240 MCGRAW-HILL’S SAT If You Can’t Find What You Want, Find What You Can! If you can’t find what you want right away, just look at the parts of the problem one at a time, and find what you can. Often, going step by step and noticing the relationships among the parts will lead you eventually to the answer you need. Lesson 2: Analyzing Problems Break Complicated Problems into Simple Ones Analyzing is key to solving many SAT math problems. Analyzing a problem means simply looking at its parts and seeing how they relate. Often, a complicated problem can be greatly simplified by looking at its individual parts. If you’re given a geometry diagram, mark up the angles and the sides when you can find them. If you’re given algebraic expressions, notice how they relate to one another. 6 feet 6 feet6 feet 6 feet For a certain fence, vertical posts must be placed 6 feet apart with supports in between, as shown above. How many vertical posts are needed for a fence 120 feet in length? You may want to divide 120 by 6 and get 20, which seems reasonable. But how can you check this with- out drawing a fence with 20 posts? Just change the question to a much simpler one to check the relation- ship between length and posts. How many posts are needed for a 12-foot fence? The figure above provides the answer. Obviously, it’s 3. But 12 ÷ 6 isn’t 3; it’s 2. What gives? If you think about it, you will see that di- viding only gives the number of spaces between the posts, but there is always one more post than spaces. So a 120-foot fence requires 20 + 1 = 21 vertical posts. Look for Simple Relationships Once you see the parts of a problem, look for simple relationships between them. Simple relationships usually lead to simple solutions. If 2x 2 + 5y = 15, then what is the value of 12x 2 + 30y? Don’t worry about solving for x and y. You only need to see the simple relationship between the ex- pressions. The expression you’re looking for, 12x 2 + 30y, is 6 times the expression you’re given, 2x 2 + 5y. So, by substitution, 12x 2 + 30y must equal 6 times 15, or 90. AB C D E F G In the figure above, ABCD is a rectangle with area 60, and AB = 10. If E, F, and G are the midpoints of their respective sides, what is the area of the shaded region? This looks complicated at first, but it becomes much simpler when you analyze the diagram. You probably know that the formula for the area of a rec- tangle is a = bh, but the shaded region is not a rec- tangle. So how do you find its area? Analyze the diagram using the given information. First, write the fact that AB = 10 into the diagram. Since the area of the rectangle is 60 and its base is 10, its height must be 6. Then, knowing that E, F, and G are midpoints, you can mark up the diagram like this: AB CD E F G 10 3 333 3 5 5 Notice that the dotted lines divide the shaded re- gion into three right triangles, which are easy to work with. The two bottom triangles have base 5 and height 3 (flip them up if it helps you to see), and the top triangle has base 10 and height 3. Since the formula for the area of a triangle is a = bh, the areas of the triangles are 7.5, 7.5, and 15, for a total area of 30. 1 2 Concept Review 2: Analyzing Problems 1. What does it mean to analyze a problem? 2. Analyze the diagram above by indicating the measures of as many angles as possible. 3. If $20,000 is divided among three people in the ratio of 2:3:5, how much does each person get? 4. If (x)(x − 1)(x − 2) is negative and x is greater than 0, then what can be concluded about x − 1 and x − 2? 95° 20° l 1 l 3 l 2 Given: l 1 || l 2 || l 3 CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 241 242 MCGRAW-HILL’S SAT SAT Practice 2: Analyzing Problems 3. In the sophomore class at Hillside High School, the ratio of boys to girls is 2 to 3. The junior class contains as many boys as the sophomore class does, and the ratio of boys to girls in the junior class is 5 to 4. If there are 200 students in the sophomore class, how many students are there in the junior class? Note: Figure not drawn to scale. 