took 3/4 of what was left, and Chuck got the rest. How many pieces of candies did Chuck get?

8 A C B D Question 2. Answer for Question 2: 2 3 1 3 = 2 3 13 –1 1 3 2 3 –2 1 3

9 Solution to Q2 The correct answer is. 2 3 –1 2 3 1 3 = 13 2 3 1 3 1 3

10 Link to review Next Question Solution to Q2 The correct answer is. 2 3 –1 2 3 1 3 = 13 2 3 1 3 1 3 = 2 3 1 3 2 = 2 3 1 hence 2 3 1 3 13= 2 3 –1

11 A C B D Question 2. 2 3 1 3 = 2 3 13 –1 1 3 2 3 –2 1 3 Correct! Next Question

12 A B C D Answer for Question 3: Question 3. After Joe got a 8% raise he earns $1,350 per week. How much was his weekly wage before the raise? $1,200 $1,250 $1,225 $1,175

13 Solution to Q3 Suppose Joe’s wage was x before the raise, then a raise of 8% is 0.08x. The answer is $1,250.

14 Solution to Q3 Suppose Joe’s wage was x before the raise, then a raise of 8% is 0.08x. Hence his new wage is x + 0.08x or 1.08x, which is $1,350. The answer is $1,250.

15 Link to review Next Question Solution to Q3 Suppose Joe’s wage was x before the raise, then a raise of 8% is 0.08x. Hence his new wage is x + 0.08x or 1.08x, which is $1,350. The answer is $1,250. Therefore 1.08x = 1350 so that x = 1350 1.08 = 1250.

16 Correct! Question 3. After Joe got a 8% raise he earns $1,350 per week. How much was his weekly wage before the raise? A B C D $1,200 $1,250 $1,225 $1,175 Next Question

17 A B C D Question 4. Answer for Question 4: 4,000 x (7.5 x 10 –9 ) x (8 x 10 3 ) = 240. 2.4 0.24 24

18 Solution to Q4 The answer is 0.24. 4,000 x (7.5 x 10 –9 ) x (8 x 10 3 ) = 4 x 10 3 x (7.5 x 10 –9 ) x (8 x 10 3 )

19 Next Question Solution to Q4 The answer is 0.24. Link to review 4,000 x (7.5 x 10 –9 ) x (8 x 10 3 ) = 4 x 10 3 x (7.5 x 10 –9 ) x (8 x 10 3 ) = 4 x 7.5 x 8 x 10 3 x 10 –9 x 10 3

20 Next Question Solution to Q4 The answer is 0.24. Link to review 4,000 x (7.5 x 10 –9 ) x (8 x 10 3 ) = 4 x 10 3 x (7.5 x 10 –9 ) x (8 x 10 3 ) = 4 x 7.5 x 8 x 10 3 x 10 –9 x 10 3 = 240 x 10 -3 = 0.24

21 Correct! A B C D 240. 2.4 0.24 24 Question 4. 4,000 x (7.5 x 10 –9 ) x (8 x 10 3 ) = Next Question

22 A B C D Answer for Question 5: Question 5. Given the right triangle as shown with c = 12, a = 5, which is the closest approximated value of x? 12 11 9 10

23 Solution to Q5 The answer is 11. By the Pythagorean Theorem it must be that x 2 + 5 2 = 12 2

24 Solution to Q5 Link to review Next Question Solution to Q5 The answer is 11. By the Pythagorean Theorem it must be that x 2 + 5 2 = 12 2 x 2 + 25 = 144 x 2 = 119 or that x = √119 ≈ 11

25 Correct! Question 5. Given the right triangle as shown with c = 12, a = 5, which is the closest approximated value of x? A B C D 12 11 9 10 Next Question

