Sat - MC Grawhill part 27 pot

250 MCGRAW-HILL’S SAT SAT Practice 4: Simplifying Problems 1. If 3y − 4z = 6 and 2y + z = 10, then y − 5z = (A) 12 (B) 6 (C) 4 (D) 0 (E) −4 2. If a + b = 3, a + c = 5, and b + c = −6, then a + b + c = (A) 4 (B) 2 (C) 1 (D) 0 (E) −1 3. If and , what is the value of ? (A) (B) (C) (D) 2 (E) 8 5 3 3 5 1 2 b c ac c + = 5 ab b + = 3 Questions 4 and 5 pertain to the following definition: For all non-zero real numbers k, let . 4. Which of the following is equivalent to ^3 − ^2? (A) (B) (C) ^1 (D) (E) ^2 5. Other than 0, what is the only value of k for which ^^k is undefined? ^ 6 5 ∧ 5 6 ∧ 1 6 ∧ =−k k 1 1 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 251 4. D Begin by simplifying ^3 − ^2 by substitution: But be careful not to pick (A) because Notice that the choices must be evaluated first be- fore we can see which one equals 1/6. Notice that choice (D) is 5. 1 Begin by simplifying ^^k by substitution: Yikes! That doesn’t look simple! But think about it: why is it that k can’t be 0? Because division by 0 is undefined, and k is in a denominator. But notice that is also in a denominator, so can’t be 0, either! Solving: Add : Multiply by k: k ≠ 1 Then check by noticing that ^^1 is undefined: ^^1 = ^(1 −1/1) = ^0 =1 − 1/0, and 1/0 is undefined. 1 1 − k 1 k 1 1 0−≠ k 1 1 − k 1 1 − k ∧∧ ∧ =− ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =− − k k k 1 1 1 1 1 1 ∧ =− =65 1 56 16. ∧ =− =−16 1 6 5 ∧ 16 ∧∧ −=− () −− () =−=3 2 1 13 1 12 23 12 16 Concept Review 4 1. −2n − (6 − 5n) − n =−2n − 6 + 5n − n = 2n − 6 2. 3. 4. 5. 6. Substitute y = 1 − x into 2y + x = 5 to get 2(1 − x) + x = 5 xx x x x 634 =× = 6410 2 6 2 4 2 10 2 325 32 3 2 2 xx x x x x x x x x xx ++ =++=++ 21 2 4 22 4 2xx x x x x +=+= + 218 23 23 3 31 23 2 2 x xx xx xx x x − +− = − () + () + () − () = − () −− () 1 Distribute: 2 − 2x + x = 5 Simplify: 2 − x = 5 Subtract 2: −x = 3 Multiply by −1: x =−3 7. 3 + m + n = n 2 + m 2 Subtract n and m: 3 = (n 2 − n) + (m 2 − m) Substitute: 8. <<4>> Substitute for <4>: = <(1 − 4) 2 > Simplify: = <(−3) 2 > Simplify: = <9> Substitute for <9>: = (1 − 9) 2 Simplify: = (−8) 2 Simplify: = 64 nnmm 22 6 3 6 1 2 − () +− () == Answer Key 4: Simplifying Problems SAT Practice 4 1. E Subtract the equations: 3y − 4z = 6 2. C Add the equations: (a + b) + (a + c) + (b + c) = 3 + 5 +−6 Simplify: 2a + 2b + 2c = 2 Divide by 2: a + b + c = 1 3. D Start by simplifying the expressions: Substituting into the original equations gives Subtract 1: Divide the fraction: Simplify: Simplify: b c = 2 a c b a ×=2 a cc a b = 4 2 a b a c ==24and a b a c += +=13 15and ab b a b b b a b ac b a c c c a c + =+=+ + =+=+ 1 1 −+= ( ) −=− 210 54 yz yz 252 MCGRAW-HILL’S SAT Lesson 5: Connecting to Knowledge Know What You Need Some SAT math questions require you to use special formulas or know the definitions of special terms. Fortunately, you won’t need to memorize very many formulas (none of that trig stuff, for instance), and some of the most important ones are given to you right on the test! Reference Information Every SAT math section gives you this reference information. Check it out and use it when you need it. The arc of a circle measures 360°. Every straight angle measures 180°. The sum of the measures of the angles in a triangle is 180°. Memorize the Key Formulas They DON’T Give You It’s awfully nice of the SAT to give you those formu- las, but those are not quite all you’ll need. Fortu- nately, we can fit the other key formulas on a single page. Here they are: Rate formula (Chapter 9, Lesson 4): Distance (or work) = rate × time Average (arithmetic mean) formulas (Chapter 9 Lesson 2): Average = Sum = average × number of things Slope formula (Chapter 10, Lesson 4): Slope = Midpoint formula (Chapter 10, Lesson 4): Midpoint = Percent change formula (Chapter 7, Lesson 5): Percent change = final starting starting − × %100 xxyy 1212 22 ++ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ , rise run yy xx = − − 21 21 sum number of things Memorize the Key Definitions You’ll also want to memorize the definitions of some key terms that show up often: Mode = the number that appears the most fre- quently in a set. Remember that mode and most both begin with mo (Chapter 9, Lesson 2). Median = the “middle number” of a set of numbers when they are listed in order. If there are an even number of numbers, the median is the average of the two middle numbers (Chapter 9, Lesson 2). Remainder = the whole number left over when one whole number has been divided into an- other whole number a whole number of times (Chapter 7, Lesson 7). Absolute value = the distance a number is from 0 on the number line (Chapter 8, Lesson 6). Prime number = an integer greater than 1 that is divisible only by itself and 1 (Chapter 7, Lesson 7). Factor = a number or expression that is part of a product. (Product = result of a multiplication.) r A = πr 2 C = 2πr ᐉ w A = ᐉw V = ᐉwh V = πr 2 h Special right trianglesc 2 = a 2 + b 2 A = 1 / 2 bh h b ᐉ w h r h b c a 2x x x s s s 3 2 30° 60° 45° 45° CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 253 Concept Review 5: Connecting to Knowledge Write out each formula, theorem, definition, or property. 1. The Pythagorean theorem 2. The zero product property 3. The parallel lines theorem 4. The rate formula 5. The average (arithmetic mean) formula 6. The definition of the median 7. The definition of the mode 8. The circumference formula 9. The circle area formula 10. The triangle area formula 254 MCGRAW-HILL’S SAT SAT Practice 5: Connecting to Knowledge 1. If x is the average (arithmetic mean) of k and 10, and y is the average (arithmetic mean) of k and 4, what is the average of x and y, in terms of k? (A) (B) (C) (D) 7k (E) 14k 2. If, on average, x cars pass a certain point on a highway in y hours, then, at this rate, how many cars should be expected to pass the same point in z hours? (A) xyz (B) (C) (D) (E) Note: Figure not drawn to scale. 3. A straight 8-foot board is resting on a rectangular box that is 3 feet high, as shown in the diagram above. Both the box and the board are resting on a horizontal surface, and one end of the board rests on the ground 4 feet from the edge of the box. If h represents the height, in feet, of the other end of the board from the top of the box, what is h? x y z xz y z x y xy z k + 7 2 k +14 2 k +14 4 Key formula(s): Key formula(s): Key formula(s): 3 feet 4 feet 8 feet h feet 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 255 3. 1.8 Key formula: The Pythagorean theorem: c 2 = a 2 + b 2 . Key theorem: In similar triangles, cor- responding sides are proportional. Notice that the figure has two right triangles, and they are similar. The hypotenuse of the bottom triangle is 5 because 3 2 + 4 2 = 5 2 . Therefore, the hypotenuse of the top triangle is 8 − 5 = 3. Since the two triangles are similar, the corresponding sides are proportional: Cross-multiply: 5h = 9 Divide by 5: h = 1.8 3 53 = h Concept Review 5 1. The Pythagorean theorem: In a right triangle, if c is the length of the hypotenuse and a and b are the lengths of the two legs, then c 2 = a 2 + b 2 (Chapter 10, Lesson 3). 2. The zero product property: If a set of numbers has a product of zero, then at least one of the numbers is zero. Conversely, if zero is multiplied by any number, the result is zero (Chapter 8, Lesson 5). 3. The parallel lines theorem: If a line cuts through two parallel lines, then all acute angles formed are congruent, all obtuse angles formed are con- gruent, and any acute angle is supplementary to any obtuse angle (Chapter 10, Lesson 1). 4. The rate formula: distance (or work) = rate × time (Chapter 9, Lesson 4). 5. The average (arithmetic mean) formula: Average = sum ÷ number of things (Chapter 9, Lesson 2). 