4. In the figure above, ᐉ 1 ʈ ᐉ 2 , AD = 8, EF = 4, GF = 3, and BG = 9. What is the total area of the shaded regions? (A) 32 (B) 36 (C) 40 (D) 42 (E) 44 1. How many odd integers are there between 1 and 99, not including 1 and 99? (A) 46 (B) 47 (C) 48 (D) 49 (E) 50 2. In the figure above, equilateral triangle ABC has an area of 20, and points D, E, and F are the mid- points of their respective sides. What is the area of the shaded region? A B C D E F 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 A B C D EF G l 1 l 2 CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 243 Concept Review 2 1. To analyze a problem means to look at its parts and find how they relate to each other. 2. Your diagram should look like this: 3. If the total is divided in the ratio of 2:3:5, then it is divided into 2 + 3 + 5 = 10 parts. The individual parts, then, are 2/10, 3/10, and 5/10 of the total. Multiplying these fractions by $20,000 gives parts of $4,000, $6,000, and $10,000. 4. If (x)(x − 1)(x − 2) is negative, and x is greater than 0, then (x − 1)(x − 2) must be negative, which means that one of the factors is positive and the other negative. Since x − 2 is less than x − 1, x − 2 must be negative and x − 1 must be positive. Answer Key 2: Analyzing Problems SAT Practice 2 1. C You might start by noticing that every other number is odd, so that if we have an even number of consecutive integers, half of them will always be odd. But this one is a little trickier. Start by solving a simpler problem: How many odd numbers are be- tween 1 and 100, inclusive? Simple: there are 100 consecutive integers, so 50 of them must be odd. Now all we have to do is remove 1, 99, and 100. That removes 2 odd numbers, so there must be 48 left. 2. 5 Don’t worry about finding the base and height of the triangle and using the formula area = (base × height)/2. This is needlessly complicated. Just notice that the four smaller triangles are all equal in size, so the shaded region is just 1/4 of the big triangle. Its area, then, is 20/4 = 5. 3. 144 If the ratio of boys to girls in the sophomore class is 2 to 3, then 2/5 are boys and 3/5 are girls. If the class has 200 students, then 80 are boys and 120 are girls. If the junior class has as many boys as the sophomore class, then it has 80 boys, too. If the ratio of boys to girls in the junior class is 5 to 4, then there must be 5n boys and 4n girls. Since 5n = 80, n must be 16. Therefore, there are 80 boys and 4(16) = 64 girls, for a total of 144 stu- dents in the junior class. 4. C Write what you know into the diagram. Because the lines are parallel, ∠GEF is congruent to ∠GCB, and the two triangles are similar. (To review simi- larity, see Lesson 6 in Chapter 10.) This means that the corresponding sides are proportional. Since GF and BG are corresponding sides, the ratio of corre- sponding sides is 3/9, or 1/3. Therefore, EF is 1/3 of BC, so BC = 12. To find the areas of the triangles, you need the heights of the triangles. The sum of the two heights must be 8, and they must be in a ratio of 1:3. You can guess and check that they are 2 and 6, or you can find them algebraically: if the height of the smaller triangle is h, then the height of the larger is 8 − h. Cross-multiply: 3h = 8 − h Add h: 4h = 8 Divide by 4: h = 2 So the shaded area is (4)(2)/2 + (12)(6)/2 = 4 + 36 = 40. h h8 1 3− = 95° 20° 20° 20° 20° 20° 20° 95° 95° 95° 95° 95° 160° 160° 160° 160° 160° 160° 85° 85° 85° 85° 85° 85° 75° 75° A B C D E F A B C D EF G l 1 l 2 8 4 3 9 2 6 12 244 MCGRAW-HILL’S SAT Lesson 3: Finding Patterns Repeating Patterns Finding patterns means looking for simple rules that relate the parts of a problem. One key to simplifying many SAT math problems is exploiting repetition. If something repeats, you usually can cancel or substitute to simplify. If 5x 2 + 7x + 12 = 4x 2 + 7x + 12, then what is the value of x? This question is much simpler than it looks at first because of the repetition in the