– THE SAT MATH SECTION – An isosceles triangle has two sides congruent and two angles opposite

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– THE SAT MATH SECTION – An isosceles triangle has two sides congruent 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: Coordinate 1: (0,2) y=x+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...

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5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 161 – THE SAT MATH SECTION – An isosceles triangle has two sides con- y = x2 + 2 Coordinate 5: (4,18) gruent and two angles opposite these sides 18 = (4)2 + 2 congruent. 18 = 16 + 2 So, both statements are true. The 18 = 18 (True) answer is choice a. Therefore, all the coordinates make 9. b. One way that you could ﬁnd your answer is equation b true. The answer is choice b. to substitute the values of x and y into each 10. d. You have to solve for the variable, x, so you equation. The equation that is true in each need to get x by itself in the problem. There- ﬁve cases is the answer. The method of doing fore, eliminate the other terms on the same this is shown below. side of the equation as x by doing the inverse operation on both sides of the equal sign. Choice a: y=x+2 This is demonstrated below—ﬁrst add 2 to Coordinate 1: (0,2) 2=0+2 both sides: 2 = 2 (True) bx – 2 = K Coordinate 2: (1,3) 3=1+2 +2 +2 3 = 3 (True) bx = K + 2 You may think, at this point, choice a is Next, you have to divide both sides by b. the answer, but you should try the third bx K+2 = coordinate. b b K+2 Coordinate 3: (2,6) 6=2+2 x= b 6 = 4 (False) The answer is choice d. Therefore, by trying all the points, you can 11. b. You should remember that the formula for see that choice a is not the answer. the slope of a line is equal to: the change in the y-coordinate m= y = x2 + 2 Choice b: the change in the x-coordinate y = x2 + 2 Coordinate 1: (0,2) You should also remember the formula 2 = 02 + 2 (y2 – y1) m= (x2 – x1) 2 = 2 (True) If you look at the graph, you will see y = x2 + 2 Coordinate 2: (1,3) that the line crosses through exactly two 3 = (1)2 + 2 points. They are: 3=1+2 Point 1 (0,2) 3 = 3 (True) Point 2 (3,0) y = x2 + 2 Coordinate 3: (2,6) Now, all you have to do is substitute the 6 = (2)2 + 2 values of point 1 and point 2 into the for- 6=4+2 mula for slope. 6 = 6 (True) x1 = 0 y1 = 2 y = x2 + 2 Coordinate 4: (3,11) x2 = 3 y2 = 0 11 = (3)2 + 2 2 m = (0 – 2) or – 3 11 = 9 + 2 (3 – 0) The answer is choice b. 11 = 11 (True) At this point, you may believe that choice b is the answer. You should check in order to be sure. 161
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 162 – THE SAT MATH SECTION – ΔXYZ is a right triangle and, likewise, 12. a. This problem is difﬁcult if you make it difﬁ- XY is perpendicular to YZ because the cult, but it’s easy if you make it easy. The eas- Pythagorean theorem is true for Figure 1. iest way to do this problem is to calculate the In Figure 2, you are given the two mean, median, and mode for the data set. angles of ΔXYZ. If a third angle measures Remember: 90°, then ∠Y is a right triangle. Thus, XY is ■ The mean is the same as the average. perpendicular to YZ. ■ The median is the middle number of m∠X + m∠Y + m∠Z = 180°, since data. First, you must order the num- the sum of the angles of a triangle = 180. bers from least to greatest. 25° + x + 65° = 180° ■ The mode is the most frequently 90° + x = 180° occurring number. Therefore, x = 90°. ∠Y is a right angle and XY is perpendicular to YZ. 5+7+6+5+7 So, the mean equals: =6 5 Thus, XY is perpendicular to YZ in The median, if found by rearranging both ﬁgures. The answer is choice c. the numbers in the data set as shown, is {5, 5, 14. d. The ﬁrst thing that you should realize is that 6, 7, 7}. Therefore, the median is 6. x and y are both greater than 0, but less than The mode is the most frequently occur- 1. So, x and y are going to be between 0 and 1 ring number. In this data set, there are two on the number line. numbers that appear most frequently: {5, 7}. Next, you see the formula for d; d = x – y. Now, inspect the answers. To solve for d, you must substitute a You will quickly see that choice a is cor- value in for x and y. However, you do not have rect: The mean = median, because 6 = 6. a value. You should recognize however that x 13. c. You have to ﬁgure out if XY and YZ are per- is less than y. Thus, whatever value you choose pendicular. The key thing to remember here is for x, the answer for d is going to be negative. that perpendicular lines intersect to form right Therefore, the answer is choice d. angles. If you can ﬁnd a right angle at the 15. a. This question involves calculating distance. point that XY and YZ intersect, then you know The pieces of information that you are given that the two segments are perpendicular. or have to calculate are rate of speed and time. In Figure 1, if XY and YZ are perpendi- The formula for distance (with these cular, then ΔXYZ is a right triangle because it speciﬁc given pieces of information) is: contains a right angle at Y. Distance = Rate × Time In ΔXYZ, you are given three sides. If ΔXYZ is a right triangle, then the Pythago- The ﬁrst step is to calculate the rate rean theorem should hold true for these traveled at by the car. three sides. Solving for rate, you have (Leg 1)2 + (Leg 2)2 = (Hypotenuse)2 Distance 110 miles Rate = Time = 2 hours = 55 mph (6)2 + (8)2 = (10)2 Now, all you have to do is substitute Note: 10 is the hypotenuse because it is into the formula above using the rate you across from the largest angle of the triangle. just solved for. 36 + 64 = 100 100 = 100 162
