Answer After reading question 17, you’re likely to come up with the equation in answer a. Since a is correct, it is not the right choice. Now manipulate the equation to see whether you can find an equivalent equation. If you subtract 3 from each side, answer b will result. From there, dividing both sides by 10, you come up with c. All those are equivalent equations. Choice d can be derived by using b and subtracting v from both sides. Choice e is not an equivalent and is therefore the correct answer. Distance, Rate, and Time Problems One type of problem made simpler by algebra are those involving distance, rate, and time. Your math review would not be complete unless you had at least one problem about trains leaving the station. Sample Distance Problem 18. A train left the station near your home and went at a speed of 50 miles per hour for 3 hours. How far did it travel? a. 50 miles b. 100 miles c. 150 miles d. 200 miles e. 250 miles Answer Use the three Success Steps to work through the problem. 1. D = R × T 2. D = 50 × 3 3. 50 × 3 = 150 Practice Try these: 19. How fast does a dirt bike go if it goes 60 miles every 3 hours? 20. How long does it take to go 180 miles at 60 miles per hour? Answers 19. R = 20 20. T = 3 HOT TIP Another way to look at the distance formula is When you’re working out a problem, cross out the let- ter that represents the value you need to find. What remains will tell you the operation you need to perform to get the answer: the horizontal line means divide and the vertical line means multiply. For example, if you need to find R, cross it out. You’re left with D and T. The line between them tells you to divide, so that’s how you’ll find R. This is a handy way to remember the formula, especially on tests, but use the method that makes the most sense to you. Three Success Steps for Distance, Rate, and Time Problems 1. First, write the formula. Don’t skip this step! The formula for Distance, Rate, and Time is D = R × T. Remember this by putting all the let- ters in alphabetical order and putting in the equal sign as soon as possible. Or think of the word DIRT where the I stands for is, which is always an equal sign. 2. Fill in the information. 3. Work the problem. –CBEST MINI-COURSE– 118 D R T Math 8: Averages, Probability, and Combinations In this lesson, you’ll have a chance to do sample aver- age questions as well as problems on probability and on the number of possible combinations. They may be a little more advanced than those you did in school, but they will not be difficult for you if you master the information in this section. Averages You probably remember how you solved average prob- lems way back in elementary school. You added up the numbers, divided by the number of numbers, and the average popped out. Here’s this process in algebraic form: = Average What makes CBEST average problems more dif- ficult is that not all the numbers will be given for you to add. You’ll have to find some of the numbers. Sample Average Question 1. Sean loved to go out with his friends, but he knew he’d be grounded if he didn’t get 80% for the semester in his English class. His test scores were as follows: 67%, 79%, 75%, 82%, and 78%. He had two more tests left to go. One was tomor- row, but his best friend Jason had invited him to his birthday party tonight. If he studied very hard and got 100% on his last test, what could he get by with tomorrow and still have a chance at the 80%? a. 65% b. 72% c. 76.2% d. 79% e. 80.2% Answer Use the four Success Steps to solve the problem. 1. Draw the horizontal line: ᎏᎏ 2. Write in the information: ᎏ 7 ᎏ = 80% 3. Multiply the number of numbers by the average to obtain the sum of the numbers: 7 × 80 = 560 4. 560 has to be the final sum of the numbers. So far, if you add up all the scores, Sean has a total of 381. With the 100 he plans to get on the last test, his total will be 481. Since he needs a sum of 560 for the average of the seven tests to come out 80, he needs 79 more points. The answer is d. Sample Average Question 2. On an overseas trip, Jackie and her husband are allowed five suitcases that average 110 pounds each. They want to pack in all the peanut butter and mango nectar they can carry to their family in Italy. They weighed their first four suitcases and the weights were as follows: 135, 75, 90, and Four Success Steps for Average Problems 1. In order to use the formula above, draw the horizontal line that is under the sum and over the number of numbers in the average formula. 2. Write in all the information you know. Put the number of numbers under the line and the average beside the line. Unless you know the whole sum, leave the top of the line blank. 