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This is a sample test for the elementary algebra diagnostic testat 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 here
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
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
Solution to Q1 The correct answer is 4.The statement “(fraction or %) of an amount” is translated to the multiplication operation.
Solution to Q1
14
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
Link to review
Next Question
Solution to Q1
14
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 12,13i.e. (12) = 4 pieces.
4
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?
Link to review
Next Question
Solution to Q2
The correct answer is .
23–1
23
13
=
13
23
13
13
= 23
13
2
=23
1
hence 23
13
1 3 =23–1
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
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.
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.
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 = 13501.08
= 1250.
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
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)
Next Question
Solution to Q4
The answer is 0.24.
Link to review
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
Solution to Q4
The answer is 0.24.
Link to review
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
A
B
C
DAnswer 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
Solution to Q5
Link to review
Next Question
Solution to Q5
The answer is 11.
By the Pythagorean Theorem it must be that x2 + 52 = 122
x2 + 25 = 144
x2 = 119 or that x = √119 ≈ 11
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
Question 6. Which of the following percentages is closest to 2/7?
D
B
C
A 30%
Answer for Question 6:
25%
20%
15%
Link to review
Next Question
Solution to Q6
The answer is 30%.
2/7 = 0.285.. ≈ 30%
Correct!
D
B
C
A 30%
25%
20%
15%
Question 6. Which of the following percentages is closest to 2/7?
Next Question
Link to review
Next Question
Solution to Q7The 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
Link to review
Next Question
Solution to Q8
The answer is 9.
is approximately 3(4) = 12,
is approximately 3
Solution to Q8
The answer is 9.
is approximately 3(4) = 12,
is approximately 3 so that
≈ 12 – 3 = 9.
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:
Link to review
Next Question
Solution to Q9The answer is -1/2.
If x = 3, then xy – x – y = –4 is 3y – 3 – y = –4 so that
2y = –1or y = –1/2
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
DNext Question
Link to review
Next Question
Solution to Q10The answer is 8.
If x = –3, then –x2 – 2x + 3 is –(–3)2 – 2(–3) + 3
= –9 + 6 + 3= 0
A
D
C
B
Question 11. If a = –2, b = –1, c = 3, then b2 – 4ac is
Answer for Question 11:
25
–25
–23
23
Link to review
Next Question
Solution to Q11The answer is 25.
If a = –2, b = –1, c = 3, then (–1)2 – 4(–2)(3) = 1 + 24 = 25
Correct!
Question 11. If a = –2, b = –1, c = 3, then b2 – 4ac is
A
D
C
B 25
–25
–23
23
Next Question
Link to review
Next Question
Solution to Q12
The answer is .𝑎−𝑐𝑏
If a – bx = cthen a – c = bx
𝑎−𝑐𝑏 = xso
D
B
C
A
Answer for Question 13:
29A – 22B
29A – 38B
–41A – 22B
–41A – 38B
Question 13. –2(3A – 4B) – 5(–7A + 6B) =
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
Correct!
D
B
C
A 29A – 22B
–41A – 22B
–41A – 38B
29A – 38B
Question 13. –2(3A – 4B) – 5(–7A + 6B) =
Next Question
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
Link to review
Next Question
Solution to Q14The answer is x = –2.
3x – y = –12x – y = 1
Subtract the two equations
– )
x = –2
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
Solution to Q15
= x 2–(–3) y –7–(–4)
The answer is x5/y3
By the Divide–Subtract rule for the exponents:
Link to review
Next Question
Solution to Q15
= x 2–(–3) y –7–(–4) = x5y–3 = x5/y3
The answer is x5/y3
By the Divide–Subtract rule for the exponents:
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
Solution to Q16The answer is 130 days.
Let x be the number of sunny days, using proportion,
x182
5The number of sunny days=
7Total number of days
Solution to Q16The answer is 130 days.
Let x be the number of sunny days, using proportion,
x182
5The number of sunny days=
7Total number of dayscross–multiplying so that 7x = 182(5)
Link to review Next Question
Solution to Q16The answer is 130 days.
