Practice Problems for Pre-Algebra and Elementary Algebra - Building Your Pre-Algebra and Elementary Algebra Skills - ACT Math For Dummies

ACT Math For Dummies (2011)

Part II. Building Your Pre-Algebra and Elementary Algebra Skills

Chapter 6. Practice Problems for Pre-Algebra and Elementary Algebra

Ready for some pre-algebra and elementary algebra practice? You’ve come to the right place! In this chapter, I provide 30 practice problems that include everything covered in Chapters 4 and 5. If you get stuck, each answer is completely worked out in the later section titled (you guessed it) “Solutions to Practice Problems.” Now get started!

Practice Problems

Following are 30 problems for you to practice your pre-algebra and elementary algebra skills. If you need to brush up on any of the concepts, refer to Chapters 4 and 5.

1. If the first six numbers of a sequence are, in order, 6, 12, 9, 18, 15, 30, what is the eighth number in the sequence?

(A) 36

(B) 42

(C) 45

(D) 48

(E) 54

2. Danielle spends her summers working as a baby sitter for families in the neighborhood. She began one day with $20, and then she worked for 4 hours in the morning at $10 per hour and received a $5 tip. In the afternoon, she worked for 3 more hours at $12 per hour but needed to pay $7 to take a cab home. How much money did she have at the end of the day?

(F) $74

(G) $94

(H) $104

(J) $108

(K) $118

3. A charitable organization sells books of raffle tickets as a fundraiser. The following graph shows the number of books that five different people have sold so far. What is the average number of books sold among these five people?

9781118001547-un0601.eps

(A) 14

(B) 14.2

(C) 14.4

(D) 14.6

(E) 15

4. If 5(x – 2) = 3(2x – 6), then x =

(F) 7

(G) 8

(H) 9

(J) 11

(K) 13

5. What is the value of n4 + 5knk3 if k = –2 and n = 3?

(A) 51

(B) 53

(C) 55

(D) 57

(E) 59

6. If you subtract 1 from a real number and then multiply by 5, the result is the same as if you multiply the same number by 4 and then add 0.5. What is the real number?

(F) 4.5

(G) 5.5

(H) 6.25

(J) 7.25

(K) 8

7. If 9781118001547-EQ06001.eps, then 13n – 2n2 =

(A) 6

(B) –8

(C) 10

(D) 12

(E) –14

8. If ax + by = c, then x =

(F) aby + ac

(G) acaby

(H) 9781118001547-EQ06002.eps

(J) 9781118001547-EQ06003.eps

(K) 9781118001547-EQ06004.eps

9. What is the value of 9781118001547-EQ06005.eps if jk = 100?

(A) 0.01

(B) 0.1

(C) 1

(D) 10

(E) 100

10. While packing for a business trip, Chet found that he had room for 4 suits, 6 shirts, and 12 ties. Assuming that he can wear any combination of one suit, one shirt, and one tie, how many combinations are possible?

(F) 4

(G) 12

(H) 22

(J) 288

(K) 4,096

11. If 9781118001547-EQ06006.eps, which of the following is the value of n?

(A) 9781118001547-EQ06065.eps

(B) 9781118001547-EQ06066.eps

(C) 9781118001547-EQ06067.eps

(D) 9781118001547-EQ06068.eps

(E) 9781118001547-EQ06069.eps

12. The following graph shows the results of a poll asking people how many pets they have. What percentage of the people asked have three or more pets?

9781118001547-un0602.eps

(F) 20 percent

(G) 21 percent

(H) 22 percent

(J) 22.5 percent

(K) 25 percent

13. If you double a number and then subtract 6, the result is the same as if you add 18 to the same number and then multiply what you get by –4. What is the result when you add 14 to the number?

(A) –13

(B) –3

(C) 3

(D) 13

(E) 23

14. If 9781118001547-EQ06007.eps, then m =

(F) 9781118001547-EQ06008.eps

(G) 9781118001547-EQ06009.eps

(H) 9781118001547-EQ06010.eps

(J) 9781118001547-EQ06011.eps

(K) 9781118001547-EQ06012.eps

15. If 9781118001547-EQ06013.eps, what is the value of y?

(A) 6

(B) –6

(C) 15

(D) –15

(E) Cannot be determined from the information given

16. What is the positive value of p if |17 – 2p| = 25?

(F) 4

(G) 8

(H) 15

(J) 21

(K) 42

17. How many factors does the number 84 have?

(A) 6

(B) 8

(C) 9

(D) 10

(E) 12

18. What is the greatest common factor among 8x3y2z4, 12x2y3z5, and 20xy6z7?

(F) 2y2z4

(G) 2x2y2z4

(H) 4xy2z4

(J) 4xy3z4

(K) 4x3y6z7

19. Eleven children took a test and scored 78, 83, 83, 84, 88, 91, 93, 93, 93, 95, and 96. Which of the following statements is true?