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 163 – THE SAT MATH SECTION – 55 miles Distance = Rate × Time = ( hour ) × (h nonrepeating decimal. Choices b and e are hours) = 55h incorrect. The answer is choice a. Choice a is 8. This can be simpliﬁed If you solve the formula above incor- to 2 2. The 2 is an irrational number. It rectly, the other choices might seem to be is nonterminating and nonrepeating. There- correct. Therefore, double-check that you are fore, 8 is not rational. The same reasoning using the correct formula and you are solv- informs you that choice d cannot be rational; ing exactly what the question is asking for. 6 2 contains the number, 2. It is irra- 16. b. Inequalities can be solved just like equations. tional. Choice c is 5 9 and 9 = 3. There- fore, 5 9 = 5 × 3 = 15. The difference between equations and 15 inequalities is that equations have an equal 15 can be written as 1 . Thus, it is a sign and inequalities have a greater than (>) rational number. or less than (
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 164 – THE SAT MATH SECTION – alternative. For example, V = l × w × h. Setting all Example: three quantities equal to one yields a volume of 6x3 = 2x2 3x1 1 (one times one times one is one). Now if you (6 ÷ 3 = 2 AND x3 – 1 = x2) double the length, the length is now two and tripling the width makes it three. Using the You should see that by following these 9x2 equation again with these new quantities gives: rules, the answer to = 3x. 3x V = 2 × 3 × 1 = 6, the answer to the question. 3x Now, what is 3x ? 20. a. This problem requires you to read carefully Easy, what is anything divided by itself? and determine what is actually given and what The answer is 1. However, if you aren’t you are really trying to solve. You are told that careful, you may simply cancel out these last the association is charged the following: terms. You cannot do this because you are Given: dividing. $20 charge for rental of the dining room. The answer is choice a. If you haven’t $2.50 charge for each dinner plate. realized this yet, choice b looks like the answer Also, the association invited four non- if you made a mental error and crossed out paying guests and they must have enough the 3x term. Don’t make this mistake! money to pay the entire bill to the hotel. 19. c. First, you have to remember the formula for Four nonpaying guests cost the associa- the volume of a rectangular prism. tion $10 because 4 × $2.50 = $10. The formula is: The association incurs the following Volume = length × width × height costs: $30 + $2.50 × (# of paying people V=l×w×h attending). The $30 comes from the $20 charge for Now, you have to interpret and write, in the dining room and $10 fee for inviting the algebraic terms, what is happening to the four nonpaying guests. dimensions of the prism. This is best The association charges: $3.00 × (# of achieved by using a key or legend. paying people attending). KEY: So, if the association must have enough Let 2 × l = the length is doubled. money to pay the hotel, what the association Let 3 × w = the width is tripled. charges must be equal to what the hotel Let h = the height remains the same. charges the association. Next, interpret the new volume based Amount the association is charged = on the new dimensions. Amount the association charges guests New Volume = (2l) × (3w) × (h) $30 + $2.50 (# of paying people) = $3.00 = 6(l × w × h) (# of paying people) You should see that the original volume Let x = # of paying people. was equal to l × w × h. The new volume, 6(l × w × h) is six times the original volume. Thus: $30 + $2.50 x = $3.00x Therefore, the answer is choice c. $30 = $.50x Trying the formula for volume with sim- 60 = x ple numbers inserted into it, like 1, then recalcu- Therefore, 60 guests must attend. The lating the new volume using the changes answer is choice a. mentioned in the problem may be an easy 164
