39. Marguerite is remodeling her bathroom floor. Each imported tile measures 1 ᎏ 2 7 ᎏ inch by 1 ᎏ 4 5 ᎏ inch. What is the area of each tile? a. 1 ᎏ 3 8 5 ᎏ square inches b. 1 ᎏ 1 3 1 5 ᎏ square inches c. 2 ᎏ 1 3 1 5 ᎏ square inches d. 3 ᎏ 3 3 5 ᎏ square inches e. 4 ᎏ 3 1 2 ᎏ square inches 40. If Deirdre walks from Point A to Point B to Point C at a constant rate of 2 mph without stopping, what is the total time she takes? a. (x + y) × 2 b. 2x + 2y c. xy Ϭ 2 d. (x + y) Ϭ 2 e. xy 2 A BC x miles y miles –THE SAT MATH SECTION– 157 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 157 Five-Choice Answers 1. c. After reading the problem, you realize that the amount of robins at the feeder is related to the number of cardinals. When this is the case, you have to set up a legend or a key defining the number of cardinals. Then, relate it to the number of robins. Step 1: Let x = number of cardinals Let 3x = number of robins Step 2: Now, you have to come up with a for- mula to solve for x. (number of cardinals) + (number of robins) = total number of birds x + 3x = 20 4x = 20 x = 5 Step 3: Take your answer for x and substitute it back into the legend. x = number of cardinals 3x = number of robins 5 = number of cardinals 3(5) = number of robins 15 = number of robins The answer is choice c. 2. b. You should realize that since the question asks for a factor of 4x 2 – 9, you have to begin factoring this expression. The question you should ask yourself is: Which method of factoring should I use? Looking at the expression 4x 2 – 9, you should notice that there are no common fac- tors between the two terms. Therefore, you cannot factor out a common factor. Also, this is not a trinomial. Thus, you cannot factor this expression like a trinomial. The method that you have to use in fac- toring is the difference of two perfect squares. There are a couple of hints in the prob- lem that clue you in on this. First, there are only two terms. Second, the operation between the two terms is subtraction. Remember, the word difference means subtraction. Now, we have to factor 4x 2 – 9. 4x 2 – 9 = (2x + 3) (2x – 3) This is now factored and we can see that one of the factors is choice b,(2x + 3). The answer is choice b. 3. a. The first thing you should remember is that the sum of the angles of a triangle equals 180°. m∠A + m∠B + m∠C = 180° Now, you can substitute the values of each angle into the formula and solve for y. Step 1: 90 + y + 40 + 3y – 10 = 180 Step 2: 4y + 120 = 180 – 120 – 120 Step 3: ᎏ 4 4 y ᎏ = ᎏ 6 4 0 ᎏ Step 4: y = 15 Next, you have to substitute the y value back into your angle measure in order to find out the degree measure of each angle. m∠A = 90 m∠B = y + 40 = (15) + 40 = 55 m∠C = 3y – 10 = 3(15) – 10 =35 The three angle measures are 90, 55, and 35, respectively, and their sum is 180. Finally, you have to look at your answer choices and determine what type of right tri- angle this is. Choice a is scalene. You remember that a scalene right triangle is a right triangle that has three sides of different length and three angles of different measure. The triangle in your problem fits this definition, but check the other three choices before you settle on choice a. –THE SAT MATH SECTION– 158 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 158 Choice b is isosceles. An isosceles right triangle has two base angles that are equal. This is not the correct answer. Choice c is equilateral. This is not an equilateral triangle since the three angles are not equal. Choice d is obtuse. An obtuse right tri- angle cannot exist since the angles of a trian- gle add up to 180°. An obtuse angle is between 90° and 180°, and if you add this to