26 Question 6. Which of the following percentages is closest to 2/7? D B C A 30% Answer for Question 6: 25% 20% 15%

27 Link to review Next Question Solution to Q6 The answer is 30%. 2/7 = 0.285.. ≈ 30%

28 Correct! D B C A 30% 25% 20% 15% Question 6. Which of the following percentages is closest to 2/7? Next Question

29 7.2 –0.9 10 –4.8 A D C B Question 7. 4.5 x 0.2 – (3.5 – 2 x 0.85) = Answer for Question 7:

30 Solution to Q7 The answer is 7.2. 4.5 x 0.2 – (3.5 – 2 x 0.85) = 0.9 – (3.5 – 1.7)

31 Link to review Next Question Solution to Q7 The answer is 7.2. 4.5 x 0.2 – (3.5 – 2 x 0.85) = 0.9 – (3.5 – 1.7) = 0.9 – 1.8 = –0.9

32 A D C B Correct! Question 7. 4.5 x 0.2 – (3.5 – 2 x 0.85) = 7.2 –0.9 10 –4.8 Next Question

33 7 13 11 9 A B D C Question 8. Answer for Question 8:

34 Link to review Next Question Solution to Q8 The answer is 9.

35 Solution to Q8 The answer is 9.

36 Correct! Question 8. A B D C 7 13 11 9 Next Question

37 3/2 –3/2 1/2 –1/2 A B C D Question 9. If x = 3, and that xy – x – y = –4, what is y? Answer for Question 9:

38 Solution to Q9 The answer is -1/2. If x = 3, then xy – x – y = –4 is 3y – 3 – y = –4.

39 Link to review Next Question Solution to Q9 The answer is -1/2. If x = 3, then xy – x – y = –4 is 3y – 3 – y = –4 so that 2y = –1 or y = –1/2

40 Correct! Question 9. If x = 3, and that xy – x – y = –4, what is y? 3/2 –3/2 1/2 –1/2 A B C D Next Question

41 0 D B C A Question 10. If x = –3, then –x 2 – 2x + 3 is Answer for Question 10: 18 6 12

42 Link to review Next Question Solution to Q10 The answer is 8. If x = –3, then –x 2 – 2x + 3 is –(–3) 2 – 2(–3) + 3 = –9 + 6 + 3 = 0

43 Correct! Question 10. If x = –3, then –x 2 – 2x + 3 is 0 D B C A 18 6 12 Next Question

44 A D C B Question 11. If a = –2, b = –1, c = 3, then b 2 – 4ac is Answer for Question 11: 25 –25 –23 23

45 Link to review Next Question Solution to Q11 The answer is 25. If a = –2, b = –1, c = 3, then (–1) 2 – 4(–2)(3) = 1 + 24 = 25

46 Correct! Question 11. If a = –2, b = –1, c = 3, then b 2 – 4ac is A D C B 25 –25 –23 23 Next Question

47 A B C D Question 12. If a – bx = c, then x = Answer for Question 12:

48 Link to review Next Question Solution to Q12 The answer is. If a – bx = c then a – c = bx = x so

49 Correct! Question 12. If a – bx = c, then x = A B C D Next Question

50 D B C A Answer for Question 13: 29A – 22B 29A – 38B –41A – 22B –41A – 38B Question 13. –2(3A – 4B) – 5(–7A + 6B) =

51 Link to review Next Question Solution to Q13 The answer is 29A –22B. –2(3A – 4B) – 5(–7A + 6B) = –6A + 8B + 35A – 30B = 29A – 22B

52 Correct! D B C A 29A – 22B –41A – 22B –41A – 38B 29A – 38B Question 13. –2(3A – 4B) – 5(–7A + 6B) = Next Question

53 A B D 14. Given the system 2x – y = 1 3x – y = –1 the solution for x is: Answer for Question 14: C 2 –2 –4 0

54 Link to review Next Question Solution to Q14 The answer is x = –2. 3x – y = –1 2x – y = 1 Subtract the two equations – ) x = –2

