1. This is a sample test for the elementary algebra diagnostic test
at the Harbor College.
This is an interactive PowerPoint file.
For the best viewing, download it and run it in the viewing–mode.
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 here

2. How many pieces of candies did Chuck get? 8
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 A 6 B 4 C 3 D
Answer for Question 1:

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

4. The statement “(fraction or %) of an amount”
Solution to Q1
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, 1 12
i.e.
(48) = 12 pieces were left. 4

5. The statement “(fraction or %) of an amount”
Solution to Q1
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, 1 12
i.e.
(48) = 12 pieces were left. 4
Mary took 2/3 so Chuck got 1/3 of 12, 1 4
i.e.
(12) = 4 pieces. 3
Link to review
Next Question

6. How many pieces of candies did Chuck get? 8
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 A 6 B 4
Correct! C 3 D
Next Question

7. 2 1
Question 2. 1 3 = 3 3 1 –1 A 3 2 –1 B 3 1 C –2 3 2 –2 D 3
Answer for Question 2:

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

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

10. 2 1 Question 2. 1 3 = 3 3 1 –1 3 2 –1 Correct! 3 1 –2 3 2 –2 3A 3 2 –1
Correct! B 3 1 C –2 3 2 –2 D 3
Next Question

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

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

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

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

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

16. Question 4.
4,000 x (7.5 x 10–9) x (8 x 103) =
240. A 24 B C
2.4 D
0.24
Answer for Question 4:

17. Solution to Q4
The answer is 0.24.
4,000 x (7.5 x 10–9) x (8 x 103)
= 4 x 103 x (7.5 x 10–9) x (8 x 103)

18. Solution to Q4
The answer is 0.24.
4,000 x (7.5 x 10–9) x (8 x 103)
= 4 x 103 x (7.5 x 10–9) x (8 x 103)
= 4 x 7.5 x 8 x 103 x 10–9 x 103
Next Question

19. Solution to Q4
The answer is 0.24.
4,000 x (7.5 x 10–9) x (8 x 103)
= 4 x 103 x (7.5 x 10–9) x (8 x 103)
= 4 x 7.5 x 8 x 103 x 10–9 x 103
= 240 x 10-3
= 0.24
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Next Question

20. Question 4.
4,000 x (7.5 x 10–9) x (8 x 103) =
240. A 24 B C
2.4 D
0.24
Correct!
Next Question

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

22. Solution to Q5
The answer is 11.
By the Pythagorean Theorem it must be that
x = 122

23. Solution to Q5
Solution to Q5
The answer is 11.
By the Pythagorean Theorem it must be that
x = 122
x = 144
x2 = 119 or that
x = √119 ≈ 11
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Next Question

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

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

26. Solution to Q6
The answer is 30%.
2/7 = ≈ 30%
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Next Question

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

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

29. 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)

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)
= 0.9 – 1.8
= –0.9
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Next Question

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

32. 3 15 – 3 26 is approximately Question 8. 13 11 9 7B 11 C 9 D 7
Answer for Question 8:

33. Solution to Q8
The answer is 9.
is approximately 3(4) = 12,
and is approximately 3
Next Question

34. Solution to Q8
The answer is 9.
is approximately 3(4) = 12,
and is approximately 3 so that
– 3 26
≈ 12 – 3 = 9.
Link to review

35. 3 15 – 3 26 is approximately Question 8. 13 11 Correct! 9 7B 11
Correct! C 9 D 7
Next Question

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

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

38. 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
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Next Question

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

40. Question 10. If x = –3, then –x2 – 2x + 3 isA 6 B 12 C 18 D
Answer for Question 10:

41. Solution to Q10
The answer is 8.
If x = –3, then –x2 – 2x + 3 is
–(–3)2 – 2(–3) + 3
= –
= 0
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Next Question

42. Question 10. If x = –3, then –x2 – 2x + 3 is
Correct! A 6 B 12 C 18 D
Next Question

43. Question 11. If a = –2, b = –1, c = 3, then b2 – 4ac is23 A 25 B C
–25
–23 D
Answer for Question 11:

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

45. Question 11. If a = –2, b = –1, c = 3, then b2 – 4ac is23 A 25
Correct! B C
–25
–23 D
Next Question

46. Question 12. If a – bx = c, then x =
𝑐−𝑎+𝑏
𝑐 𝑎–𝑏 C
𝑎−𝑐 𝑏 D
Answer for Question 12:

47. Solution to Q12 The answer is . If a – bx = c then a – c = bx so = x
𝑎−𝑐 𝑏
If a – bx = c
then a – c = bx so
𝑎−𝑐 𝑏
= x
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Next Question