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 165 – THE SAT MATH SECTION – 21. c. If you want to ﬁnd the roots of an equation Choice a: 16 questions. The boy got 50% of the questions correct. An easy math algebraically, you have to factor the equation calculation shows that he got 8 correct if and solve for the variable term. there were only 16 questions on the test. So, looking at the trinomial 2x2– x –15, you However, you know that he had 10 out of the should notice the following: ﬁrst 12 correct. This answer is not possible (1) There are no common factors and cannot be true. between the three terms. You can rule out choice e, 18, using the (2) There are three terms. This elimi- same logic: Half of 18 is 9, and you know that nates the technique of factoring by the boy got at least 10 questions correct, so difference of two perfect squares. choice e is also incorrect. (3) You can always use the quadratic formula to ﬁnd the roots. This is Choice b: 24 questions on the test. You sometimes difﬁcult, especially if have to set up a proportion in order to check you do not remember the formula. this answer. The proportion is: Let’s try factoring into two binomials. # of questions correct % = After trial and error, you will see that 100 # of total questions the expression can be factored into The percentage that he got correct is (2x + 5)(x – 3) = 0. 50%. Thus, the formula for choice b is: Now, you are multiplying two binomi- x correct 50 = als together and the product is equal to zero. 24 100 Thus, one of the binomial terms, if not both, If you solve for x by cross multiplying, equals zero. the answer is x = 12. So, let’s set each term equal to zero and The boy got 10 out of the ﬁrst 12 cor- solve for x. rect. This means that he only had 2 out of the 2x + 5 = 0 x – 3 = 0 2 next 12 remaining questions correct; 12 is –5 –5 +3 +3 1 equal to .1666. . . . and this is not equal to 4 ; 2x 5 1 = –2 x=3 4 is the fraction of remaining questions cor- 2 rect. Thus, choice b is incorrect. 5 x = – 2 and x = 3 The answer is choice c. However, watch Choice c: 26 questions on the test. The out for the other choices because they are proportion for choice c is: 5 there to trick you; x = – 2 is an answer, how- 5 x correct 50 ever, it is not listed. Only 2 is listed and that = 26 100 is not the same answer. After solving the proportion, you ﬁnd 22. d. This question requires a different type of that x = 13. Once again, the boy had 10 out problem-solving technique. The most effec- of the ﬁrst 12 correct. Therefore, he had only tive way to solve this question is through trial 3 questions correct out of the next 14 if there and error. You start to eliminate wrong 3 were 26 questions on the test; 14 = .214 . . . answers by testing their validity. Here is what 1 This answer is not equal to 4 or .25. There- that means. fore, choice c is incorrect. 165
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 166 – THE SAT MATH SECTION – Choice d: 28 questions on the test. x = 36 Hopefully, by process of elimination, this is Step 2: Since x = 36, the base angles are both the answer. You should still check it however. 36° and the vertex angle is 3(36) = 108°. The proportion is: Step 3: This triangle is an obtuse triangle x correct 50 = 28 100 since there is one angle contained in the tri- You ﬁnd that x = 14 after cross multi- angle that is obtuse. The obtuse angle is the plying. Therefore, the boy had 4 correct out vertex angle. of the 16 remaining questions. You know this The answer is choice d. 24. d. Real numbers have many properties. You because he had 10 out of the ﬁrst 12 correct; 4 1 16 = 4 = .25. This is the answer. need to remember a few of them. Let’s take a There were 28 questions on the test. look at each one of the ﬁve choices in order The answer is choice d. to determine which one is the distributive 23. d. As a point of reference: property. A scalene triangle has three unequal sides. 1 1 1 1 Choice a: 3 + 2 = 2 + 3 An acute triangle contains an angle less Does this look familiar to you? It than 90 degrees. should. This is the commutative property. If A right triangle contains an angle equal to the order of the terms is switched, but you 90 degrees. still have the same answer when the opera- The ﬁrst thing you should do when you tion is performed, then the commutative encounter a word problem involving geome- property exists. try is to draw a diagram and create a legend. Choice b: 3 + 0 = 3 Legend: This is known as the identity property Let x = base angle. for addition. Sometimes, it is called the zero Let 3x = the vertex angle. property of addition. Either way, this is not Now that you have deﬁned the angles, it is the distributive property. time to draw a diagram similar to the one Choice c: (1.3 × 0.07) × 0.63 = 1.3 × (0.07 × below. 0.63) D This is the associative property. The 3x parenthesis may be placed around different groups of numbers but the answer does not x° x° O G change. Multiplication is associative. You will see that in an isosceles triangle the base angles are equal. Next, in order to Choice d: –3(5 + 7) = (–3)(5) + (–3)(7) classify the triangle, you need to ﬁnd out the This is the distributive property. You exact angle measures. can multiply the term outside the parenthe- This is done by remembering the fact ses by each term inside the parentheses. The that the sum of the angles of a triangle is 180°. left side of the equation is equal to the right side. This is the answer and it is an impor- Step 1: x + x + 3x = 180 tant property to remember. 5x 180 = The answer is choice d. 5 5 166