a right angle, you have a sum over 180°. Therefore, choice d is incorrect. The answer is choice a. 4. d. Remember, when multiplying radical expres- sions, multiply terms outside the radical together and multiply terms inside the radical together. The formula is demonstrated below. a͙b ෆ × c͙d ෆ = (a × c)͙b ෆ × ͙d ෆ Therefore, (͙x ෆ )(͙2x ෆ ) = ͙x × 2x ෆ = ͙2x 2 ෆ Of course, you remember that x × 2x = 2x 2 because when multiplying terms with like bases, you add the exponents. This is your answer, ͙2x 2 ෆ ;however,it is not one of the choices. Therefore, you must reduce this expression to simplest form in order to match your answer to one of the choices. Remember, you can break a radical down into separate terms: ͙2x 2 ෆ = ͙2 ෆ × ͙x 2 ෆ ͙2 ෆ cannot be simplified, but ͙x 2 ෆ can be simplified to x. ͙2 ෆ × ͙x 2 ෆ = ͙2 ෆ × x = ͙2x ෆ This answer is not listed as one of the four choices. Yet, x͙2 ෆ is choice d. Multipli- cation is commutative, so ͙2 ෆ (x) = x͙2 ෆ . Important: Many times, the answer that you calculate will not be one of the answers listed in a multiple-choice problem. Trust your work if you have done it correctly. You may have to manipulate your answer or sim- plify it so that it matches one of the answers. The answer is choice d. 5. c. This involves multiple probabilities. The first thing you should try to figure out is what you are trying to find. The spinning of the spinner is an inde- pendent event each time. This means that the outcome on the first trial does not influ- ence the outcome of the second trial. Now, you must figure out the probabili- ties of each event. When trying to figure out the probabilities of multiple events, a tree diagram is most helpful. This is shown below. The first two branches represent the first event, G or not G. The second set of branches represents what can happen on the second spin of the spinner, B or not B. You are interested in the events of the top branch, G then B. These probabilities are P(G then B) = ᎏ 1 4 ᎏ × ᎏ 1 2 ᎏ = ᎏ 1 8 ᎏ . You can calculate the P(G) by noticing that G is ᎏ 1 4 ᎏ of the circle. Likewise, you can find the P(B) by noticing B is ᎏ 1 2 ᎏ of the circle. Since you want the event (G then B), you multiply the probabilities to get ᎏ 1 8 ᎏ . You can also find the probabilities of G and B, respectively, by subdividing area B into two sections as shown on the following page. G B ~B B ~B ~G –THE SAT MATH SECTION– 159 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 159 The P(G) = ᎏ 1 4 ᎏ . However, P(B) = ᎏ 2 4 ᎏ . If you do the problem this way, you will find that: P(G then B) = P(G) × P(B). ᎏ 1 4 ᎏ × ᎏ 2 4 ᎏ = ᎏ 1 2 6 ᎏ or ᎏ 1 8 ᎏ . The answer is choice c. 6. d. Method 1: There are two different ways you can approach this problem. The first way is to identify the four choices and see which one looks like a mathematical property that you remember. If you can do this, you will clearly see that choice d is the commutative property and this is always true. If you can’t remember or don’t see that choice d is the commutative property, don’t worry. You can always solve the problem another way. Method 2: You can solve this problem by sub- stituting values in for a and b. Then, see which answer choice is true. Let’s see how this works: Let a and b equal 2 and 3, respectively. Choice a: ᎏ 2 3 ᎏ ≠ ᎏ 3 2 ᎏ Choice b: 2 + 2(3) = 3 + 2(2) 2 + 5 = 3 + 4 7 = 7 This looks like this answer choice might be true. You should try two different values just to be sure. Let a and b equal 1 and 2, respectively. 