48. Question 12. If a – bx = c, then x =
𝑐−𝑎+𝑏
𝑐 𝑎–𝑏 C
𝑎−𝑐 𝑏
Correct! D
Next Question

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

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

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

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

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

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

55. Question 15. 𝑥 2 𝑦 –7 𝑥 –3 𝑦 –4 = x y3 y3/x x5 y3 x5/y3A
y3/x B
x5 y3 C
x5/y3 D
Answer for Question 15:

56. Solution to Q15
The answer is
x5/y3
By the Divide–Subtract rule for the exponents:
𝑥 2 𝑦 –7 𝑥 –3 𝑦 –4
= x 2–(–3) y –7–(–4)

57. Solution to Q15
The answer is
x5/y3
By the Divide–Subtract rule for the exponents:
𝑥 2 𝑦 –7 𝑥 –3 𝑦 –4
= x 2–(–3) y –7–(–4)
= x5y–3
= x5/y3
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Next Question

58. Question 15. 𝑥 2 𝑦 –7 𝑥 –3 𝑦 –4 = x y3 y3/x x5 y3 x5/y3 Correct!A
y3/x B
x5 y3 C
x5/y3
Correct! D
Next Question

59. 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? A
130 days
125 days B C
120 days D
115 days
Answer for Question 16:

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

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

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

63. 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? A
130 days
Correct!
125 days B C
120 days D
115 days
Next Question

64. Question 17. y is inversely proportional to x2, if x = 2 then y = 9, what is x if y = 4?6 B 4 C 3 2 D
Answer for Question 17:

65. Solution to Q17
The answer is 3. k
y is inversely proportional to x2 means
y = x2

66. Solution to Q17
The answer is 3. k
y is inversely proportional to x2 means
y = x2 k
if x = 2, y = 9 then
9 =
so k = 36, 22 36
and that y = x2

67. Solution to Q17
The answer is 3. k
y is inversely proportional to x2 means
y = x2 k
if x = 2, y = 9 then
9 =
so k = 36, 22 36
and that y = x2 36
so if y = 4, then 4 = x2

68. Solution to Q17
The answer is 3. k
y is inversely proportional to x2 means
y = x2 k
if x = 2, y = 9 then
9 =
so k = 36, 22 36
and that y = x2 36
so if y = 4, then 4 = x2
4x2 = 36
x2 = 9 and that x = 3 or –3.
Link to review
Next Question

69. Question 17. y is inversely proportional to x2, if x = 2 then y = 9, what is x if y = 4?6 B 4
Correct! C 3 2 D
Next Question

70. Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) =A B
–2x2 + 3x + 14
–2x2 – 17x – 2 C D
–2x2 – 3x + 14
Answer for Question 18:

71. Solution to Q18
The answer is –2x2 + 3x + 14.
(x – 2)(2x – 3) – (3x + 2)(x – 4)
= (x – 2)(2x – 3) + (–3x – 2)(x – 4)

72. Solution to Q18
The answer is –2x2 + 3x + 14.
(x – 2)(2x – 3) – (3x + 2)(x – 4)
= (x – 2)(2x – 3) + (–3x – 2)(x – 4)
= x2 – 7x + 6 – 3x2 + 10x + 8
= –2x2 + 3x + 14
Link to review
Next Question

73. Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) =A B
–2x2 + 3x + 14
Correct!
–2x2 – 17x – 2 C D
–2x2 – 3x + 14
Next Question

74. 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? A
4 hr
3¾ hr B C
3½ hr
3¼ hr D
Answer for Question 19:

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

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

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

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

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

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

81. 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? A
4 hr
3¾ hr
Correct! B C
3½ hr
Next Question
3¼ hr D

82. Question 20. Which of the following is a factor of 3x – 3y + ax – ayB
a – 3
y + x C
x – y D
Answer for Question 20:

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

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

85. Question 20. Which of the following is a factor of 3x – 3y + ax – ayB
a – 3
y + x C
x – y D
Correct!
Next Question

86. Question 21. 2x – 1 x – 1 Combine and simplify – x – 3 x + 2
Answer for Question 21:

87. Solution to Q21
The answer is
x2 + 7x – 5
(x – 3)(x + 2)
Multiply the expression by 1 in the form of LCD/LCD distribute and simplify:
2x – 1
x – 1
(x – 3)(x + 2) –
(x – 3)(x + 2)
x – 3
x + 2

88. Solution to Q21 The answer is x2 + 7x – 5 (x – 3)(x + 2)
Multiply the expression by 1 in the form of LCD/LCD distribute and simplify:
(x + 2)
(x – 3)
2x – 1
x – 1
(x – 3)(x + 2) –
(x – 3)(x + 2)
x – 3
x + 2