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 167 – THE SAT MATH SECTION – 25. c. You have to factor this expression accord- The ground and the house meet at a right angle because you are told that it is level ingly. Notice that there are only two terms ground. This makes the diagram a right tri- and there is a subtraction sign between them. angle. Thus, in order to solve for x, you have Sometimes, that is a clue to try to factor to use the Pythagorean theorem. Remember, using the difference of two perfect squares the Pythagorean theorem is a2 + b2 = c2, technique. However, in this case, 3x2 and 27 where c is the hypotenuse, or longest side, of are not perfect squares. Therefore, you have a right triangle. It can also be written as to try a different method. (leg 1)2 + (leg 2)2 = (hypotenuse)2. First, notice that there is a common In this case, the ladder is 5 feet from the factor of 3 in both terms. Factor this term house. This distance is leg 1 or a. out of both terms. Once you do, the expres- sion is 3(x2 – 9). The ladder is across from the right angle. This makes it the hypotenuse. The job is not done. You have to factor The hypotenuse, or c, is 13 feet. COMPLETELY! Look at the expression (x2 – 9). This is a binomial with two perfect Thus, you have to solve for leg 2, or b, the following way. squares separated by a subtraction sign. Thus, 52 + b2 = 132 this binomial can be factored according the 25 + b2 = 169 difference of two perfect squares. The expres- b2 = 144 sion now becomes: 3(x + 3)(x – 3). b = 12 feet The answer is choice c. If you are not The answer is choice d. careful, you may select one of the alternate Note: You could easily solve this equa- choices. Remember, factor completely and tion if you recognize that this right triangle is do not stop factoring until each term is sim- a Pythagorean triplet. It is a 5-12-13 right tri- pliﬁed to lowest terms. 26. d. This is a word problem involving geometry angle and 12 feet had to be the length of leg 2 once you saw that 5 feet was leg 1’s length and and ﬁgures. The best way of solving a prob- 13 feet was the length of the hypotenuse. lem like this is to read it carefully and then 27. b. You can outline all the possibilities that can try to draw a diagram that best illustrates occur. First, you have either boys or girls at what is being described. the party. You also know that they are either You should draw a diagram similar to wearing a mask or not wearing a mask. the one below. Therefore, you can start outlining the possi- ble events. You are told that 20 students did not 13 feet wear masks. In addition, you know that 9 boys x feet did not wear masks. Therefore, calculations tell you that 11 girls did not wear masks. 5 feet Now, if 11 girls did not wear masks and 7 girls did wear masks, then 18 girls attended You are trying to ﬁnd out the height of the party. the ladder as it rests against the house. This height is represented as x. 167
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 168 – THE SAT MATH SECTION – If 18 girls attended the party and 15 Thus, the product of bc is even. Since a is odd, a × (even #) = even number. Therefore, boys were at the party, then 33 students attended the school costume party overall. this expression is even and not the answer The answer is choice b. you are searching for. 28. c. You have to be able to read and interpret the Another way you could try this prob- lem (if you do not remember the even × odd wording in this problem in order to develop an equation to solve. = even number rule) is to substitute num- Let x = the number. bers for a, b, c. Let’s say a = 5; b = 6; c = 9. 1 Then a(bc) = 5(6 × 9) = 5(54) = 270. Now, “One-half of a number” is 2 x. The word “is” means equals. So, you have This is an even number and not the answer 1 written 2 x = . that you are looking for. The last phrase is “8 less than two-thirds of Choice b: acb0 the number.” The phrase “less than” means This expression requires that you eval- uate b0 ﬁrst. This is an important rule to to subtract and switch the order of the num- remember. Any term raised to the zero bers. The reason for reversing the order of power is 1. Well, a × c is an odd number 2 the terms is that 8 is deducted from 3 of the times an odd number. The product of any 2 number. Thus, the last part is 3 x – 8. two odd numbers is an odd number. Thus, 1 2 The equation to solve is 2 x = ( 3 )x – 8. an odd number times 1 is an odd number. Choice b is an odd number. There is no need Finally, you have to solve the equation. to try the other expressions. 1 2 2 x = ( 3 )x – 8 30. a. This question fortunately, or unfortunately, 1 1 – 2x – 2x requires simple memorization. You must Find a common → denominator in order to 0 = 64x – 8 remember the properties of a parallelogram in order to get this question correct. There are subtract the like terms. 3 – 6x six basic properties of every parallelogram. 1 0 = 6x – 8 They are: +8 +8 1. The opposite sides of a parallelogram 6 → Multiply both sides by 1 8 = 6x 1 are congruent. in order to solve for x. 2. The opposite sides of a parallelogram 48 = x are parallel. The answer is choice c. 3. The opposite angles of a parallelogram 29. b. This problem can be difﬁcult if you simply are congruent. look at it and try to guess. It becomes easier 4. The consecutive angles of a parallelo- if you try each answer by substituting into gram are supplementary. the expression. Here is a way of doing it. 5. The diagonals of a parallelogram bisect Choice a: a(bc) each other. You are told that a and c are odd and b 6. The diagonal of a parallelogram is even. Following order of operations, you divides the parallelogram into two multiply bc ﬁrst. Remember, that an even × congruent triangles. odd = even number. This is always true. 168