1 + 2(2) = 2 + 2(1) 1 + 4 ≠ 2 + 2 5 ≠ 4 So, by double-checking your choice, you can see that choice b is not true. Choice c: 2 – 3 × 3 – 2 Therefore, choice c is false. Now, you know that choice d must be the answer. You should check it to be sure. Choice d: 2 + 3 = 3 + 2 1 + 2 = 2 + 1 5 = 5 3 = 3 Therefore, the answer is choice d. 7. a. The first thing that you have to realize is that the question requires you to simplify the expression. If you are simplifying rational expres- sions (rational expressions look like frac- tions), you have to factor the numerator and denominator if possible. Step 1: Factor the numerator, x 2 + 2x. There is a common factor of x between the two terms. Then, factor out an x and place it out- side a pair of parentheses. Then, divide x 2 and 2x, respectively, by x in order to find out what terms are on the inside of the parentheses. So, x 2 + 2x becomes x(x + 2). Step 2: The denominator cannot be fac- tored. Therefore, you can now cancel out the like terms between the numerator and denominator. ᎏ x(x x +2) ᎏ = (x + 2) The answer is choice a. 8. a. You should recognize this first statement. Think to yourself: Is there a triangle that has two sides congruent and, thus, two angles opposite the sides congruent? The answer is yes. It is an isosceles triangle. Now, look at the converse statement. If two angles are congruent in a trian- gle, are the sides opposite these angles also congruent? The answer is yes. –THE SAT MATH SECTION– 160 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 160 An isosceles triangle has two sides con- gruent and two angles opposite these sides congruent. So, both statements are true. The answer is choice a. 9. b. One way that you could find your answer is to substitute the values of x and y into each equation. The equation that is true in each five cases is the answer. The method of doing this is shown below. Choice a: y = x + 2 Coordinate 1: (0,2) 2 = 0 + 2 2 = 2 (True) Coordinate 2: (1,3) 3 = 1 + 2 3 = 3 (True) You may think, at this point, choice a is the answer, but you should try the third coordinate. Coordinate 3: (2,6) 6 = 2 + 2 6 = 4 (False) Therefore, by trying all the points, you can see that choice a is not the answer. Choice b: y = x 2 + 2 Coordinate 1: (0,2) y = x 2 + 2 2 = 0 2 + 2 2 = 2 (True) Coordinate 2: (1,3) y = x 2 + 2 3 = (1) 2 + 2 3 = 1 + 2 3 = 3 (True) Coordinate 3: (2,6) y = x 2 + 2 6 = (2) 2 + 2 6 = 4 + 2 6 = 6 (True) Coordinate 4: (3,11) y = x 2 + 2 11 = (3) 2 + 2 11 = 9 + 2 11 = 11 (True) At this point, you may believe that choice b is the answer. You should check in order to be sure. Coordinate 5: (4,18) y = x 2 + 2 18 = (4) 2 + 2 18 = 16 + 2 18 = 18 (True) Therefore, all the coordinates make equation b true. The answer is choice b. 10. d. You have to solve for the variable, x, so you need to get x by itself in the problem. There- fore, eliminate the other terms on the same side of the equation as x by doing the inverse operation on both sides of the equal sign. This is demonstrated below—first add 2 to both sides: bx – 2 = K + 2 + 2 bx = K + 2 Next, you have to divide both sides by b. ᎏ b b x ᎏ = ᎏ K b +2 ᎏ x = ᎏ K b +2 ᎏ The answer is choice d. 11. b. You should remember that the formula for the slope of a line is equal to: m = You should also remember the formula m = If you look at the graph, you will see that the line crosses through exactly two points. They are: Point 1 (0,2) Point 2 (3,0) Now, all you have to do is substitute the values of point 1 and point 2 into the for- mula for slope. x 1 = 0 y 1 = 2 x 2 = 3 y 2 = 0 m = ᎏ ( ( 0 3 – – 2 0 ) ) ᎏ or – ᎏ 2 3 ᎏ The answer is choice b. (y 2 – y 1 ) ᎏ (x 2 – x 1 ) the change in the y-coordinate ᎏᎏᎏᎏ the change in the x-coordinate –THE SAT MATH SECTION– 161 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 161 12. a. This problem is difficult