equation. If you subtract the repetitive terms from both sides of the equation, it reduces to 5x 2 = 4x 2 . Subtracting 4x 2 from both sides then gives x 2 = 0, so x = 0. Patterns in Geometric Figures Sometimes you need to play around with the parts of a problem until you find the patterns or relationships. For instance, it often helps to treat geometric figures like jigsaw puzzle pieces. The figure above shows a circle with radius 3 in which an equilateral triangle has been inscribed. Three diameters have been drawn, each of which in- tersects a vertex of the triangle. What is the sum of the areas of the shaded regions? This figure looks very complicated at first. But look closer and notice the symmetry in the figure. No- tice that the three diameters divide the circle into six congruent parts. Since a circle has 360°, each of the central angles in the circle is 360° ÷ 6 = 60°. Then no- tice that the two shaded triangles fit perfectly with the other two shaded regions to form a sector such as this: Moving the regions is okay because it doesn’t change their areas. Notice that this sector is 1/3 of the entire circle. Now finding the shaded area is easy. The total area of the circle is a = r 2 = (3) 2 = 9. So the area of 1/3 of the circle is 9/3 = 3. Patterns in Sequences Some SAT questions will ask you to analyze a sequence. When given a sequence question, write out the terms of the sequence until you notice the pattern. Then use whole-number di- vision with remainders to find what the ques- tion asks for. 1, 0, −1, 1, 0, −1, 1, 0, −1, . . . If the sequence above continues according to the pattern shown, what will be the 200th term of the sequence? Well, at least you know it’s either 1, 0, or −1, right? Of course, you want a better than a one-in-three guess, so you need to analyze the sequence more deeply. The sequence repeats every 3 terms. In 200 terms, then the pattern repeats itself 200 ÷ 3 = 66 times with a remainder of 2. This means that the 200th term is the same as the second term, which is 0. What is the units digit of 27 40 ? The units digit is the “ones” digit or the last digit. You can’t find it with your calculator because when 27 40 is expressed as a decimal, it has 58 digits, and your calculator can only show the first 12 or so. To find the units digit, you need to think of 27 40 as a term in the sequence 27 1 , 27 2 , 27 3 , 27 4 , . . . . If you look at these terms in decimal form, you will notice that the units digits follow a pattern: 7, 9, 3, 1, 7, 9, 3, 1, . . . . The sequence has a repeating pattern of four terms. Every fourth term is 1, so the 40th term is also 1. Therefore, the units digit of 27 40 is 1. 3 60° 60° CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 245 Concept Review 3: Finding Patterns Solve the following problems by taking advantage of repetition. 1. If 5 less than 28% of x 2 is 10, then what is 15 less than 28% of x 2 ? 2. If m is the sum of all multiples of 3 between 1 and 100, and n is the sum of all multiples of 3 between 5 and 95, what is m − n? 3. How much greater is the combined surface area of two cylinders each with a height of 4 cm and a radius of 2 cm than the surface area of a single cylinder with a height of 8 cm and a radius of 2 cm? Solve each of the following problems by analyzing a sequence. 4. What is the units digit of 4 134 ? 5. The first two terms of a sequence are 1 and 2. If every term after the second term is the sum of the previous two, then how many of the first 100 terms are odd? 246 MCGRAW-HILL’S SAT SAT Practice 3: Finding Patterns 1. (A) 5 (B) 15 (C) 20 (D) 40 (E) 60 2. Every term of a sequence, except the first, is 6 less than the square of the previous term. If the first term is 3, what is the fifth term of this sequence? (A) 3 (B) 15 (C) 19 (D) 30 (E) 43 −4, 0, 4, −4, 0, 4, −4, 0, 4, . . . 3. If the sequence above continues according to the pattern shown, what is the sum of the first 200 terms of the sequence? (A) −800 (B) −268 (C) −4 (D) 0 (E) 268 If 3 then 6 y x y x+= += 2 10, 4. In the figure above, ABCD is a square and BC = 10. What is the total area of the shaded regions? 