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 169 – THE SAT MATH SECTION – Every parallelogram has these six proper- 36. c. The total amount of proﬁt according to the ties. However, speciﬁc types of parallelograms, graph is 9% of the year’s income. Therefore, 225,198 × .09 = 20,267.2. such as rectangles, rhombus, and squares, have additional properties. One of the properties 37. b. First, solve for x: shared by both rectangles and squares happens x2 – 1 = 36 Add 1 to both sides. to be that the diagonals are congruent. So, the +1 +1 answer is choice a. Not every parallelogram x2 = 37 has this property, only speciﬁc parallelograms x2 = 37 such as rectangles or squares. Take the square root of both 31. d. 52% is the same as .52 (drop the % sign and sides. x2 = move the decimal point two places to the 37 13 26 52 left); 25 = 50 = 100 ; 52 × 100 = .52; And 52 × x= 37 10–2 = 52 × .01 = .52. Obviously, .052 does not equal .52, so your answer is d. 37 is an irrational number. Irrational 32. c. The mean is the average. First, you add 80 + numbers cannot be expressed as a ratio of 85 + 90 + 90 + 95 + 95 + 95 + 100 + 100 = two integers. (Simply put, irrational num- 830. Divide by the number of tests: 830 ÷ 9 = bers have decimal extensions that never ter- 92.22, which shows that statement I is false. minate or extensions that never repeat.) The median is the middle number, which is A prime number has only two positive 95. And the mode is the number that appears factors, itself and 1. Rational numbers can be most frequently, which is also 95; therefore, expressed as a ratio of two integers. The set statement II is correct. of integers is: { . . . –3, –2, –1, 0, 1, 2, 3, . . . }. 33. d. An obtuse angle measures greater than 90°. A 37 is not prime, rational, or an integer. square has four angles that are 90° each, as You can use your calculator to see that it is 6 does a rectangle and cube. The angles inside a with a decimal extension that neither termi- triangle add up to 180°, and one angle in a nates nor repeats. right triangle is 90°, so the other two add up to 38. c. An effective way ﬁgure out this question is to 90°, so there cannot be one angle that alone plug in some low, easy numbers to see what has more than 90 degrees. Therefore, the will happen. Below we picked (1,7) as our answer is d. point A and (5,15) as our point B. (Note that 5 20 34. c. Set up a proportion: 100 = x . Cross multi- the x-coordinate of our point B is 4 greater ply: 5x = 2,000. Then divide both sides by 5 than the x-coordinate of our point A.) to get x = 400. This is only the ﬁrst part of x y the problem. If you chose answer d, you for- 0 5 got to do the next step, which is to ﬁnd what 1 7 pick as A 50 x number is 50% of 400; 100 = 400 , or reduce 2 9 1 x to 2 = 400 . Then again, cross multiply: 400 = 3 11 2x. Divide both sides by 2 to get x = 200. 4 13 35. d. If you look at the pattern, you will see it is 5 15 pick as B 3x – 1. Plug in some numbers, like 3(1) – 1 = 6 17 2, 3(2) – 1 = 5, 3(3) – 1 = 8, etc. You can see that since every other number is even, of the As you can see, the y-coordinate of B is ﬁrst 100 terms, half will be even. 8 greater than the y-coordinate of A. 169
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 170 – THE SAT MATH SECTION – 39. c. Converting mixed numbers into improper denominator (35) into the numerator (81). Any remainder becomes part of the mixed fractions is a two-step process. First, multiply number (35 goes into 81 twice with a the whole number by the denominator (bot- 11 remainder of 11, hence 2 35 ). tom number) of the fraction. Then add that 40. d. We use D = RT, and rearrange for T. Divid- number to the numerator of the fraction. So 2 9 4 9 ing both sides by R, we get T = D ÷ R. The 1 7 becomes 7 and 1 5 becomes 5 . Since Area total distance, D = (x + y), and R = 2 mph. 9 9 81 11 = length × width, 7 × 5 = = 2 35 . Remem- 35 Thus, T = D ÷ R becomes T = (x + y) ÷ 2. 81 ber, to convert the improper fraction ( 35 ) back into a mixed number, you divide the 170
