Page 1. Math Module 2: Negative Numbers, Proportions, and the Dreaded Fractions

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1 Page 1 Math Module 2: Negative Numbers, Proportions, and the Dreaded Fractions

Copyright 2001-2013 by Steve Kirshenbaum and Celia Barranon, SAT Preparation Group, LLC. www.satprepgroup.com. All rights reserved.

2 Copyright by Steve Kirshenbaum and Celia Barranon, SAT Preparation Group, LLC. All rights reserved. No part of this book may be reproduced in any form or by any electronic or mechanical means including information storage and retrieval systems without express written permission from SAT Preparation Group, LLC. This publication is designed to provide accurate and authoritative information in regard to the subject matter within this book. DISCLAIMER: Peak performance strategies for SAT/ACT prep include physical exercise, stretching, and optimal nutrition for maximum results. It is recommended that participants in this program consult with a trained medical professional prior to engaging in such activities as they relate to the services offered by SAT Preparation Group, LLC. SAT Preparation Group, LLC is not responsible for illness or injury due to these activities. *SAT is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. Your comments and corrections are welcome. Please send them to: SAT Preparation Group, LLC Heron Bay Blvd, Unit Coral Springs, FL Printed and bound in the United States of America Page 2

3 Math Module 2: Negative Numbers, Proportions, and the Dreaded Fractions The fearsome nature of negative numbers, proportions, and fractions will be thoroughly addressed in this module. Again, it s not that this information is so difficult to comprehend; the combination of spectacularly inadequate preparation by the school system, both public and private, and the College Board s proclivity for asking fringe type questions makes these types of questions difficult for the beginning SAT student. Have no fear SATPG is here! Effectively Dealing with Negative Numbers There are really only two important issues concerning negative numbers on the SAT and ACT. The first has to do with greater and lesser values and the second involves proper addition and subtraction. Lesser Greater Let s knock out the simpler of the two concepts first. Using the number line above, we can see that as positive numbers move from left to right they become greater. For example: 10 > 5 (10 is greater than 5; the open side of an inequality signifies the greater quantity.) The same is true of negative numbers: the numbers that are further to the right are greater than the numbers that are more to the left. For example: - 5 > From right to left along the number line the quantities become smaller. For example: - 8 < - 3. Be mindful of this as you make decisions on the SAT or ACT. Here s a very simple but often ignored rule to follow when adding or subtracting with negative signs: ALWAYS USE PARENTESES WHEN SUBTRACTING MULTIPLE TERMS TO DISTRIBUTE THE NEGATIVE SIGN PROPERLY. Suppose you had to deal with the following question: when x - y + z - a is subtracted from b the result is thirteen; what would the properly written equation look like? Page 3

4 Reverse order because of the subtracted from b - (x - y + z - a) = 13 Distribute properly b - x + y - z + a = 13 Now you try (Please print these pages out so you can write it out by hand) Put a correct inequality sign in the appropriate blanks ,000-10,000,000 Write the proper equations/fragments for the following: 6. Twice a number minus four is subtracted from the same number is 12: 7. Eighty-two is decreased by the sum of x and y: 8. h minus the difference between g and k: 9. Thirty-eight decreased by x minus fourteen: 10. Four added to z less than w: Answers located on page 12 Page 4

5 Creating Comfortable Fractions On the SAT or ACT, discomfort can often be averted by doing one simple thing: dividing by 1. When fractions are mixed with whole numbers it can be confusing, but dividing by 1 can save the day! Let s suppose that you are were asked to add What could you do? 4 You could change the dynamics of the problem by simply dividing 2 by 1: Now that there are two fractions to deal with in this sample addition problem, there are several non-calculator options to solve this sum. The Same Denominator Method is one of my favorites because it s fast and easy. How do you add these two fractions? 3 + 2, (4) = 8 Change the 2 to 8 to create the same denominator to make adding a breeze. 1 (4) = 11 Remember that when adding fractions with the same denominator, only the top numbers are added. The Bowtie Method is another quick way of adding or subtracting two fractions, particularly ones with different denominators. Let s use the same example as above first and then an additional example with unlike denominators. The procedure for the Bowtie Method is 1) Begin in the upper left of the two fractions 2) Multiply the top left number with the bottom right number and place the result in the numerator Page 5 after an equal sign 3) Carry the sign and place that next to the previous result 4) Multiply the top right element with the bottom left and place that to the right of the carried sign 5) Multiply the two denominators and place that result in the denominator 6) Add/subtract and simplify if possible. Visual representation on next page