(A) The median is two points higher than the mode.

(B) The median is one point higher than the mode.

(C) The median and the mode are the same.

(D) The median is one point lower than the mode.

(E) The median is two points lower than the mode.

20. Marv bought a flat-screen TV that was marked at 20 percent off its original price. The store added 10 percent tax to the marked-down price, so Marv ended up paying $880. What was the original price of the TV?

(F) $800

(G) $880

(H) $968

(J) $1,000

(K) $1,080

21. If m is the lowest common denominator of the fractions 9781118001547-EQ06014.eps, 9781118001547-EQ06015.eps, and 9781118001547-EQ06016.eps, which of the following is true?

(A) m < 50

(B) 50 ≤ m < 60

(C) 60 ≤ m < 70

(D) 70 ≤ m < 80

(E) m ≥ 80

22. What product results when you multiply the two solutions of |7n + 1| = 3n – 11?

(F) 3

(G) –3

(H) –6

(J) 9

(K) –12

23. A school has a 2:3:35 ratio of administrators, teachers, and students, respectively. If the school has exactly 15 teachers, what is the total number of administrators, teachers, and students?

(A) 160

(B) 175

(C) 200

(D) 320

(E) 375

24. If you divide a number by 3 and then add 13, the number you end up with is the same as if you subtract 36 from the same number and then multiply by 2. What is the number?

(F) 51

(G) 63

(H) 75

(J) 81

(K) 99

The following graph shows the number of houses Tina sold in each of the first six months of the year. Use this graph to answer Questions 25 and 26.

9781118001547-un0603.eps

25. What is the total number of houses Tina sold from January through March, inclusive?

(A) 5

(B) 7

(C) 8

(D) 10

(E) 12

26. What is the average number of houses that Tina sold over the six months?

(F) 4

(G) 4.5

(H) 5

(J) 5.5

(K) 6

27. If 9781118001547-EQ06021.eps, then x =

(A) 9781118001547-EQ06022.eps

(B) 9781118001547-EQ06023.eps

(C) 9781118001547-EQ06024.eps

(D) 9781118001547-EQ06025.eps

(E) 9781118001547-EQ06026.eps

28. If you toss a penny, a nickel, a dime, and a quarter, what is the probability that exactly two of the coins will land heads up and exactly two will land tails up?

(F) 9781118001547-EQ06027.eps

(G) 9781118001547-EQ06028.eps

(H) 9781118001547-EQ06029.eps

(J) 9781118001547-EQ06030.eps

(K) 9781118001547-EQ06031.eps

29. If 9781118001547-EQ06032.eps, what is the value of 4tu?

(A) 9781118001547-EQ06033.eps

(B) 9781118001547-EQ06034.eps

(C) 9781118001547-EQ06035.eps

(D) 9781118001547-EQ06036.eps

(E) Cannot be determined from the information given

30. In his first four basketball games this season, Jerome has scored 13, 17, 19, and 21 points. How many points does he need to score in his next game to bring his average score for the first five games up to 20 points per game?

(F) 10

(G) 20

(H) 26

(J) 27

(K) 30

Solutions to Practice Problems

So, how did you do? Find out here where I provide the answers to the 30 practice questions from the preceding section, complete with the worked-out solutions.

1. E. Begin by noticing that 6 is half of 12, 9 is half of 18, and so on. Then, you need to find out why 12 goes to 9, 18 goes to 15, and so on. As you probably noticed, subtracting 3 is the answer. So the sequence is generated by an alternating rule: Multiply by 2, subtract 3, multiply by 2, subtract 3, and so on. Thus, the sequence continues as follows:

6, 12, 9, 18, 15, 30, 27, 54 . . .

The eighth number in the sequence is 54, so the correct answer is Choice (E).

