Success step english 8 ppsx

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 ■ Subtract: difference, minus, decreased by, reduced by, diminished by, less, take away, subtract, low- ered by, dropped by, shortened by, lightened by, less, less than, subtracted from, take from, deducted from. Note: The words in bold are backwardswords. (See below.) ■ Multiply: product, times, of, multiplied by, twice, thrice, squared, cubed, doubled, tripled, rows of, columns of ■ Divide: quotient of, ratio of, halved, per, split, equal parts of, divided by, divided into, recipro- cal. Note: The words in bold are backwardswords. (See below.) ■ Equals: is, equal to, the same as, amounts to, equivalent to, gives us, represents Backwardswords Backwardswords are words in a word problem that tend to throw off test takers; they indicate the opposite of which the numbers appear in the problem. Only subtraction and division have backwardswords. Addi- tion and multiplication come out the same no matter which number is written first: 2 + 6 is the same as 6 + 2, but 2 – 6 is not the same as 6 – 2. Using the numbers 10 and 7, notice the following translations: Subtraction: 10 minus 7 is the same as 10 – 7 10 take away 7 is 10 – 7 10 less 7 is the same as 10 – 7 But 10 less than 7 is the opposite, 7 – 10 10 subtracted from 7 is also 7 – 10 Division: 10 over 7 is written ᎏ 1 7 0 ᎏ The quotient of 10 and 7 is ᎏ 1 7 0 ᎏ But 10 divided into 7 is written ᎏ 1 7 0 ᎏ And the reciprocal of ᎏ 1 7 0 ᎏ is ᎏ 1 7 0 ᎏ Writing Word Problems in Algebraic Form Sample Word Conversion Questions The following are simple problems to rewrite in alge- braic form. Using N for a number, try writing out the problems below. Remember to add parentheses as needed to avoid order of operation problems. 1. Three added to a number represents 6. 2. Six subtracted from a number is 50. Answers Use the four Success Steps to find the answer to ques- tion 1. 1. “Represents” is the verb. Put in an equal sign: = 2. 3, 6, and N are the numbers: 3 N =6 3. Added means +: 3 + N = 6 4. No parentheses are needed. Follow the Success Steps for question 2. Four Success Steps for Converting Words to Algebra In order to make an equation out of words use these steps: 1. Find the verb. The verb is always the = sign. 2. Write in the numbers. 3. Write in the symbols for the other code words. Be careful of backwardswords. 4. If necessary, add parentheses. HOT TIP When setting up division problems in algebra, avoid using the division sign: ÷. Instead, use the division line: ᎏ 3 4 ᎏ . –CBEST MINI-COURSE– 124 1. “Is” is the verb. Put in an equal sign: = 2. 6, N, and 50 are the numbers: 6 N = 50. 3. Subtracted from means –, but it is a back- wardsword: N – 6 = 50. 4. No parentheses are needed. Practice Underline the backwardswords, then write the equations. 3. A number subtracted from 19 is 7. 4. 3 less a number is 5. 5. 3 less than a number is 5. 6. 9 less a number is –8. 7. A number taken from 6 is –10. 8. 30 deducted from a number is 99. 9. The quotient of 4 and a number equals 2. 10. The reciprocal of 5 over a number is 10. 11. 6 divided into a number is 3. Change the following sentences into algebraic equations. 12. The sum of 60 and a number all multiplied by 2 amounts to 128. 13. Forty combined with twice a number is 46. 14. $9 fewer than a number costs $29. 15. 7 feet lengthened by a number of feet all divided by 5 is equivalent to 4 feet. 16. 90 subtracted from the sum of a number and one gives us 10. 17. Half a number plus 12 is the same as 36. Answers 3. subtracted from, 19 – N = 7 4. 3 – N = 5 5. less than, N – 3 = 5 6. 9 – N = –8 7. taken from, 6 – N = –10 8. deducted from, N – 30 = 99 9. ᎏ N 4 ᎏ = 2 10. reciprocal of, ᎏ N 5 ᎏ = 10 11. divided into, ᎏ N 6 ᎏ = 3 12. (60 + N)2 = 128 or 2(60 + N) = 128 13. 40 + 2N = 46 14. N – $9 = $29 15. ᎏ 7+ 5 N ᎏ = 4 16. (N +1) – 90 = 10 17. ᎏ N 2 ᎏ +12 = 36 or ᎏ 1 2 ᎏ N + 12 = 36 Words or Numbers? Try these two problems and determine which is easier for you. 1. Three more than five times a number equals 23. 2. Jack had three more than five times the number of golf balls than Ralph had. If Jack had 23 golf balls, how many did Ralph have? Answers: 1. 23 = 3 + 5N 2. 