55 Correct! 14. Given the system 2x – y = 1 3x – y = –1 the solution for x is: A B D C 2 –2 –4 0 Next Question

56 A B D Answer for Question 15: C y 3 /x

57 Solution to Q15 The answer is By the Divide–Subtract rule for the exponents:

58 Link to review Next Question Solution to Q15 The answer is By the Divide–Subtract rule for the exponents:

59 Correct! A B D C y 3 /x Next Question

60 C B D Question 16. From experience, it’s sunny 5 out of 7 days at Sunny Hills. In a period of 182 days, how many cloudy days are expected? Answer for Question 16: A 130 days 120 days 115 days 125 days

61 Solution to Q16 The answer is 130 days. Let x be the number of sunny days, using proportion, x 182 5 The number of sunny days = 7 Total number of days

62 Solution to Q16 The answer is 130 days. Let x be the number of sunny days, using proportion, x 182 5 The number of sunny days = 7 Total number of days cross–multiplying so that 7x = 182(5)

63 Link to reviewNext Question Solution to Q16 The answer is 130 days. Let x be the number of sunny days, using proportion, x 182 5 The number of sunny days = 7 Total number of days cross–multiplying so that 7x = 182(5) x = 182(5) 7 26 x = 130

64 Correct! C B D A 130 days 120 days 115 days 125 days Question 16. From experience, it’s sunny 5 out of 7 days at Sunny Hills. In a period of 182 days, how many cloudy days are expected? Next Question

65 A B D Question 17. y is inversely proportional to x 2, if x = 2 then y = 9, what is x if y = 4? Answer for Question 17: C 4 6 2 3

66 Solution to Q17 The answer is 3. x2x2 k y = y is inversely proportional to x 2 means

67 Solution to Q17 The answer is 3. x2x2 k y = y is inversely proportional to x 2 means if x = 2, y = 9 then and that y = 2 k 9 = so k = 36, x2x2 36

68 Solution to Q17 The answer is 3. x2x2 k y = y is inversely proportional to x 2 means if x = 2, y = 9 then and that y = 2 k 9 = so k = 36, x2x2 36 so if y = 4, then 4 = x2x2 36

69 Solution to Q17 The answer is 3. x2x2 k y = Link to review Next Question y is inversely proportional to x 2 means if x = 2, y = 9 then and that y = 2 k 9 = so k = 36, x2x2 36 so if y = 4, then 4 = x2x2 36 4x 2 = 36 x 2 = 9 and that x = 3 or –3.

70 Correct! A B D C 4 6 2 3 Question 17. y is inversely proportional to x 2, if x = 2 then y = 9, what is x if y = 4? Next Question

71 A C D Answer for Question 18: B Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) = –2x 2 – 3x + 14 –2x 2 – 17x – 2 –2x 2 + 3x + 14 –2x 2 + 3x – 2

72 Solution to Q18 The answer is –2x 2 + 3x + 14. (x – 2)(2x – 3) – (3x + 2)(x – 4) = (x – 2)(2x – 3) + (–3x – 2)(x – 4)

73 Solution to Q18 The answer is –2x 2 + 3x + 14. Link to review Next Question (x – 2)(2x – 3) – (3x + 2)(x – 4) = (x – 2)(2x – 3) + (–3x – 2)(x – 4) = x 2 – 7x + 6 – 3x 2 + 10x + 8 = –2x 2 + 3x + 14

74 Correct! Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) = A C D B –2x 2 – 3x + 14 –2x 2 – 17x – 2 –2x 2 + 3x – 2 –2x 2 + 3x + 14 Next Question

75 A C D Question 19. Starting with a full tank of gas, we drove at a constant speed for 2½ hr and we used up 2/5 of the tank. How much more time we can travel at the same speed before we run out of gas? Answer for Question 19: B 3¼ hr 3¾ hr 3½ hr 4 hr

76 Solution to Q19 The answer is Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Use proportion to solve the problem. 3 hrs. 3 4