2. G. Danielle started the day with $20. In the morning, she earned ($10 × 4) + $5 = $45, so she had $65. In the afternoon, she earned $12 × 3 = $36, giving her a total of $101. The cab ride cost Danielle $7, so she ended the day with $101 – $7 = $94.

3. B. The graph shows that the five people have sold 16, 12, 19, 14, and 10 books of tickets, so you can plug these numbers into the formula for the mean:

9781118001547-EQ06037.eps

4. G. Begin by distributing to remove the parentheses:

5(x – 2) = 3(2x – 6)

5x – 10 = 6x – 18

Isolate x and solve:

–10 = x – 18

8 = x

5. E. Begin by plugging in –2 for k and 3 for n:

n4 + 5knk3 = 34 + 5(–2)(3) – (–2)3

Evaluate the expression using the order of operations: Start with the powers, then do multiplication, and finally work any addition and subtraction. Here’s what the calculations look like:

= (3)(3)(3)(3) + 5(–2)(3) – (–2)(–2)(–2) = 8130 + 8 = 59

6. G. Let x be the number. Then you can make the following equation:

5(x – 1) = 4x + 0.5

Simplify and solve:

5x – 5 = 4x + 0.5

x – 5 = 0.5

x = 5.5

7. A. Cross-multiply to remove the fractions:

9781118001547-EQ06038.eps

Distribute on both sides of the equation and simplify:

9781118001547-EQ06070.eps

Thus, substitute 9781118001547-eq06077.eps for n in 13n – 2n2:

9781118001547-EQ06071.eps

8. J. First isolate the x term on one side of the equation:

ax + by = c

ax = cby

Next divide both sides of the equation by a:

9781118001547-EQ06039.eps

9. B. Simplify by factoring both the numerator (top number) and denominator (bottom number) and then canceling out the common factor j + k:

9781118001547-EQ06040.eps

Substitute 100 for jk and simplify:

9781118001547-EQ06041.eps

10. J. Chet packed 4 suits, 6 shirts, and 12 ties, and he can wear any combination of those. Multi-plying these three numbers gives you 4 × 6 × 12 = 288.

11. C. To remove the fractions from the equation 9781118001547-EQ06042.eps, multiply each term by the common denominator of 10:

9781118001547-EQ06043.eps

Simplify by dividing 10 by the denominator and then multiplying by the numerator:

1(n) – 5(1 – n) = 2(n + 3)

n – 5 + 5n = 2n + 6

Solve for n:

9781118001547-EQ06072.eps

12. H. To begin, you need to find out how many people participated in the poll: 225 + 185 + 175 + 90 + 75 = 750. Exactly 90 of these people have 3 pets, and 75 have more than 3. So 165 people have 3 or more pets. To find the percentage, make a fraction as follows:

9781118001547-EQ06044.eps

13. C. Let x equal the number, and then translate the words into the following equation:

2x – 6 = –4(x + 18)

Solve for x:

2x – 6 = –4x – 72

6x – 6 = –72

6x = –66

x = –11

Thus, when you add 14 to the number, the result is –11 + 14 = 3.

14. K. To find the value of m, you need to express both sides of the equation in terms of the same base. The base 3 works well because:

32 = 9 and9781118001547-EQ06045.eps

Thus, you can change the equation as follows:

9781118001547-EQ06046.eps

Because the bases are equal, the exponents must be equal. As a result, you can drop the bases and solve for m:

9781118001547-EQ06047.eps

15. A. Multiply both sides of the equation by 2x – 9 to get rid of the fraction:

9781118001547-EQ06048.eps

Distribute on the right side of the equation:

4x – 3y = 4x – 18

Subtract 4x from both sides:

–3y = –18

The x terms drop out, allowing you to solve for y:

y = 6

16. J. Remove the absolute value bars by separating the equation |17 – 2p| = 25 into two equations:

17 – 2p = 25 17 – 2p = –25

–2p = 8 –2p = –42

p = –4 p = 21

The only positive value of p is 21, so the correct answer is Choice (J).

17. E. Begin by writing down 1 and 84 with space between them:

9781118001547-EQ06073.eps

The number 84 is even as noted by 84 ÷ 2 = 42. So put these two numbers at the beginning and end of the list:

9781118001547-EQ06074.eps

The number 84 is divisible by 3 (84 ÷ 3 = 28), so add these two numbers to the list:

9781118001547-EQ06075.eps

Continue in this fashion until you reach the middle factors:

9781118001547-EQ06076.eps

As you can see, 84 has exactly 12 factors, so the correct answer is Choice (E).