if you make it diffi- cult, but it’s easy if you make it easy. The eas- iest way to do this problem is to calculate the mean, median, and mode for the data set. Remember: ■ The mean is the same as the average. ■ The median is the middle number of data. First, you must order the num- bers from least to greatest. ■ The mode is the most frequently occurring number. So, the mean equals: ᎏ 5+7+6 5 +5+7 ᎏ = 6 The median, if found by rearranging the numbers in the data set as shown, is {5, 5, 6, 7, 7}. Therefore, the median is 6. The mode is the most frequently occur- ring number. In this data set, there are two numbers that appear most frequently: {5, 7}. Now, inspect the answers. You will quickly see that choice a is cor- rect: The mean = median, because 6 = 6. 13. c. You have to figure out if XY and YZ are per- pendicular. The key thing to remember here is that perpendicular lines intersect to form right angles. If you can find a right angle at the point that XY and YZ intersect, then you know that the two segments are perpendicular. In Figure 1, if XY and YZ are perpendi- cular, then ΔXYZ is a right triangle because it contains a right angle at Y. In ΔXYZ, you are given three sides. If ΔXYZ is a right triangle, then the Pythago- rean theorem should hold true for these three sides. (Leg 1) 2 + (Leg 2) 2 = (Hypotenuse) 2 (6) 2 + (8) 2 = (10) 2 Note: 10 is the hypotenuse because it is across from the largest angle of the triangle. 36 + 64 = 100 100 = 100 ΔXYZ is a right triangle and, likewise, X ෆ Y ෆ is perpendicular to YZ because the Pythagorean theorem is true for Figure 1. In Figure 2, you are given the two angles of ΔXYZ. If a third angle measures 90°, then ∠Y is a right triangle. Thus, X ෆ Y ෆ is perpendicular to Y ෆ Z ෆ . m∠X + m∠Y + m∠Z = 180°, since the sum of the angles of a triangle = 180. 25° + x + 65° = 180° 90° + x = 180° Therefore, x = 90°. ∠Y is a right angle and X ෆ Y ෆ is perpendicular to Y ෆ Z ෆ . Thus, X ෆ Y ෆ is perpendicular to Y ෆ Z ෆ in both figures. The answer is choice c. 14. d. The first thing that you should realize is that x and y are both greater than 0, but less than 1. So, x and y are going to be between 0 and 1 on the number line. Next, you see the formula for d; d = x – y. To solve for d, you must substitute a value in for x and y. However, you do not have a value. You should recognize however that x is less than y. Thus, whatever value you choose for x, the answer for d is going to be negative. Therefore, the answer is choice d. 15. a. This question involves calculating distance. The pieces of information that you are given or have to calculate are rate of speed and time. The formula for distance (with these specific given pieces of information) is: Distance = Rate × Time The first step is to calculate the rate traveled at by the car. Solving for rate, you have Rate = ᎏ D T is i t m an e ce ᎏ = ᎏ 1 2 10 ho m u i r le s s ᎏ = 55 mph Now, all you have to do is substitute into the formula above using the rate you just solved for. –THE SAT MATH SECTION– 162 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 162 . into two sections as shown on the following page. G B ~B B ~B ~G THE SAT MATH SECTION 159 5658 SAT2 006[ 04] (fin).qx 11/21/05 6 :44 PM Page 159 The P(G) = ᎏ 1 4 ᎏ . However, P(B) = ᎏ 2 4 ᎏ . If. – ᎏ 2 3 ᎏ The answer is choice b. (y 2 – y 1 ) ᎏ (x 2 – x 1 ) the change in the y-coordinate ᎏᎏᎏᎏ the change in the x-coordinate THE SAT MATH SECTION 161 5658 SAT2 006[ 04] (fin).qx 11/21/05 6 :44 PM. y miles THE SAT MATH SECTION 157 5658 SAT2 006[ 04] (fin).qx 11/21/05 6 :44 PM Page 157 Five-Choice Answers 1. c. After reading the problem, you realize that the amount of robins at the feeder