5. What is the units digit of 3 40 ? (A) 1 (B) 3 (C) 6 (D) 7 (E) 9 A B CD 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 247 4. 50 Move the shaded regions around, as shown above, to see that they are really half of the square. Since the area of the square is (10)(10) = 100, the area of the shaded region must be half of that, or 50. 5. A The number 3 40 is so big that your calculator is useless for telling you what the last digit is. Instead, think of 3 40 as being an element in the sequence 3 1 , 3 2 , 3 3 , 3 4 , . If you write out the first six terms or so, you will see that there is a clear pattern to the units digits: 3, 9, 27, 81, 243, 729, . . . . So the pattern in the units digits is 3, 9, 7, 1, 3, 9, 7, 1, . . . . The sequence repeats every four terms. Since 40 is a mul- tiple of 4, the 40th term is the same as the 4th and the 8th and the 12th terms, so the 40th term is 1. Concept Review 3 1. Don’t worry about the percent or about finding x. Translate: 5 less than 28% of x 2 is 10 means .28(x 2 ) − 5 = 10 Subtract 10: .28(x 2 ) − 15 = 0 So 15 less than 28% of x 2 is 0. 2. m = 3 + 6 + 9 + + 93 + 96 + 99 n = 6 + 9 + + 93 When you subtract n from m, all the terms cancel except 3 + 96 + 99 = 198. extra bases. Each base has an area of π(2) 2 = 4π, so the surface area of the smaller cylinders is 2(4π) = 8π greater than that of the larger cylinder. 4. Your calculator is no help on this one because 4 134 is so huge. Instead, think of 4 134 as a term in the sequence 4 1 , 4 2 , 4 3 , 4 4 , . . . . What is the units digit of 4 134 ? If you write out the first few terms, you will see a clear pattern to the units digits: 4, 16, 64, 256, . . . . Clearly, every odd term ends in a 4 and every even term ends in a 6. So 4 134 must end in a 6. 5. The first few terms are 1, 2, 3, 5, 8, 13, 21, . . . . Since we are concerned only about the “evenness” and “oddness” of the numbers, think of the sequence as odd, even, odd, odd, even, odd, odd, even, . . . . No- tice that the sequence repeats every three terms: (odd, even, odd), (odd, even, odd), (odd, even, odd), . . . . In the first 100 terms, this pattern re- peats 100/3 = 33 1 ⁄3 times. Since each pattern con- tains 2 odd numbers, the first 33 repetitions contain 66 odd numbers and account for the first 99 terms. The last term must also be odd because each pattern starts with an odd number. There- fore, the total number of odds is 66 + 1 = 67. Answer Key 3: Finding Patterns SAT Practice 3 1. 2. A If every term is 6 less than the square of the previous term, then the second term must be (3) 2 − 6 = 9 − 6 = 3. The third term, then, is also (3) 2 − 6 = 3, and so on. Every term, then, must be 3, including the fifth. 3. C The sequence repeats every three terms: (−4, 0, 4), (−4, 0, 4), (−4, 0, 4), . . . . Each one of the groups has a sum of 0. Since 200/3 = 67 2 ⁄3, the first 200 terms contain 67 repetitions of this pattern, plus two extra terms. The 67 repetitions will have a sum of 67(0) = 0, but the last two terms must be −4 and 0, giving a total sum of −4. C 6 y x y x += + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = () =2 3 2 210 20 3. Don’t calculate the total surface area. Instead, just notice that the two small cylinders, stacked to- gether, are the same size as the large cylinder. But remember that you are comparing surface areas, not volumes. The surface areas are almost the same, except that the smaller cylinders have two 2 2 8 4 4 A B CD 248 MCGRAW-HILL’S SAT Lesson 4: Simplifying Problems Beeline, Substitute, Combine, and Cancel When a problem seems overwhelming, try one of these four simplification strategies: beelining, substituting, combining, and canceling. Look for the Beeline—The Direct Route Many SAT problems have “beelines”—direct paths from the given information to the answer. We sometimes miss the “beeline” because we get trapped in a knee-jerk response—for