6 Step 2 Step = (3)(1) + (2)(4) = = 11 Step Now let s try another example: Step 2 Step = (7)(9) -- (5)(8) = = 23 Step (8)(9) C. When taking the SAT/ACT, choose the method you are most comfortable using. Now you try (Please print these pages out so you can write it out by hand) 1. 31/2 + 5/3: 2. 9/4 -- 1/8: 3. 12/7 + 3/5: 4. 7/8 -- 1/4: 5. 22/5 + 4/15: 6. 8/3 -- 1: 7. 4/7 + 4/21: /2: /8: /6 + 14/48: Answers located on page 12 Page 6

7 Multiplying and Dividing Fractions Let s begin with multiplying fractions, which is easier than adding or subtracting fractions with different denominators. When multiplying two or more fractions, simply multiply all the numerators across the top and multiply all of the denominators across the bottom to produce the desired result. Here s an example of multiplication: 7 x 4 = 28 = 14_ When dividing two fractions (example A) or a fraction and a whole number (example B) flip right-hand fraction and multiply. Let s do a couple of fractional division examples: A) 3 5 = 3 x 6 = 18 = 9_ B) 4 9 = 4 9 = 4 x 1 = 4_ Now you try (Please print these pages out so you can write it out by hand) 17/3 x 6/34: 9/2 4/5: 1/5 4/7: 12/11 2/5: 8/13 20/19: Answers located on page 12 Page 7

8 Proportions A proportion is simply two fractions with an equal sign between them: a = c_ b d Of course, at times, test writers like to combine a fraction on one side of and equation with just a number (in non-fraction form) on the other side of the equation: a = c b The solution to this dilemma is to simply divide the non-fraction side of the equation by 1 in order to make both sides fractions: a = c_ b 1 Here s a very simple rule and process to follow WHENEVER you see or can create ANY proportion: automatically cross-multiply. Proportions can take many forms on the SAT or ACT: averages, midpoints, similar triangles, circle graphs, and ordinary proportions. Just remember to ALWAYS cross-multiply WHENEVER you have or can create ANY proportion. Just in case there is any confusion with the term cross-multiplying, here is an example. Suppose you are faced with cross-multiplying: a = c_. b d 1) multiply top left and bottom right: a x d 2) follow that with an equal sign: a x d = 3) multiply top right with bottom left: a x d = c x b The result would look like the following: ad = cb or da = bc What happens in the case of a more complicated proportion? Try this: Absolutely nothing changes. Automatically cross-multiply. x + 1 = 2y (x + 1) = 3(2y --8) distributing 5x + 5 = 6y combining like terms 5x + 29 = 6y Page 8

9 Complex Ratios Complex ratios is a term that I invented to handle questions involving colons (:) or the word to between numbers. This could look like either the ratio of the three angles of a triangle is 2:3:5 or the ratio of the three angles of a triangle is 2 to 3 to 5. In either case, the correct procedure would be to write the following equation: 2x + 3x + 5x = 180 (You are free to choose any variable that you desire by the way 2n + 3n + 5n = 180 would work equally as well.) Now let s bring this to the next level. Suppose, taking the problem above, we are asked to solve for something more specific: if the ratio of the three angles of a triangle is 2:3:5, what is the difference between the second greatest angle and the least greatest angle? L M G Total 1. Set up and label the problem 2x + 3x + 5x = 180 L = least, M = middle or second greatest, G = greatest) 2. Solve for x 10x = 180 or x = Find the second greatest angle by multiplying 18 x 3 = 54 and the least angle by multiplying 18 x 2 = Solve the problem = 18 Now you try (Please print these pages out so you can write it out by hand) The ratio of a certain chemical is 3 parts carbon to 8 parts hydrogen to 1 part oxygen; out of every 144 atoms how many of them are hydrogen atoms?: The ratio of palm trees to orange trees in a certain grove is 4:3; If there are 84 total trees, how many are orange trees?: The ratio of kittens to puppies to ferrets in a certain pet shop is 5:4:2, respectively (meaning in that same order); if there are 121 kittens, puppies, and ferrets in the store, how many are kittens?: Answers located on page 12 Page 9