Let x be the number of sunny days, using proportion,
x182
5The number of sunny days=
7Total number of dayscross–multiplying so that 7x = 182(5)
x = 182(5)7
26
x = 130
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
A
B
D
Question 17. y is inversely proportional to x2, if x = 2 then y = 9, what is x if y = 4?
Answer for Question 17:
C
4
6
2
3
Solution to Q17The answer is 3.
x2
ky = y is inversely proportional to x2 means
if x = 2, y = 9 then
and that y =22
k9 = so k = 36,
x236
Solution to Q17The answer is 3.
x2
ky = y is inversely proportional to x2 means
if x = 2, y = 9 then
and that y =22
k9 = so k = 36,
x236
so if y = 4, then 4 = x2
36
Solution to Q17The answer is 3.
x2
ky =
Link to review
Next Question
y is inversely proportional to x2 means
if x = 2, y = 9 then
and that y =22
k9 = so k = 36,
x236
so if y = 4, then 4 = x2
36
4x2 = 36 x2 = 9 and that x = 3 or –3.
Correct!
A
B
D
C
4
6
2
3
Question 17. y is inversely proportional to x2, if x = 2 then y = 9, what is x if y = 4?
Next Question
A
C
D
Answer for Question 18:
B
Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) =
–2x2 – 3x + 14
–2x2 – 17x – 2
–2x2 + 3x + 14
–2x2 + 3x – 2
Solution to Q18The answer is –2x2 + 3x + 14.
(x – 2)(2x – 3) – (3x + 2)(x – 4) = (x – 2)(2x – 3) + (–3x – 2)(x – 4)
Solution to Q18The answer is –2x2 + 3x + 14.
Link to review
Next Question
(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
Correct!
Question 18. (x – 2)(2x – 3) – (3x + 2)(x – 4) =
A
C
D
B
–2x2 – 3x + 14
–2x2 – 17x – 2
–2x2 + 3x – 2
–2x2 + 3x + 14
Next Question
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
Solution to Q19The 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.34
Solution to Q19The 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:
x2/55/2=
3/5
3 hrs.34
Solution to Q19The 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:
x2/55/2=
3/52x
53=
5 25
3 hrs.34
cross–multiply
Solution to Q19The 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:
x2/55/2=
3/52x
53=
5 25
23=
3 hrs.34
cross–multiply
Solution to Q19The 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:
x2/55/2=
3/52x
53=
5 25
23=
3 hrs.34
cross–multiply
Solution to Q19The 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:
x2/55/2=
3/52x
53=
5 25
23=
4x = 15x = 15/4 = 3 4
3
3 hrs.34
cross again
cross–multiply
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
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
Solution to Q20The 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)
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
B
C
A
Question 21.
Answer for Question 21:
D
x – 32x – 1
x + 2x – 1– Combine and simplify
(x – 3)(x + 2)x2 – x – 5
(x – 3)(x + 2)x2 + 7x + 1
(x – 3)(x + 2)x2 + 7x – 5
(x – 3)(x + 2)x2 – x + 1
Solution to Q21The answer is
Multiply the expression by 1 in the form of LCD/LCD distribute and simplify:
(x – 3)(x + 2)x2 + 7x – 5
x – 32x – 1
x + 2x – 1– (x – 3)(x + 2)(x – 3)(x + 2)
Solution to Q21The answer is
Multiply the expression by 1 in the form of LCD/LCD distribute and simplify:
(x – 3)(x + 2)x2 + 7x – 5
x – 32x – 1
x + 2x – 1– (x – 3)(x + 2)(x – 3)(x + 2)
(x + 2) (x – 3)
Solution to Q21The 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)x2 + 7x – 5
x – 32x – 1
x + 2x – 1– (x – 3)(x + 2)(x – 3)(x + 2)
= (2x – 1) (x + 2) – (x – 1)(x – 3)(x – 3)(x + 2)
=(x – 3)(x + 2)
x2 + 7x – 5
(x + 2) (x – 3)
Correct!
Question 21.