6. The definition of the median: The “middle num- ber” of a set of numbers when they are listed in order. If there are an odd number of numbers, the median is the “middle number,” and if there are an even number of numbers, it is the average of the two middle numbers (Chapter 9, Lesson 2). 7. The definition of the mode: The number that ap- pears the most frequently in a set (Chapter 9, Lesson 2). 8. The circumference formula: Circumference = 2πr (Chapter 10, Lesson 5). 9. The circle area formula: Area =πr 2 (Chapter 10, Lesson 5). 10. The triangle area formula: Area = base × height/2 (Chapter 10, Lesson 5). Answer Key 5: Connecting to Knowledge SAT Practice 5 1. C Key formula: Average = sum ÷ number of things. So if x is the average of k and 10, then . And if y is the average of k and 4, then y = . The average of x and y, then, is 2. D Key formulas: Number of cars = rate ×time, and Since x is the number of cars and y is the time in hours, the rate is x/y cars per hour. Using the first formula, then, the number of cars that would pass in z hours is You should notice, too, that simply plugging in values for x, y, and z can make the problem easier to think about. Say, for instance, that x = 10 cars pass every y = 2 hours. In z = 4 hours, then, it should be clear that 20 cars should pass by. Plug- ging these numbers into the choices, you will see that (D) is the only one that gives an answer of 20. xy z xz y cars per hour hours () () = rate number of cars time = . kk kk kk + + + = + + + = + = + 10 2 4 2 2 10 4 4 4 214 4 7 2 k + 4 2 x k = +10 2 3 feet 4 feet 8 feet h feet 3 feet 5 feet 256 MCGRAW-HILL’S SAT Keep Your Options Open There are often many good ways to solve an SAT math problem. Consider different strate- gies. This gives you a way to check your work. If two different methods give you the same an- swer, you’re probably right! Numerical Analysis—Plugging In Let’s come back to the problem we saw in Lesson 4: If 3x 2 + 5x + y = 8 and x≠ 0, then what is the value of ? Back in Lesson 4 we solved this using substitution, an algebraic method. Now we’ll use a numerical method. Notice that the equation contains two unknowns. This means that we can probably find solutions by guessing and checking. Notice that the equation works if x = 0 and y = 8. But—darn it—the problem says x ≠ 0! No worries— notice that x = 1 and y = 0 also work. (Check and see.) Now all we have to do is plug those numbers in for x and y: . Same answer, whole different approach! “Plugging in” works in two common situations: when you have more unknowns than equations and when the answer choices contain un- knowns. Always check that your numbers sat- isfy the conditions of the problem. Then solve the problem numerically, and write down the answer. If the answer choices contain un- knowns, plug the values into every choice and eliminate those that don’t give the right answer. If more than one choice gives the right answer, plug in again with different numbers. If 3m = mn + 1, then what is the value of m in terms of n? (A) n + 1 (B) n − 2 (C) (D) (E) Because the choices contain unknowns, you can plug in. Pick a simple number for m to start, such as 1. Plugging into the equation gives 3 = n + 1, which has the 2 3 + n 1 3 + n 1 3 − n 16 2 35 16 2 0 31 51 16 8 2 22 − + = − + == y xx () () () 16 2 35 2 − + y xx solution n = 2. Now notice that the question asks for m, which is 1. Write that down and circle it. Now substitute 2 for n in the choices and see what you get: (A) 3 (B) 0 (C) 1 (D) 1/5 (E) 2/5 Only (C) gives the right answer. Algebraic Analysis You can also solve the problem above algebraically: 3m = mn + 1 Subtract mn:3m − mn = 1 Factor: m(3 − n) = 1 Divide by (3 − n): m = Testing the Choices Some SAT math questions can be solved just by “testing” the choices. Since numerical choices are usually given in order, start by testing choice (C). If (C) is too big, then (D) and (E) are too big, also, leaving you with just (A) and (B). This means that you have only one more test to