10 Reciprocal Relationships Reciprocal relationships is another term that I coined because these types of problems are completely foreign to the typical math classroom. The concept is fairly simple: when a certain quantity of items is multiplied by a constant amount every regular time period, then in the reverse direction the multiplier is the reciprocal of the original multiplier. Let s do an example that illustrates this. Suppose, starting in 1980, that the population of fish in a certain pond has tripled every seven years; if the population of fish in this particular pond was 33,750 in 2011, what was the population of fish in 1980? 1. Set up a timeline Place the total amount of fish under the appropriate year 33, Realize that the total number of fish was tripling (each 7 years times 3) 4. Realize that going backwards is one-third the final number for each step 5. Multiply each backward step times 1/3 until you get to the 1980 value For 1994: x 1/3 = For 1987: x 1/3 = 3750 For 1980: 3750 x 1/3 = 1250, which is the correct answer Now you try (Please print these pages out so you can write it out by hand) If the number of voters in Plant City has quadrupled every three years such that the number of voters in 2011 was 212,480, how many voters were in Plant City in 1999?: Answers located on page 12 Page 10

11 Remainders Questions Remainders are something you might remotely remember doing in the third grade maybe. The SAT remainder questions are quite simple once you know the strategy. Let s begin with a sample question and then analyze it. If a number n is divided by 6 and the remainder is 5, then what is remainder when n + 13 is divided by 6? OK then, let s have at it. 1. Here s how you write it down: n r 5 6 That s right: n over six r 5 (not equals 5) 1. Add the remainder to the denominator to get the least possible value of n 2A. Continuously adding chunks of the denominator to n yields an infinite set of numbers with the same remainder. So, adding results in 11. Easy enough, you might say. Now if 6, the denominator in this case, is added to 11 the result is 17, which also generates a remainder of 5 because 6 goes into 12 with 5 left over. Similarly, when 6 is added to 17 the result is 23; 6 goes into 18 with 5 left over, ad infinitum. 3. Use the first available number, in this case, 11 to set up the next step 4. Write out the new equation: = 24 remainder What if the numerator was a sufficiently large enough number that you may have difficulty dealing with mentally? The answer: do long division; whatever the leftover number at the bottom is the remainder. Now you try (Please print these pages out so you can write it out by hand) When a number y is divided by 7 the remainder is 2, what is the remainder when y + 17 is divided by 7?: Answers located on page 12 Page 11

12 Answers Negative numbers < > > < > x - (2x - 4) = 12 x - 2x + 4 = x = 12 x = (x + y) 82 - x - y h - (g - k) h - g + k 38 - (x - 14) 38 - x + 14 x + 52 z - w + 4 Creating Comfort 103/6 68/32 = 17/8 81/35 20/32 = 5/8 14/3 Multiplying and Dividing 1 45/8 7/20 30/11 38/65 Complex Ratios 3x + 8x + 1x =144 12x = 144 x = 12 (12)(8) = 96 hydrogen atoms 4x + 3x = 84 7x = 84 x = 12 (12)(3) = 36 orange trees 5x + 4x + 2x = x = 121 x = 11 (11)(5) = 55 kittens Reciprocal Relationships 830 Remainders 5 5/3 16/21 5/2 61/8 17/8 Page 12

13 Negative Numbers, Proportions, and Fractional SAT Questions 1: What is the difference between the sum of all positive even integers less than or equal to 60 and the sum of all positive odd integers less than 60? 3: If x = 1/m and y = 1/n and if m = 3 and n = 4, what is the value of 1/x + 1/y? 2: If x/y = ½ and y/z = 2/3, then x/z = (A) 2/3 (B) 1/3 (C) 3/2 (D) 7/6 (E) 6/7 4: If x = 1 + ½ + ¼ + 1/8, and y = 1 + ½a, then how much greater than x is y? (A) ½ (B) ¼ (C) 1/8 (D) 1/16 (E) 1/256 Page 13

14 Negative Numbers, Proportions, and Fractional SAT Questions 5: If y = 5, what is the value of: 2y(y - 3) + 2y(y + 3) 2y (A) 5 (B) 100 (C) 28 (D) 50 (E) 10 7: What is the value of n, if n + 5 = 47 n 42 (A) 210 (B) 52 (C) 47 (D) 42 (E) 215 6: p/q = 9, p/r = 27, and q = 6. What is the value of r? (A) 27 (B) 54 (C) 3 (D) 6 (E) 2 8: If y/n = 3y/18, what is the value of n? (A) 6 (B) 18 (C) 3 (D) 6 (E) 9 Page 14