77 Solution to Q19 The answer is Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Use proportion to solve the problem. Hours of traveling time: Fraction of the tank: x 2/5 5/2 = 3/5 3 hrs. 3 4

78 Solution to Q19 The answer is Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Use proportion to solve the problem. Hours of traveling time: Fraction of the tank: x 2/5 5/2 = 3/5 2x 5 3 = 5 2 5 3 hrs. 3 4 cross–multiply

79 Solution to Q19 The answer is Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Use proportion to solve the problem. Hours of traveling time: Fraction of the tank: x 2/5 5/2 = 3/5 2x 5 3 = 5 2 5 2 3 = 3 hrs. 3 4 cross–multiply

80 Solution to Q19 The answer is Let x be the number of hours that we may travel on the remaining 3/5 tank of gas. Use proportion to solve the problem. Hours of traveling time: Fraction of the tank: x 2/5 5/2 = 3/5 2x 5 3 = 5 2 5 2 3 = 3 hrs. 3 4 cross–multiply

81 Solution to Q19 The answer is Link to review Next Question Let x be the number of hours that we may travel on the remaining 3/5 tank of gas, Use proportion to solve the problem. Hours of traveling time: Fraction of the tank: x 2/5 5/2 = 3/5 2x 5 3 = 5 2 5 2 3 = 4x = 15 x = 15/4 = 3 4 3 3 hrs. 3 4 cross again cross–multiply

82 A C D B Correct! Question 19. Starting with a full tank of gas, we drove at a constant speed for 2½ hr and we used up 2/5 of the tank. How much more time we can travel at the same speed before we run out of gas? 3¼ hr 3¾ hr 3½ hr 4 hr Next Question

83 B C A Question 20. Which of the following is a factor of 3x – 3y + ax – ay Answer for Question 20: D y + x a – 3 x – y 3 – a

84 Solution to Q20 The answer is (x – y) Factoring by grouping: 3x – 3y + ax – ay = 3(x – y) + a(x – y)

85 Solution to Q20 The answer is (x – y) Link to review Next Question Factoring by grouping: 3x – 3y + ax – ay = 3(x – y) + a(x – y) = (x – y) (3 + a)

86 Correct! B C A D y + x a – 3 x – y 3 – a Question 20. Which of the following is a factor of 3x – 3y + ax – ay Next Question

87 B C A Question 21. Answer for Question 21: D x – 3 2x – 1 x + 2 x – 1 –Combine and simplify (x – 3)(x + 2) x 2 – x – 5 (x – 3)(x + 2) x 2 + 7x + 1 (x – 3)(x + 2) x 2 + 7x – 5 (x – 3)(x + 2) x 2 – x + 1

88 Solution to Q21 The answer is Multiply the expression by 1 in the form of LCD/LCD distribute and simplify: (x – 3)(x + 2) x 2 + 7x – 5 x – 3 2x – 1 x + 2 x – 1 – (x – 3)(x + 2)

89 Solution to Q21 The answer is Multiply the expression by 1 in the form of LCD/LCD distribute and simplify: (x – 3)(x + 2) x 2 + 7x – 5 x – 3 2x – 1 x + 2 x – 1 – (x – 3)(x + 2) (x + 2) (x – 3)

90 Solution to Q21 The answer is Link to review Next Question Multiply the expression by 1 in the form of LCD/LCD distribute and simplify: (x – 3)(x + 2) x 2 + 7x – 5 x – 3 2x – 1 x + 2 x – 1 – (x – 3)(x + 2) = (2x – 1) (x + 2) – (x – 1)(x – 3) (x – 3)(x + 2) = x 2 + 7x – 5 (x + 2) (x – 3)

91 Correct! Question 21. B C A D (x – 3)(x + 2) x 2 – x – 5 (x – 3)(x + 2) x 2 + 7x + 1 (x – 3)(x + 2) x 2 + 7x – 5 (x – 3)(x + 2) x 2 – x + 1 x – 3 2x – 1 x + 2 x – 1 –Combine and simplify Next Question