do, at most, until you find the answer. If 3(2) n+1 – 3(2) n = 24, what is the value of n? (A) 2 (B) 3 (C) 4 (D) 5 (E) 6 Here you can take an algebraic or a numerical ap- proach. That is, you can solve the equation for n or you can “test” the choices to see if they work. For this les- son, we’ll try the “testing” strategy. Since the choices are given in ascending order, we’ll start with the mid- dle number, (4). Substituting 4 for n gives us 3(2) 5 – 3(2) 4 on the left side, which equals 48, not 24. (It’s okay to use your calculator!) Since that doesn’t work, we can eliminate choice (C). But since it’s clearly too big, we can also rule out choices (D) and (E). That’s why we start with (C)—even if it doesn’t work, we still narrow down our choices as much as possible. Now just test ei- ther (A) or (B). Notice that (B) gives us 3(2) 4 – 3(2) 3 , which equals 24, the right answer. Now try solving the problem algebraically, and see if it’s any easier! 1 3 − n Lesson 6: Finding Alternatives CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 257 Concept Review 6: Finding Alternatives 1. When can a multiple-choice problem be solved by just “testing the choices”? 2. When solving by testing the choices, why is it often best to start with choice (C)? 3. When testing the choices, when is it not necessarily best to start with choice (C)? 4. When can you simplify a multiple-choice question by plugging in values? 5. What are the four steps to solving by plugging in values? 6. Why is it best to understand more than one way to solve a problem? 258 MCGRAW-HILL’S SAT SAT Practice 6: Finding Alternatives Try to find at least two different ways of solving each of the following problems, and check that both meth- ods give you the same answer. 1. If m = 2x − 5 and n = x + 7, which of the follow- ing expresses x in terms of m and n? (A) (B) m − n + 2 (C) (D) m − n + 12 (E) 2(m − n + 12) 2. Three squares have sides with lengths a, b, and c. If b is 20% greater than a and c is 25% greater than b, then by what percent is the area of the largest square greater than the area of the small- est square? (A) 20% (B) 50% (C) 75% (D) 125% (E) 225% 3. Jim and Ellen together weigh 290 pounds. Ellen and Ria together weigh 230 pounds. All three to- gether weigh 400 pounds. What is Ellen’s weight? (A) 110 lbs (B) 120 lbs (C) 130 lbs (D) 140 lbs (E) 170 lbs mn−+12 2 mn−+2 2 4. A painter used one-fourth of her paint on one room and one-third of her paint on a second room. If she had 10 gallons of paint left after painting the second room, how many gallons did she have when she began? (A) 19 (B) 24 (C) 28 (D) 30 (E) 50 5. If and 4r = 7t, what is the value of s in terms of t? (A) 35t (B) (C) 35t − 4 (D) 31t (E) 70t 35 4 t r s = 5 CHAPTER 6 / WHAT THE SAT MATH IS REALLY TESTING 259 test choice (B) next. If Ellen weighs 120 pounds, then Jim weighs 290 − 120 = 170 pounds and Ria weighs 230 − 120 = 110 pounds. In total, they weigh 120 + 170 + 110 = 400 pounds. Bingo! The answer is (B). Method 2: Use algebra. Let e = Ellen’s weight, j = Jim’s weight, and r = Ria’s weight. Translate the problem into equations: e + j = 290 e + r = 230 e + j + r = 400 Add first two equations: 2e + j + r = 520 Subtract third equation: −(e + j + r = 400) e = 120 4. Method 1: Use algebra. Let x be the number of gal- lons of paint that she starts with. Translate the problem into an equation: x − (1/4)x − (1/3)x = 10 Simplify: (5/12)x = 10 Multiply by 12/5: x = 24 The answer is (B). Method 2: Test the choices. Look at the problem carefully. She uses one-fourth of her paint, then one-third of her paint, and is left with 10 gallons of paint, a whole number. This suggests that she started with a quantity that is divisible by 3 and 4. Since 24 is divisible by 3 and 4, it’s a good choice to test. One-fourth of 24 is 6, and one-third of 24 is 8. This means she would be left with 24 − 6 − 8 = 10. Bingo! 