15 Solutions to Negative Numbers, Proportions, and Fractional SAT Questions 1: What is the difference between the sum of all positive even integers less than or equal to 60 and the sum of all positive odd integers less than 60? Solution (level 4) From 1 to 60, there are 30 even integers and 30 odd. The lowest positive even integer is 2 and the lowest positive odd integer is 1. For each odd integer from 1 to 60, there is an even integer that is one greater. The difference of each of the 25 pairs of integers (2-1, 4-3, etc.) is 1. 1 x 30 = 30 3: If x = 1/m and y = 1/n and if m = 3 and n = 4, what is the value of 1/x + 1/ y? Solution (level 4) To begin, substitute the values for m and n into x and y. Then, make the x and y into equations with fractions on both sides. x/1 = 1/3 and y/1 = 1/4. Cross multiply to get 3x = 1 and 4y = 1. Then divide both sides of each equation by the variables to get 1/x = 3 and 1/y = 4. 1/x + 1/y = = 7 1: If x/y = ½ and y/z = 2/3, then x/z = 4: If x = 1 + ½ + ¼ + 1/8, and y = 1 + ½x, then how much greater than x is y? a) 2/3 b) 1/3 c) 3/2 d) 7/6 e) 6/7 Solution (level 4) To answer this question, begin by cross multiplying to cancel out the fractions. This leaves you with y = 2x and 3y = 2z Since we are looking for x/z. a) ½ b) ¼ c) 1/8 d) 1/16 e) 1/256 Solution (level 4) To begin this problem, convert all of the fractions of x so that they have the same common denominator: 8. x adds up to be 15/8. Therefore, y is Divide 2x = y by 2z. Since 1 + ½(15/8), or 2z = 3y, this comes out to 2x/2z = y/3y. Cancel out the 2 s on the left and the y s on the right to end up with x/z = 1/3, answer (B) 16/ /16. Adding this together gives us y = 31/16. To convert x into an equivalent fraction, multiply by 2/2 2/2(15/8) = 30/16. y - x = 31/16-30/16 = 1/16, answer (D) Page 15

16 Solutions to Negative Numbers, Proportions, and Fractional SAT Questions 4: If y = 5, what is the value of: 2y(y - 3) + 2y(y + 3) 7: What is the value of n, if n + 5 = 47 2y Solution (level 3) n 42 a) 5 To solve this problem, simply a) 210 Solution (level 2) b) 100 c) 28 d) 50 plug in the known value for y and solve the expression. Substituting in the known value for y, we get b) 52 c) 47 d) 42 To answer this question, simply look at the equation and pick out the variable you are looking for. You can split up e) 10 2(5)(5-3) + 2(5)(5 + 3) e) 215 the numerator and the 2(5) Distribute out to get: denominator of the fractions and solve either part independently. The denominator 10(2) + 10(8) of the fractions, n = 42, is 10 Which simplifies to: exactly what the question asks for, so the answer is (D) : If y/n = 3y/18, what is the value of n? = 100. Divide by 10 to get 10, answer (E) 6: p/q = 9, p/r = 27, and q = 6. What is the value of r? a) 27 b) 54 c) 3 d) 6 e) 2 Solution (level 4) Whenever you see proportions and are asked to find the value of a variable, automatically crossmultiply the proportions. In this problem, cross-multiply the equations to get 9q = p and 27r = p. Plug in the known value of q to get 9(6) = p, or p = 54. Plug this value into the other equation to get 27r = 54. Divide both sides by 27 to get r = 2, answer (E) a) 6 b) (18 c) 3 d) 6 e) 9 Solution (level 3) In this problem you must cancel the terms that you can in order to solve for the desired variable. Because there is a y in the denominator of both sides of the equation, the y s cancel out, leaving 1/n = 3/18. Now, cross multiply to get 3n = 18. Divide both sides by 3 to get n = 6, answer (A) Page 16

Chapter 1 Review of Equations and Inequalities

Chapter 1 Review of Equations and Inequalities Part I Review of Basic Equations Recall that an equation is an expression with an equal sign in the middle. Also recall that, if a question asks you to solve