92 B D A Question 22. The solutions of 3x(x – 1) = x 2 + 9 are Answer for Question 22: C x = –2/3, 3 x = 2/3, –3 x = –3/2, 2 x = 3/2, –2

93 Solution to Q22 The answer is x = –3/2, 3 3x(x – 1) = x 2 + 9 3x 2 – 3x = x 2 + 9 2x 2 – 3x – 9 = 0

94 Solution to Q22 The answer is x = –3/2, 3 Link to review Next Question 3x(x – 1) = x 2 + 9 3x 2 – 3x = x 2 + 9 2x 2 – 3x – 9 = 0 (2x + 3)(x – 3 ) = 0 so x = –3/2, 3

95 B D A Question 22. The solutions of 3x(x – 1) = x 2 + 9 are C x = –3/2, 2 x = 3/2, –2 Correct! x = –2/3, 3 x = 2/3, –3 Next Question

96 B D C Question 23. Solve for x if Answer for Question 23: A 1 – a 1 1 + a 1 –1 a – 1 1 x 1 a =1 –

97 Solution to Q23 The answer is x = Link to review Next Question 1 – a 1 x 1 a =1 – x 1 =1 – a x = 1 – a 1

98 Correct! B D C A 1 – a 1 1 + a 1 –1 a – 1 1 Question 23. Solve for x if x 1 a =1 – Next Question

99 A D C Question 24. Answer for Question 24: B x + 1 (x 2 – 3x) x2x2 (1 – x 2 ) (x 2 – 4x + 3) * Simplify x –x – 1 x x – 1 x 1 – x x

100 Solution to Q24 The answer is –x – 1 x (x 2 – 3x) x2x2 (1 – x 2 ) (x 2 – 4x + 3) * = x(x – 3)(1 – x)(1 + x) x 2 (x – 3)(x – 1)

101 Solution to Q24 The answer is Link to review Next Question –x – 1 x (x 2 – 3x) x2x2 (1 – x 2 ) (x 2 – 4x + 3) * = x(x – 3)(1 – x)(1 + x) x 2 (x – 3)(x – 1) = –1 x –x – 1

102 A D C B Correct! Question 24. x + 1 x –x – 1 x x – 1 x 1 – x x (x 2 – 3x) x2x2 (1 – x 2 ) (x 2 – 4x + 3) * Simplify Next Question

103 A D B Question 25. The vertices of a rectangle are (1, 1), (5, 1), (1, 7), and (5, 7). What is the area of the rectangle? Answer for Question 25: C 16 20 24 28

104 Solution to Q25 The answer is 24. Link to review Next Question We have (1, 1) (1, 5) (7, 1) (7, 5) So the area of the rectangle is Δx Δy = 6×4 = 24. Δy = 5 – 1 = 4 Δx = 7 – 1 = 6

105 Correct! Question 25. The vertices of a rectangle are (1, 1), (5, 1), (1, 7), and (5, 7). What is the area of the rectangle? A D B C 16 20 24 28 Next Question

106 D C B Question 26. The solution for the inequality 3 > 1 – 2x ≥ –3 is Answer for Question 26: A 2–1 2 2 2

107 Link to review Next Question Question 26. The answer is 3 > 1 – 2x ≥ –3 subtract 1 2–1 2 > – 2x ≥ –4 divide by –2 and reverse the inequality –1 < x ≤ 2 or 2–1

108 Correct! D C B A 2–1 2 2 2 Question 26. The solution for the inequality 3 > 1 – 2x ≥ –3 is Next Question

109 A C B Answer for Question 27: D ft3 1 3 3 2 3 2 ft 3 ft x 5 ft ft3 2 2 3 Question 27. Following are two prints of different sizes of the same picture of Tomo. With the given measurements, what is x?