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 171 – THE SAT MATH SECTION – P art 2: Grid-in Questions provided that the answer ﬁts. If you are entering a decimal, do not begin with a 0. For example, sim- ply enter .5 if you get 0.5 for an answer. “Grid-in”questions are also called student-response ques- Enter mixed numbers as improper fractions or ■ tions because no answer choices are given; you, the stu- decimals. This is important for you to know when dent, generate the response. Otherwise, grid-in questions working on the grid-in section. As a math stu- are just like ﬁve-choice questions. In responding to the dent, you are used to always simplifying answers grid-in questions on the SAT, there are several things you to their lowest terms and often converting will need to know about the special four-column grid. improper fractions to mixed numbers. On this Become familiar with the answer grid below. section of the test, however, just leave improper fractions as they are. For example, it is impossible 1 3 to grid 1 2 in the answer grid, so simply grid in 2 / / instead. You could also grid in its decimal form of 1.5. Either answer is correct. • • • • If the answer ﬁts the grid, do not change its form. ■ 0 0 0 If you get a fraction that ﬁts into the grid, do not 1 1 1 1 waste time changing it to a decimal. Changing the 2 2 2 2 form of an answer can result in a miscalculation 3 3 3 3 4 4 4 4 and is completely unnecessary. 5 5 5 5 Enter the decimal point ﬁrst, followed by the ﬁrst ■ 6 6 6 6 three digits of a long or repeating decimal. Do not 7 7 7 7 round the answer. It won’t be marked as wrong if 8 8 8 8 you do, but it is not necessary. 9 9 9 9 If the answer is a fraction that requires more than ■ 17 four digits, like 25 , write the answer as a decimal The above answer grid can express whole numbers 17 instead. The fraction 25 does not ﬁt into the grid from 0 to 9999, as well as some fractions and decimals. and it cannot be reduced; therefore, you must To grid an answer, write it in the top row of the column. turn it into a decimal by dividing the numerator If you need to write a decimal point or a fraction bar, by the denominator. In this case, the decimal skip a column and ﬁll in the necessary oval below it. would be .68. If a grid-in answer has more than one possibility, ■ Very important: No grid-in questions will have a ■ enter any of the possible answers. This can occur negative answer. If you get a negative number, when the answer is an inequality or the solution you have done something wrong. to a quadratic equation. For example, if the Write the answer in the column above the oval. ■ answer is x < 5, enter a 4. If the answer is x = ± 3, The answer you write will be completely disre- enter positive 3, since negative numbers cannot garded because the scoring machine will only be entered into the grid. read the ovals. It is still important to write this If you are asked for a percentage, only grid the ■ answer, however, because it will help you check numerical value without the percentage sign. your work at the end of the test and ensure that There is no way to grid the symbol, so it is simply you marked the appropriate ovals. not needed. For example, 54% should be gridded Answers that need fewer than four columns, except ■ as .54. Don’t forget the decimal point! 0, may be started in any of the four columns, 171
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 172 – THE SAT MATH SECTION – 3 4 / 7 3 4 . 7 . 3 4 7 / / / / / • • • • • • • • • • 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 Remember these important tips: Be especially careful that a fraction bar or deci- ■ If you write in the correct answer but do not mal point is not marked in the same column as a ﬁll in the oval(s), you will get the question digit. marked wrong. If you know the correct answer but ﬁll in the Now it is time to do some grid-in practice prob- wrong oval(s), you will get the question lems. Be sure to review the strategies listed above to marked wrong. ensure that you fully understand the grid system. If you do not fully erase an answer, it may be Remember: You will never be penalized for an incorrect marked wrong. answer on the grid-in questions—so go ahead and guess. Check your answer grid to be sure you didn’t Good luck! mark more than one oval per column. 172
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 173 – THE SAT MATH SECTION – ANSWER SHEET 1. 2. 3. 4. 5. / / / / / / / / / / • • • • • • • • • • • • • • • • • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 6. 7. 8. 9. 10. / / / / / / / / / / • • • • • • • • • • • • • • • • • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 11. 12. 13. 14. 15. / / / / / / / / / / • • • • • • • • • • • • • • • • • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 16. 17. 18. 19. 20. / / / / / / / / / / • • • • • • • • • • • • • • • • • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 173
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 174 – THE SAT MATH SECTION – ANSWER SHEET (continued) 21. 22. 23. 24. 25. / / / / / / / / / / • • • • • • • • • • • • • • • • • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 26. 27. 28. 29. 30. / / / / / / / / / / • • • • • • • • • • • • • • • • • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 31. 32. 33. 34. 35. / / / / / / / / / / • • • • • • • • • • • • • • • • • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 36. 37. 38. 39. 40. / / / / / / / / / / • • • • • • • • • • • • • • • • • • • • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 174