in- stance, automatically solving every equation or using the Pythagorean theorem on every right triangle. Avoid the knee-jerk response. Instead, step back and look for the “beeline.” If and what is the value of ? This problem looks tough because of all the un- knowns. You might do the knee-jerk thing and try to solve for a, b, and c. Whoa, there! Step back. The ques- tion doesn’t ask for a, b, and c. It asks for a fraction that you can get much more directly. Notice that just mul- tiplying the two given fractions gets you almost there: . This is close to what you want—all you have to do is divide by 3 to get . Substituting the given values of the fractions gives you which is the value of . Simplify by Substituting The simplest rule in algebra is also the most powerful: Anything can be substituted for its equal. When you notice a complicated expres- sion on the SAT, just notice if it equals some- thing simpler, and substitute! If 3x 2 + 5x + y = 8 and x ≠ 0, then what is the value of ? Again, take a deep breath. Both the equation and the fraction look complicated, but you can simplify by 16 2 35 2 − + y xx a c10 1 4 33 1 4 ×÷= , a c10 3 25 3 10 a b b c a c ×= a c10 b c5 3= , 3 2 1 4 a b = just remembering that anything can be substituted for its equal. Notice that 3x 2 + 5x appears in both the equation and the fraction. What does it equal? Subtract y from both sides of the equation to get 3x 2 + 5x = 8 − y. If you substitute 8 − y for 3x 2 + 5x in the fraction, you get Nice! Simplify by Combining or Canceling Many algebraic expressions can be simplified by combining or canceling terms. Always keep your eye out for like terms that can be com- bined or canceled and for common factors in fractions that can be canceled. If m and n are positive integers such that m > n and what is the value of m + n? To simplify this one, it helps to know a basic factor- ing formula from Chapter 8, Lesson 5: m 2 – n 2 = (m – n) (m + n). If you factor the numerator and denominator of the fraction, a common factor reveals itself, and it can be canceled: Since m + n must equal 9. If f(x) = 2x 2 – 5x + 3 and g(x) = 2x 2 + 5x + 3, then for how many values of x does f(x) = g(x)? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Remember the simple rule that anything can be substituted for its equal, and then cancel to simplify. Since f(x) = g(x), you can say that 2x 2 – 5x + 3 = 2x 2 + 5x + 3. Subtract 2x 2 and 3: −5x = 5x Add 5x:0 = 10x Divide by 10: 0 = x So the answer is (A) 0, right? Wrong! Remember that the question asks for how many values of x are the function values equal. Since we only got one solution for x, the answer is (B) 1. mn+ = 2 9 2 , mn mn mnmn mn mn 22 22 2 2 − − = −+ − = +()() () . mn mn 22 22 9 2 − − = , 16 2 8 28 8 2 − − = − − = y y y y () . CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 249 Concept Review 4: Simplifying Problems Simplify the following expressions. 1. −2n − (6 − 5n) − n 2. 3. 4. 5. Solve the following problems with substitution. 6. If y = 1 − x and 2y + x = 5, then what is the value of x? 7. If 3 + m + n = n 2 + m 2 , what is the value of ? 8. For all real numbers x, let <x> = (1 − x) 2 . What is the value of <<4>>? nnmm 22 6 − () +− () xx 6 6410 2 32 xx x x ++ 21 2x + 218 23 2 2 x xx − +− . 2? 95° 20° l 1 l 3 l 2 Given: l 1 || l 2 || l 3 CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 241 242 MCGRAW-HILL’S SAT SAT Practice 2: Analyzing Problems 3. In the sophomore class at Hillside. 40. h h8 1 3− = 95° 20° 20° 20° 20° 20° 20° 95° 95° 95° 95° 95° 160° 160° 160° 160° 160° 160° 85° 85° 85° 85° 85° 85° 75° 75° A B C D E F A B C D EF G l 1 l 2 8 4 3 9 2 6 12 244 MCGRAW-HILL’S SAT Lesson 3: Finding Patterns Repeating Patterns Finding patterns means looking for simple rules that relate the parts of a problem. One key to simplifying many SAT math. 240 MCGRAW-HILL’S SAT If You Can’t Find What You Want, Find What You Can! If you can’t find what you want right away, just look at the parts of the problem one at a time, and