110 Solution to Q27 The answer is 3 1/3 ft. Link to reviewNext Question The ratios of the measurements must satisfies the proportion x : 5 = 2 : 3 or that 2 ft 3 ft x 5 ft 5 = x 3 2

111 Solution to Q27 The answer is 3 1/3 ft. Link to reviewNext Question The ratios of the measurements must satisfies the proportion x : 5 = 2 : 3 or that 2 ft 3 ft x 5 ft 5 = x 3 2 x = 3 2 (5)= 3 3 1 so

112 Correct! 2 ft 3 ft x 5 ft A C B D ft3 1 3 3 2 3 3 2 2 3 Question 27. Following are two prints of different sizes of the same picture of Tomo. With the given measurements, what is x? Next Question

113 Answer for Question 28: A D Question 28. Which of the following pictures represents A, B and C on the real line most accurately if A = 3, B = 9 and C = 12? C B 0 0 0 0

114 Solution to Q28 The answer B. 0 C = 12 must be the right most point. 12

115 Solution to Q28 The answer B. 0 C = 12 must be the right most point. By dividing from 0 to 12 into 4 pieces 12

116 Solution to Q28 The answer B. Link to review Next Question 0 12639 C = 12 must be the right most point. By dividing from 0 to 12 into 4 pieces we see that A = 3, B = 9 and C = 12 must be

117 Correct! A D Question 28. Which of the following pictures represents A, B and C on the real line most accurately if A = 3, B = 9 and C = 12? C B 0 0 0 0 Next Question

118 A B Question 29. Let point A = (–10, 20), the coordinate of the point that is 40 to the right and 40 below A is Answer for Question 29: (30, –20) C D (40, –20) (30, –30) (40, –30)

119 Solution to Q29 The answer (30, –20). Link to review Next Question the point that’s 40 to the right and 40 below of (–10, 20) is (–10 + 40, 20 – 40) = (30, –20)

120 Correct! A B (30, –20) C D (40, –20) (30, –30) (40, –30) Question 29. Let point A = (–10, 20), the coordinate of the point that is 40 to the right and 40 below A is Next Question

121 A D Question 30. The slope of the line that has x–intercept at 4 and y intercept at 3 is Answer for Question 30: C B 3/4 –3/4 4/3 –4/3

122 Solution to Q30 The answer is –3/4. The slope is “rise/run”.

123 Solution to Q30 The answer is –3/4. Link to review Next Question The rise and run for (4, 0) and (0, 3) is (4, 0) (0, 3) 4,–3 run rise The slope is “rise/run”. Hence the slope is –3/4.

124 A D C B 3/4 –3/4 4/3 –4/3 Correct! Next Question Question 30. The slope of the line that has x–intercept at 4 and y intercept at 3 is

125 D Question 31. The slope of a vertical line is Answer for Question 31: C B Undefined 0 1 –1 A

126 The slope is “rise/run”. The run for two points on a vertical line is 0 hence the slope = rise/0 is not defined. Link to review Solution to Q31 The answer is “undefined”. Next Question

127 D Question 31. The slope of a vertical line is C B Undefined 0 1 –1 A Correct! Next Question

128 A D Question 32. Find the y–intercept of the line that contains the points (4, 15) and (9, 45). Answer for Question 32: C B –8 –9 –10 –11

129 Solution to Q32 The answer is –9. The slope of the line that contains the points (4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6

130 Solution to Q32 The answer is –9. The slope of the line that contains the points (4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6 so the equation of the line containing these points is y = 6(x – 4) + 15

131 Solution to Q32 The answer is –9. Link to review The slope of the line that contains the points (4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6 so the equation of the line containing these points is y = 6(x – 4) + 15 or y = 6x – 9 Hence the y–intercept is at y = –9.

132 A D C B –8 –9 –10 –11 Question 32. Find the y–intercept of the line that contains the points (4, 15) and (9, 45). Correct!


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