3. Multiply the number of numbers by the aver- age. This will give you the sum of the numbers. 4. Using this sum, solve the problem. Sum of the Numbers ᎏᎏᎏ Number of Numbers –CBEST MINI-COURSE– 119 120. How much weight are they allowed to stuff into their fifth bag? Answer 1. Draw the horizontal line: ᎏᎏ 2. Write in the information: ᎏ 5 ᎏ = 110 3. Multiply the number of numbers by the average to obtain the sum of the numbers: 5 × 110 = 550. 4. Since the total weight they can carry is 550 lb. and they already have 420 lb. (135 + 75 + 90 + 120), the fifth suitcase can weigh as much as 550 − 420, or 130 lbs. Frequency Charts Some average problems on the CBEST use frequency charts. Sample Frequency Chart Question 3. The following list shows class scores for an easy Science 101 quiz. What is the average of the scores? a. 28.5 b. 85 c. 91 d. 95 e. 100 Answer Use the four Success Steps to solve the problem. 1. In this frequency chart, the test score is given on the right, and the number of students who received each grade is on the left: 10 students got 100, 15 got a 90, etc. 2. Multiply the number of students by the score, because to find the average, each student’s grade has to be added individually: 10 × 100 = 1,000 15 × 90 = 1,350 3 × 80 = 240 2 × 70 = 140 3. Then add the multiplied scores: 1,000 + 1,350 + 240 + 140 = 2,730 4. Then divide the total number of students, 30 (10 + 15 + 3 + 2), into 2,730 to get the average: ᎏ 2, 3 7 0 30 ᎏ = 91. 10 15 3 2 10 15 3 2 10 15 3 2 100 90 80 70 # of Students Score Four Success Steps for Frequency Chart Questions 1. Read the question and look at the chart. Make sure you understand what the different columns represent. 2. If a question asks you to find the average, multiply the numbers in the first column by the numbers in the second column. 3. Add the figures you got by multiplying. 4. Divide the total sum by the sum of the left col- umn. This will give you the average. –CBEST MINI-COURSE– 120 Other Average Problems There are other kinds of averages besides the mean, which is usually what is meant when the word average is used: ■ Median is the middle number in a range. ■ Mode is the number that occurs most frequently. ■ Range is the difference between the highest and lowest number. Sample Median Question 4. What is the median of 6, 8, 3, 9, 4, 3, and 12? a. 2 b. 6 c. 9 d. 10 e. 12 Answer To get a median, put the numbers in order—3, 3, 4, 6, 8, 9, 12—and choose the middle number: 6. If there are an even number of numbers, average the middle two (you probably won’t have to do that on the CBEST). Mode The mode is the number used most frequently in a series of numbers. In the above example, the mode is 3 because 3 appears twice and all other numbers are used only once. Look again at the frequency table from the frequency chart sample question. Can you find the mode? More students (15) earned a score of 90 than any other score. Therefore the mode is 15. Range To obtain the range, subtract the smallest number from the largest number. The range in the median sample question is 12 − 3 or 9. The range in the frequency chart sample question is 100 − 70, or 30. Probability Suppose you put one entry into a drawing that had 700 entrants. What would be your chances of winning? 1 in 700 of course. Suppose you put in two entries. Your chances would then be 2 in 700, or reduced, 1 in 350. Probabilities are fairly simple if you remember the few tricks that are explained in this section. Sample Probability Question 5. If a nickel were flipped thirteen times, what is the probability that heads would come up the thir- teenth time? a. 1:3 b. 1:2 c. 1:9 d. 1:27 e. 1:8 Answer Use the four Success Steps to solve the problem. 1. Form a fraction. 2. Each time a coin is flipped, there are 2 possibili- ties—heads or tails—so 2 goes on the bottom of the fraction. The thirteenth time, there are still going to be only two possibilities. 3. The number of chances given is 1. There is only one head on a coin. Therefore, the fraction is ᎏ 1 2 ᎏ . Four Success Steps for Probability Questions 1. Make a fraction. 2. Place the total number of different possibilities on the bottom. 3. Place the number of the chances given on the top. 4. If the answers are in a:b form, place the numerator of the fraction first, and the denom- inator second. –CBEST MINI-COURSE– 121 “Thirteen times” is extra information and does not have a bearing on this case. 4. The answer is b, 1:2. The numerator goes to the left of the colon and the denominator to the right. Sample Probability Question 6. A spinner is divided into 6 parts. The parts are numbered 1–6. When a player spins the spinner, what are the chances the player will spin a num- ber less than 3? Answer Once again, use the four Success Steps. 