23 = 3 + 5N Did you notice that the two problems were the same, but the second one was more wordy? If question 1 was easier, you can work word problems more easily by eliminating non-essential words. If question 2 was easier, you can work out problems more easily by pic- turing actual situations. If they were both equally easy, –CBEST MINI-COURSE– 125 then you have mastered this section. Go on to the sec- tion on two-variable problems, which is a little more difficult. Practice If you found wordy word problems difficult, here are some more to try: 18. Sally bought 6 less than twice the number of boxes of CDs that Raphael (R) bought. If Sally bought 4 boxes, how many did Raphael buy? 19. A 1-inch by 13-inch rectangle is cut off a piece of linoleum that was made up of three squares; each had N inches on a side. This left 62 square inches left to the original piece of linoleum. How long was each of the sides of the squares? 20. Six was added to the number of sugar cubes in a jar. After that, the number was divided by 5. The result was 6. How many sugar cubes were in the jar? Answers 18. Sally = 2R – 6. Substitute 4 for Sally: 4 = 2R – 6 19. (N 2 × 3) – (1 × 13) = 62. N is the side of a square so the area of the square is N 2 . There were three squares, so you have 3N 2 .1 × 13 was taken away (–). Notice that the parentheses, while not strictly necessary if you follow the order of operations, will help you keep track of the numbers. 20. ᎏ 6+ 5 N ᎏ = 6 Problems with Two Variables In solving problems with two variables, you have to watch out for another backwards phrase: as many as. Sample Two-Variable Questions The following equations require the use of two vari- ables. Choose the answers from the following: a. 2x = y b. 2y = x c. 2 + x = y d. 2 + y = x e. none of the above 21. Twice the number of letters Joey has equals the number of letters Tina has. Joey = x, Tina = y. 22. Tuli corrected twice as many homework assign- ments as tests. Homework =x, tests = y. Answers 21.a. “Equals”is the verb. Joey or x is on one side of the verb, Tina or y is on the other. A straight ren- dering will give you the answer a, or 2x = y, because Tina has twice as many letters. To check, plug in 6 for y. If Tina has 6 letters, Joey will have 6 ÷ 2, or 12. The answer makes sense. 22. b. “Corrected” is the verb. Which did Tuli correct fewer of? Tests. You need to multiply 2 times the tests to reach the homework assignments. Check: Three Success Steps for Problems with Two Variables When turning “as many as” sentences into equa- tions, consider the following steps. 1. Read the problem to decide which variable is least. 2. Combine the number given with the least variable. 3. Make the combined number equal to the larger amount. –CBEST MINI-COURSE– 126 If there are 6 tests, then there are 12 homework assignments: 2 × 6 = 12. This answer makes sense. Practice Now that you are clued in, try the following using the same answer choices as above. a. 2x = y b. 2y = x c. 2 + x = y d. 2 + y = x e. none of the above 23. Sandra found two times as many conch shells as mussel shells. Conch = x, mussel = y. 24. Sharon walked two more miles today than she walked yesterday. Today = x, yesterday = y. 25. Martin won two more chess games than his brother won. His brother = x, Martin = y. Answers 23. b. 24. d. 25. c.  Math 10: The CA Approach to Word Problems Of course, it helps to know the formula or method needed to solve a problem. But there are always those problems on the test that you don’t recognize or can’t remember how to do, and this may cause you a little anxiety. Even experienced math teachers experience that paralyzing feeling at times. But you shouldn’t allow anxiety to conquer you. Nor should you jump into a problem and start figuring madly without a careful reading and analysis of the problem. The CA SOLVE Approach When approaching a word problem, you need the skills of a detective. Follow the CA SOLVE method to uncover the mystery behind a problem that is unfa- miliar to you. C Stands for Conquer Conquer that queasy feeling—don’t let it conquer you. To squelch it, try step A. A Stands for Answer Look at the answers and see if there are any similarities among them. Notice the form in which the answers are written. Are they all in cubic inches? Do they all con- tain pi? Are they formulas? S Stands for Subject Experience Many problems are taken from real life situations or are based on methods you already know. Ask: “Do I have any experience with this subject or with this type of problem? What might a problem about the subject be asking me? Can I remember anything that might relate to this problem?” Eliminate experiences or methods of solving that don’t seem to work. But be careful; sometimes sorting through your experiences and methods memory takes a long time. O Stands for Organize the Facts Here are some ways to Organize your data: 1. Look for clue words in the problem that tell you to add, subtract, multiply, or divide. 