B
C
A
D
(x – 3)(x + 2)x2 – x – 5
(x – 3)(x + 2)x2 + 7x + 1
(x – 3)(x + 2)x2 + 7x – 5
(x – 3)(x + 2)x2 – x + 1
x – 32x – 1
x + 2x – 1– Combine and simplify
Next Question
B
D
A
Question 22. The solutions of 3x(x – 1) = x2 + 9 are
Answer for Question 22:
C
x = –2/3, 3
x = 2/3, –3
x = –3/2, 2
x = 3/2, –2
Solution to Q22The answer is x = –3/2, 3
Link to review
Next Question
3x(x – 1) = x2 + 9 3x2 – 3x = x2 + 92x2 – 3x – 9 = 0(2x + 3)(x – 3 ) = 0so x = –3/2, 3
B
D
A
Question 22. The solutions of 3x(x – 1) = x2 + 9 are
C
x = –3/2, 2
x = 3/2, –2
Correct!
x = –2/3, 3
x = 2/3, –3
Next Question
Solution to Q23
The answer is x =
Link to review
Next Question
1 – a1
x1a = 1 –
x1 = 1 – a
x =1 – a
1
A
D
C
Question 24.
Answer for Question 24:
B
x + 1
(x2 – 3x)x2
(1 – x2)(x2 – 4x + 3)*Simplify
x
–x – 1x
x – 1x
1 – xx
Solution to Q24
The answer is –x – 1x
(x2 – 3x)x2
(1 – x2)(x2 – 4x + 3)*
=x(x – 3)(1 – x)(1 + x)
x2(x – 3)(x – 1)
Solution to Q24
The answer is
Link to review
Next Question
–x – 1x
(x2 – 3x)x2
(1 – x2)(x2 – 4x + 3)*
=x(x – 3)(1 – x)(1 + x)
x2(x – 3)(x – 1)
=
–1
x–x – 1
A
D
C
B Correct!
Question 24.
x + 1x
–x – 1x
x – 1x
1 – xx
(x2 – 3x)x2
(1 – x2)(x2 – 4x + 3)*Simplify
Next Question
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
Solution to Q25The 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
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
D
C
B
Question 26. The solution for the inequality 3 > 1 – 2x ≥ –3 is
Answer for Question 26:
A 2–1
2–1
2–1
2–1
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 or2–1
Correct!
D
C
B
A 2–1
2–1
2–1
2–1
Question 26. The solution for the inequality 3 > 1 – 2x ≥ –3 is
Next Question
A
C
B
Answer for Question 27:
D ft3 13
ft3 23
2 ft
3 ft
x
5 ft
ft3
ft2 23
Question 27. Following are two prints of different sizes of the same picture of Tomo. With the given measurements, what is x?
Solution to Q27The answer is 3 1/3 ft.
Link to review Next 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
32
Solution to Q27The answer is 3 1/3 ft.
Link to review Next 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
32
x =32 (5
)= 3 3
1
so
Correct!
2 ft
3 ft
x
5 ft
A
C
B
D ft3 13
ft3 23
ft3
ft2 23
Question 27. Following are two prints of different sizes of the same picture of Tomo. With the given measurements, what is x?
Next Question
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
B0
0
0
0
Solution to Q28The answer B.
0
C = 12 must be the right most point.By dividing from 0 to 12 into 4 pieces
12
Solution to Q28The answer B.
Link to review
Next Question
0 1263 9
C = 12 must be the right most point.By dividing from 0 to 12 into 4 pieceswe see that A = 3, B = 9 and C = 12 must be
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
B0
0
0
0Next Question
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)
Solution to Q29The 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)
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
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
Solution to Q30The 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.
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
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 Q31The answer is “undefined”.
Next Question
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
Solution to Q32The answer is –9.
The slope of the line that contains the points (4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6
Solution to Q32The answer is –9.
The slope of the line that contains the points (4, 15) and (9, 45) is (45 –15)/(9 – 40) = 6so the equation of the line containing these points is y = 6(x – 4) + 15
Solution to Q32The 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) = 6so the equation of the line containing these points is y = 6(x – 4) + 15 or
y = 6x – 9Hence the y–intercept is at y = –9.
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