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 175 – THE SAT MATH SECTION – 5 G rid-in Practice Problems 14. If of x is 15, what number is x decreased by 1? 3 15. Half the difference of two positive numbers is 20. 1. The barns in a certain county are numbered con- If the smaller of the two numbers is 3, what is the secutively from 2,020 to 2,177. How many barns sum of the two numbers? are in the county? 16. In a club of 35 boys and 28 girls, 80% of the boys 2. For some ﬁxed value of x, 8(x + 4) = y. After the and 25% of the girls have been members for value of x is increased by 3, 8(x + 4) = z. What is more than two years. If n percent of the club the value of z – y? have been members for more than two years, 3. If 28 × 42 = 16x, then x = ? what is the value of n? 17. A string is cut into two pieces that have lengths in 4. If (y – 1)3 = 27, what is the value of (y + 1)2? the ratio of 3:8. If the differences between the lengths of the two pieces of string is 45 inches, 5. When a is divided by 6, the remainder is 5; and what is the length, in inches, of the shorter piece? when b is divided by 6, the remainder is 4. What is the remainder when a + b is divided by 6? 18. Fruit for a dessert costs $2.40 a pound. If 10 pounds of fruit are needed to make a dessert that 6. When a positive integer k is divided by 7, the serves 36 people, what is the cost of the fruit remainder is 2. What is the remainder when 6k is needed to make enough of the same dessert to divided by 3? serve 48 people? 7. During course registration, 28 students enroll in 19. In the ﬁgure below, if line segment AB is parallel biology. After three boys are dropped from the to line segment CD and BE is perpendicular to class, 44% of the class consists of boys. What per- ED, what is the value of y? cent of the original class did girls comprise? x A B 8. If 3x – 1 = 11 and 4y = 12, what is the value of y ? 9. If 1 – x – 2x – 4x = 7x – 1, what is the value of x? E 10. If 4x2 + 20x + r = (2x + s)2 for all values of x, F 50° what is the value of r – s? y 11. If (3p + 2)2 = 64 and p > 0, what is a possible C D value of p? 12. If (x – 1)(x – 3) = –1, what is a possible value of x? 13. For what integer value of y is y + 5 > 8 and 2y – 3 < 7? 175
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 176 – THE SAT MATH SECTION – 20. In the ﬁgure below, what is the value of x? 26. Through how many degrees does the minute hand of a clock move from 12:25 a .m. to 12:40 B a.m. of the same day? 3 25° 27. A line with a slope of 14 passes through points (7,3k) and (0,k). What is the value of k? E 3x° 5 28. In the ﬁgure below, the slope of line l1 is 6 and 1 the slope of line l2 is 3 . What is the distance from x° 35° point A to point B? A C D 21. In the ﬁgure below, what is the length of line seg- l1 y ment AD? B A l2 (3,y1) 10 10 B 30° (3,y2) 15° 15° x 7 A D E C 22. If the lengths of two sides of an isosceles triangle 29. The dimensions of a rectangular box are integers are 9 and 20, what is the perimeter of the triangle? greater than 1. If the area of one side of this box is 12 and the area of another side is 15, what is 23. If the integer lengths of the three sides of a trian- the volume of the box? gle are 5, x, and 10, what is the least possible perimeter of the triangle? 30. What is the number of inches in the minimum 1 length of 4 -inch-wide tape needed to cover com- 24. If the product of the lengths of the three base pletely a cube whose volume is 8 cubic inches? sides of a triangle is 120, what is a possible perimeter of the triangle? 31. What is the greatest of four consecutive even integers whose average is 19? 25. What is the area of the square below? 32. If the average of x, y, and z is 12, what is the aver- x age of 3x, 3y, and 3z? 4x – 1 176
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 177 – THE SAT MATH SECTION – 33. The probability of selecting a green marble at ran- 36. If x + 2x + 3x + 4x = 1, then what is the value of x2? dom from a jar that contains only green, white, 1 and yellow marbles is 4 . The probability of select- 37. What is the least positive integer p for which ing a white marble at random form the same jar is 441p is the cube of an integer? 1 3 . If this jar contains 10 yellow marbles, what is the total number of marbles in the jar? 38. The average of 14 scores is 80. If the average of four of these scores is 75, what is the average of 34. If the operation is deﬁned by the equation the remaining 10 scores? x y = 2x + 3y, what is the value of a in the x 39. If 3x – 1 = 9 and 4y + 2 = 64, what is the value of y ? equation a 4 = 1 a? p+p+p 35. x, y, 22, 14, 10, . . . 40. If = 12 and p > 0, what is the value of p? p×p×p In the sequence above, each term after the ﬁrst term, x, is obtained by halving the term that comes before it and then adding 3 to that num- ber. What is the value of x – y? 177