1. Form a fraction. 2. Total number of possibilities = 6. Therefore, 6 goes on the bottom. 3. Two goes on top, since there are 2 numbers less than 3: 1 and 2. 4. The answer is ᎏ 2 6 ᎏ or reduced ᎏ 1 3 ᎏ = 1:3. Combinations Combination problems require the solver to make as many groups as possible given certain criteria. There are many different types of combination problems, so these questions need to be read carefully before attempting to solve them. One of the easiest ways to make combination problems into CBEST points is to make a chart and list in a pattern all the possibilities. The following sample question is a typical CBEST combination problem. Sample Combination Question 7. Shirley had three pairs of slacks and four blouses. How many different combinations of one pair of slacks and one blouse could she make? a. 3 b. 4 c. 7 d. 12 e. 15 Answer To see this problem more clearly, you may want to make a chart: Each pair of slacks can be matched to 4 different blouses, making 4 different outfits for each pair of the 3 pairs of slacks, 3 × 4, making a total of 12 possible combinations. Sample Combination Question 9. Five tennis players each played each other once. How many games were played? a. 25 b. 20 c. 15 d. 10 e. 5 –CBEST MINI-COURSE– 122 Answer This combination problem is a little trickier in that there are not separate groups of items as there were for the slacks and blouses. This question involves the same players playing each other. But solving it is not diffi- cult. First, take the total number of players and sub- tract one: 5 − 1 = 4. Add the numbers from 4 down: 1 + 2 + 3 + 4 = 10. To learn how this works, take a look at the following chart: Letter the five players from A to E: ■ A plays B, C, D, and E (4 games) ■ B has already played A, so needs to play C, D, E (3 games) ■ C has already played A and B, so needs to play D, E (2 games) ■ D has already played A, B, and C, so needs to play E (1 game) ■ E has played everyone Adding up the number of games played (1 + 2 + 3 + 4) gives a total of 10, choice d. This same question might be asked on the CBEST using the number of games 5 chess players played or the number of handshakes that occur when 5 people shake hands with each other once. Other Combination Problems Although the above combination problems are the most common, other kinds of problems are possible. The best way to solve other combination problems is to make a chart. When you notice a pattern, stop and multiply. For example, if you’re asked to make all the possible combinations of three letters using the letters A through D, start with A: AAA ABA ACA ADA AAB ABB ACB ADB AAC ABC ACC ADC AAD ABD ACD ADD There seem to be 16 possibilities that begin with A, so probably there are 16 that begin with B and 16 that begin with C and D, so multiplying 16 × 4 will give you the total possible combinations: 64. Math 9: The Word Problem Game The directions for the word problem game are simple: While carefully observing a word problem, find all the math words and numbers in the problem. Eliminate the nonessential words and facts in order to find your answer. Operations in Word Problems To prepare for the game, make five columns on a sheet of paper. Write one of these words on the top of each column: Add, Subtract, Multiply, Divide, Equals. Now try to think of five words that tell you to add, five that tell you to subtract, and so on. If you can think of five for each column, you win the first round. If you can’t think of five, you can cheat by looking at the list below. How did you do? 0 = keep studying 1–3 for each = good 4–6 for each = excellent 7+ for each = Why are you reading this book? ■ Add: sum, plus, more than, larger than, greater than, and, increased by, added to, in all, altogether, total, combined with, together, length- ened by –CBEST MINI-COURSE– 123 . at the 80%? a. 65% b. 72 % c. 76 .2% d. 79 % e. 80.2% Answer Use the four Success Steps to solve the problem. 1. Draw the horizontal line: ᎏᎏ 2. Write in the information: ᎏ 7 ᎏ = 80% 3. Multiply. he’d be grounded if he didn’t get 80% for the semester in his English class. His test scores were as follows: 67% , 79 %, 75 %, 82%, and 78 %. He had two more tests left to go. One was tomor- row, but. 240 2 × 70 = 140 3. Then add the multiplied scores: 1,000 + 1,350 + 240 + 140 = 2 ,73 0 4. Then divide the total number of students, 30 (10 + 15 + 3 + 2), into 2 ,73 0 to get the average: ᎏ 2, 3 7 0 30 ᎏ =