18. H. The greatest factor among 8, 12, and 20 is 4, because it divides all three of these numbers evenly; therefore, you can rule out Choices (F) and (G). The lowest exponent among the three x’s is 1, so you can rule out Choice (K). The lowest exponent among the three y’s is 2, so you can rule out Choice (J). Therefore, the correct answer is Choice (H).

19. E. The scores are 78, 83, 83, 84, 88, 91, 93, 93, 93, 95, and 96. The median is the middle value among these scores, so it’s 91. The mode is the most common value, so it’s 93. The median is two points lower than the mode, so the correct answer is Choice (E).

20. J. Let m be the marked-down price of the TV before the 10 percent tax was included. Thus, the price of the TV with tax included was 110 percent of m, and this amount was $880. So

9781118001547-EQ06049.eps

As you can see, the marked-down price before tax was $800. This amount was 20 percent off of the original price, so it represents 80 percent of the original price. Let p be the original price. So

9781118001547-EQ06050.eps

21. D. Begin by listing the multiples of 6, 8, and 9:

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72 . . .

Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80 . . .

Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 . . .

Because m = 72, you know that the correct answer is Choice (D).

22. G. Begin by splitting the equation into two separate equations:

7n + 1 = 3n – 11 7n + 1 = – (3n – 11)

Solve both equations:

4n + 1 = –11 7n + 1 = –3n + 11

4n = –12 10n + 1 = 11

n = –3 10n = 10

n = 1

–3 × 1 = –3, so the correct answer is Choice (G).

23. C. The school has 15 teachers and a 2:3 ratio of administrators to teachers, so make a proportion like this:

9781118001547-EQ06051.eps

Plug in 15 for teachers and a for administrators, and then cross-multiply and solve for a:

9781118001547-EQ06052.eps

The school has 10 administrators and a 2:35 ratio of administrators to students, so create the following proportion:

9781118001547-EQ06053.eps

Plug in 10 for administrators and s for students, and then cross-multiply and solve for s:

9781118001547-EQ06054.eps

The school has 175 students, 10 administrators, and 15 teachers, so the total of all these is 175 + 10 + 15 = 200.

24. F. Let x equal the number, and then create the following equations:

9781118001547-EQ06055.eps

To eliminate the fraction, multiply every term by 3:

x + 39 = 6(x – 36)

Simplify and solve for x:

x + 39 = 6x – 216

39 = 5x – 216

255 = 5x

51 = x

25. D. According to the graph, Tina sold 2 houses in January, 3 in February, and 5 in March, so she sold a total of 10 during those three months.

26. G. Tina sold 2 houses in January, 3 in February, 5 in March, 6 in April, 7 in May, and 4 in June. Place this information into the formula for the mean to get your answer:

9781118001547-EQ06056.eps

27. B. To isolate the square root expression, add x to both sides of the equation:

9781118001547-EQ06057.eps

Square both sides of the equation:

9781118001547-EQ06058.eps

Simplify both sides:

x2 + 1 = (x + 2)(x + 2)

x2 + 1 = x2 + 4x + 4

Subtract x2 from both sides and solve for x:

9781118001547-EQ06059.eps

28. J. Begin by counting the number of ways each coin can land as independent events. Because each coin has 2 sides, each one can land 2 different ways as independent events.

Multiply to find the number of possible outcomes: 2 × 2 × 2 × 2 = 16. Of these, count the number of ways in which exactly 2 coins come up heads:

Penny and Nickel Penny and Dime Penny and Quarter

 Nickel and Dime Nickel and Quarter

   Dime and Quarter

Thus, 6 outcomes are the target outcome. Plug these numbers into the formula for probability:

9781118001547-EQ06060.eps

29. B. To answer the question, isolate the expression 4tu on one side of the equation. Begin by multiplying both sides by 2t – 1 and simplifying:

9781118001547-EQ06061.eps

Factor out a 3 on the left side:

3(4tu) = 5

Now divide both sides of the equation by 3:

9781118001547-EQ06062.eps

30. K. Let x be the number of points Jerome needs to score in his fifth game to get an average of 20 points over 5 five games. Plug this information into the formula for an average:

9781118001547-EQ06063.eps

Simplify and solve for x:

9781118001547-EQ06064.eps