2. Try out each answer to see which one works. Look for answers to eliminate. 3. Think of formulas or methods that have worked for you in solving problems like this in the past. Write them down. There should be plenty of room on your test booklet for this. –CBEST MINI-COURSE– 127 L Stands for Live Living the problem means pretending you’re actually in the situation described in the word problem. To do this effectively, make up details concerning the events and the people in the problem as if you were part of the picture. This process can be done as you are read- ing the problem and should take only a few seconds. V Stands for View View the problem with different numbers while keep- ing the relationships between the numbers the same. Use the simplest numbers you can think of. If a prob- lem asked how long it would take a rocket to go 1,300,000 miles at 650 MPH, change the numbers to 300 miles at 30 MPH. Solve the simple problem, and then solve the problem with the larger numbers the same way. E Stands for Eliminate Eliminate answers you know are wrong. You may also spend a short time checking your answer if there is time. Sample Question Solve this problem using the SOLVE steps described above. 1. There are 651 children in a school. The ratio of boys to girls is 4:3. How many boys are there in the school? a. 40 b. 325 c. 372 d. 400 e. 468 Answer 1. Subject Experience: You know that 4 and 3 are only one apart and 4 is more. You can conclude from this that boys are a little over half the school population. Following up on that, you can cut 651 in half and eliminate any answers that are under half. Furthermore, since there are three numbers in the problem and two are paired in a ratio, you can conclude that this is a ratio problem. Then you can think about what methods you used for ratio problems in the past. 2. Organize: The clue word total means to add. In the context in which it is used, it must mean girls plus boys equals 651. Also, since boys is written before girls, the ratio should be written Boys:Girls. 3. Live: Picture a group of three girls and four boys. Now picture more of these groups, so many that the total would equal 651. 4. View: If there were only 4 boys and 3 girls in the school, there would still be a ratio of 4 to 3. Think of other numbers that have a ratio of 4:3, like 40 and 30. If there were 40 boys and 30 girls, there would be 70 students in total, so the answer has to be more than 40 boys. Move on to 400 boys and 300 girls—700 total students. Since the total in the problem is 651, 700 is too large, but it is close, so HOT TIP Don’t try to keep a formula in your head as you solve the problem. Although writing does take time and effort, jot- ting down a formula is well worth it for three reasons: 1) A formula on paper will clear your head to work with the numbers; 2) You will have a visual image of the formula you can refer to and plug numbers into; 3) The formula will help you see exactly what operations you will need to per- form to solve the problem. –CBEST MINI-COURSE– 128 . parentheses are needed. Follow the Success Steps for question 2. Four Success Steps for Converting Words to Algebra In order to make an equation out of words use these steps: 1. Find the verb. The. 9 – N = 8 7. taken from, 6 – N = –10 8. deducted from, N – 30 = 99 9. ᎏ N 4 ᎏ = 2 10. reciprocal of, ᎏ N 5 ᎏ = 10 11. divided into, ᎏ N 6 ᎏ = 3 12. (60 + N)2 = 1 28 or 2(60 + N) = 1 28 13. 40. homework assignments. Check: Three Success Steps for Problems with Two Variables When turning “as many as” sentences into equa- tions, consider the following steps. 1. Read the problem to decide