5. Method 1: Plug in. Let s = 35, so r = 35/5 = 7 and t = 4. (Check that they “fit.”) Since the question asks for the value of s, write down 35 and circle it. Plugging these values into the choices gives (A) 140 (B) 35 (C) 136 (D) 124 (E) 280 The answer is (B). Method 2: Use algebra. Solve the first equation for s: s = 5r. Then solve the second equation for r: r = (7/4)t. Then substitute: s = 5(7/4)t = 35t/4. Concept Review 6 1. There are many situations in which this is possi- ble, but perhaps the most common is where you’re asked to find the solution of an equation, and the choices are ordinary numbers. 2. Because the answer choices are usually presented in numerical order. If choice (C) doesn’t work, you may be able to tell whether it is too big or too small, and thereby eliminate two other answers as well. This way, you will only need to “test” one more choice to get the answer. 3. When it is not easy to tell whether the choice is “too big or too small,” or when there is no pattern to the choices. 4. Usually, when the answer choices contain un- knowns or represent ratios of unknowns; also when the problem contains more unknowns than equations. 5. (1) Check that the values satisfy any given equa- tions or other restrictions, (2) write down the values you are plugging in for each unknown, (3) solve the problem numerically and write down this number, and (4) plug in the values to every choice and eliminate those that don’t give the right answer. If more than one choice gives the right answer, plug in different numbers. 6. Because the two methods can provide a “check” against one another: if they both give the same an- swer, you are almost certainly right! Answer Key 6: Finding Alternatives SAT Practice 6 1. Method 1: The problem asks you to solve for x in terms of m and n. Notice that every choice con- tains the expression m − n. By substitution: m − n = (2x − 5) − (x + 7) Simplify: m − n = x − 12 Add 12: m − n + 12 = x So the answer is (D). Method 2: Just plug in simple values for the un- knowns. If x = 1, then m = (2)(1) − 5 =−3 and n = (1) + 7 = 8. Since the problem asks for x, write down its value, 1, and circle it. Then plug in m = −3 and n = 8 to every choice, and simplify: (A) −4.5 (B) −9 (C) 0.5 (D) 1 (E) 2 So the answer is (D). 2. Method 1: Plug in numbers. Let a be 100. If b is 20% greater than a, then b = 120. If c is 25% greater than b, then c = 150. The area of the largest square, then, is (150) 2 = 22,500, and the area of the smallest square is (100) 2 = 10,000. The percent difference is (22,500 − 10,000)/10,000 = 1.25 = 125% (D). Method 2: Use algebra. b = 1.2a and c = (1.25)(1.2a) = 1.5a. So the area of the smallest square is a 2 and the area of the largest square is (1.5a) 2 = 2.25a 2 . Since 2.25a 2 − a 2 = 1.25a 2 , the area of the bigger square is 125% larger. 3. Method 1: Test the choices, starting with (C). If Ellen weighs 130 pounds, then Jim weighs 290 − 130 = 160 pounds and Ria weighs 230 − 130 = 100 pounds. All together, their weight would be 130 + 160 + 100 = 390 pounds. Close, but too small. This means that our guess for Ellen’s weight is too big (because increasing Ellen’s weight decreases both Jim and Ria’s weight by the same amount, for a net decrease). This lets us eliminate choices (C), (D), and (E). Since our guess wasn’t far off, it makes sense to . feet 8 feet h feet 3 feet 5 feet 256 MCGRAW-HILL’S SAT Keep Your Options Open There are often many good ways to solve an SAT math problem. Consider different strate- gies. This gives you a way to. 15and ab b a b b b a b ac b a c c c a c + =+=+ + =+=+ 1 1 −+= ( ) −=− 210 54 yz yz 252 MCGRAW-HILL’S SAT Lesson 5: Connecting to Knowledge Know What You Need Some SAT math questions require you to use special formulas or. The circumference formula 9. The circle area formula 10. The triangle area formula 254 MCGRAW-HILL’S SAT SAT Practice 5: Connecting to Knowledge 1. If x is the average (arithmetic mean) of k and