89. Solution to Q21 The answer is x2 + 7x – 5 (x – 3)(x + 2)
Multiply the expression by 1 in the form of LCD/LCD distribute and simplify:
(x + 2)
(x – 3)
2x – 1
x – 1
(x – 3)(x + 2) –
(x – 3)(x + 2)
x – 3
x + 2
= (2x – 1) (x + 2) – (x – 1)(x – 3)
(x – 3)(x + 2)
x2 + 7x – 5 =
(x – 3)(x + 2)
Link to review
Next Question

90. Question 21. 2x – 1 x – 1 Combine and simplify – x – 3 x + 2
Correct!
(x – 3)(x + 2)
Next Question

91. Question 22. The solutions of 3x(x – 1) = x2 + 9 areB
x = –3/2, 2
x = 2/3, –3 C
x = –2/3, 3 D
Answer for Question 22:

92. Solution to Q22
The answer is x = –3/2, 3
3x(x – 1) = x2 + 9
3x2 – 3x = x2 + 9
2x2 – 3x – 9 = 0

93. Solution to Q22
The answer is x = –3/2, 3
3x(x – 1) = x2 + 9
3x2 – 3x = x2 + 9
2x2 – 3x – 9 = 0
(2x + 3)(x – 3 ) = 0
so x = –3/2, 3
Link to review
Next Question

94. Question 22. The solutions of 3x(x – 1) = x2 + 9 areB
x = –3/2, 2
x = 2/3, –3
Correct! C
x = –2/3, 3 D
Next Question

95. 1 Question 23. Solve for x if a = 1 – x 1 1 – a 1 1 + a –1 1 + a 1B
1 + a –1 C
1 + a 1 D
a – 1
Answer for Question 23:

96. Solution to Q23
The answer is x = 1
1 – a 1
a =
1 – x 1 =
1 – a x 1
x =
1 – a
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Next Question

97. 1 Question 23. Solve for x if a = 1 – x 1 Correct! 1 – a 1 1 + a –1B
1 + a –1 C
1 + a 1 D
a – 1
Next Question

98. Question 24. (1 – x2) (x2 – 3x) Simplify * x2 (x2 – 4x + 3) x + 1 xA x
–x – 1 B x
x – 1 C x
1 – x D x
Answer for Question 24:

99. Solution to Q24
The answer is
–x – 1 x
(1 – x2)
(x2 – 3x) * x2
(x2 – 4x + 3)
(1 – x)(1 + x)
x(x – 3) =
x2(x – 3)(x – 1)

100. Solution to Q24 The answer is –x – 1 x (1 – x2) (x2 – 3x) * x2–1
(1 – x)(1 + x)
x(x – 3) =
x2(x – 3)(x – 1)
–x – 1 = x
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Next Question

101. Question 24. (1 – x2) (x2 – 3x) Simplify * x2 (x2 – 4x + 3) x + 1 xA x
–x – 1
Correct! B x
x – 1 C x
1 – x D x
Next Question

102. 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 16 20 B 24 C 28 D
Answer for Question 25:

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

104. 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 16 20 B 24
Correct! C 28 D
Next Question

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

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

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

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

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

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

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

112. Which of the following pictures represents
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? A B C D
Answer for Question 28:

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

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

115. C = 12 must be the right most point.
Solution to Q28
The answer B.
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 3 6 9 12
Link to review
Next Question

116. Which of the following pictures represents
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? A
Correct! B C D
Next Question

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

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

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

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

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

122. Solution to Q30
The answer is –3/4.
The slope is “rise/run”.
The rise and run for (4, 0) and (0, 3) is
(4, 0)
(0, 3)
4,–3
run
rise
Hence the slope is –3/4.
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123. Question 30. The slope of the line that has
x–intercept at 4 and y intercept at 3 is A
3/4
–3/4
Correct! B C
4/3 D
–4/3
Next Question

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

125. Solution to Q31
The answer is “undefined”.
The slope is “rise/run”.
The run for two points on a vertical line is 0 hence the slope = rise/0 is not defined.
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126. Question 31. The slope of a vertical line is
Undefined
Correct! B C 1 D –1
Next Question

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

128. 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

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
so the equation of the line containing these points is y = 6(x – 4) + 15

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 or
y = 6x – 9
Hence the y–intercept is at y = –9.
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131. Question 32. Find the y–intercept of the line that contains the points (4, 15) and (9, 45).–8 –9 B
Correct! C
–10 D
–11