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 178 – THE SAT MATH SECTION – G rid-In Answers If (3p + 2)2 = 64 and p > 0, the expression 11. 2. inside the parentheses is either 8 or –8. Since p > 0, let 3p + 2 = 8; then 3p = 6 and p 1. 158. If you are given two numbers, A and B, then = 2. A possible value of p is 2. B – A + 1 is the formula for ﬁnding the If (x – 1)(x – 3) = –1, then x2 – 4x + 3 = –1, 12. 2. quantity of items between the two numbers. so x2 – 4x + 4 = 0. Factoring this equation Therefore, 2,177 – 2,020 + 1 = 157 + 1. gives (x – 2)(x – 2) = 0; x = 2. Thus, a possi- 2. 24. If the value of x is increased by 3, then the value of y is increased by 8 × 3 = 24. Since ble value for x is 2. 13. 4. If 2y – 3 < 7, then 2y < 10, so y < 5. Since y + after x is increased by 3, 8(x + 4) = z, the 5 > 8 and 2y – 3 < 7, then y > 3 and at the value of z – y = 24. same time y < 5. Thus, the integer must be 4. 3. 3. Find the value of x by expressing each side 5 3 If 3 of x is 15, then 5 x = 15, so x = 5 (15) = 14. 8. of the equation as a power of the same base. 3 28 × 24 = 24x 9. Therefore, x decreased by 1 is 9 – 1 = 8. 212 = 24x 15. 46. If half the difference of two positive num- bers is 20, then the difference of the two 12 = 4x, so 3 = x 3 = 27 and (y – 1)3 = 27, then y – 1 = 3, positive numbers is 40. If the smaller of the 4. 25. Since 3 so y = 4. Thus, (y + 1)2 = (4 + 1)2 = 52 = 25. two numbers is 3, then the other positive number must be 43 since 43 –3 = 40. Thus, 6. 0. Let k equal 9, then 6k = 54. When 54 is the sum of the two numbers is 43 + 3 = 46. divided by 3 the remainder is 0. Since 80% of 35 = .80 × 35 = 28 and 25% of 28 16. 55.5. 7. 50. After three boys are dropped from the class, = .25 × 28 = 7, then 35 of the 63 boys and girls 25 students remain. Of those 25, 44% are have been club members for more than two boys, so 56% are girls. Since 56% of 25 = .56 35 × 25 = 14, 14 girls are enrolled in biology. years. Since 63 = .5555 . . . , 55.5% of the club have been members for more than two years. Hence, 14 of the 28 students in the original 17. 27. Since the lengths of the two pieces of string class were girls. Thus, the number of girls in 14 are in the ratio 3:8, let 3x and 8x represent the original class comprised 28 or 50%. 8. 4 . If 3x – 1 = 11, then 3x = 12, and x = 4. Since their lengths: 3 4 4y = 12, then y = 3. Therefore, x = 3 . You 8x – 3x = 45 y 4 5x = 45 can grid this in as 3 or 1.333. 2 x=9 9. 14 . If 1 – x – 2x – 4x – 7x = 7x – 1, then 1 – 7x = Since 3x = (3)9 = 27, the length of the 7x – 1, so 1 + 1 = 7x + 7x and 2 = 14x. 2 shorter piece of string is 27 inches. Hence, 14 = x. 10 10. 20. 4x2 + 20x + r = (2x + s)2 18. 32. If 10 pounds of fruit serve 36 people, then 36 10 pound serves one person. So, 48 × 36 = 4 × = (2x + s)(2x + s) 10 40 3 = 3 pounds. Since the fruit costs $2.40 a = (2x)(2x) + (2x)(s) + (s)(2x) + (s)(s) = 4x2 + 2xs + 2xs + s2 pound, the cost of the fruit needed to serve 40 48 people is 3 × $2.40 = 40 × .80 = $32.00. = 4x2 + 4xs + s2 19. 20. The measures of vertical angles are equal, so Since the coefﬁcients of x on each side of ∠EFC = 50. In right triangle CEF, the meas- the equation must be the same, 20 = 4s, so ures of the acute angles add up to 90, so s = 5. Comparing the last terms of the poly- ∠ECF + 50 = 90 or ∠ECF = 90 – 50 = 40. nomials on the two sides of the equation makes r = s2 = 52 = 25. Therefore, r – s = 25 – 5 = 20. 178
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 179 – THE SAT MATH SECTION – Since the measures of acute angles formed 26. 90. From 12:25 a .m. to 12:40 a .m. of the same by parallel lines are equal, y + ∠ECF = 50. day, the minute hand of the clock moves 15 Hence, y + 40 = 50, so y = 10. minutes since 40 – 25 = 15. There are 60 20. 35. In triangle ABC, ∠ACB = 180 – 35 – 25 = minutes in an hour, so 15 minutes repre- 15 120. Since angles ACB and DCE form a sents 60 of a complete rotation. Since there straight line, ∠DCE = 180 – 120 = 60. are 360° in a complete rotation, the minute hand moves 15 × 360 = 15 × 6 = 90. Angle BED is an exterior angle of triangle 60 27. 3 . ECD. Therefore, 3x = 60 + x Since the line that passes through points 4 2x = 60 3 (7,3k) and (0,k) has a slope of 14 , then: x = 30 3 3k – k 7 – 0 = 14 21. 1.5. Angle B measures 15 + 30 + 15 or 60°, so the 2k 3 7 = 14 sum of the measures of angles A and C is 28k = 21 120. Since AB = BC = 10, then ∠A = ∠C = 21 3 k = 28 = 4 60, so triangle ABC is equiangular. A trian- 3 5 28. 2 . Since the slope of the line l1 is 6 , then gle that is equiangular is also equilateral, so y1 – 0 15 5 AC = 10. Angles BDE and BED each meas- 5 or y1 = = = 6 3–0 6 2 ure 60 + 15 = 75°, since they are exterior 1 The slope of line l2 is 3 , so angles of triangles ADB and CEB. Therefore, y2 – 0 1 3 triangles ADB and CEB have the same shape or y2 = 3 = 1 = 3–0 3 and size, so AD = CE. Since you are given Since points A and B have the same that DE = 7, then AD + CE = 3, so AD = 1.5. x-coordinates, they lie on the same vertical 22. 49. If the lengths of two sides of an isosceles tri- 2 line, so the distance from A to B = y1 – y angle are 9 and 20, then the third side must 5 3 be 9 or 20. Since 20 is not less than 9 + 9, the = 2 – 1 or 2 3 third side cannot be 9. Therefore, the Grid as 2 . lengths of the three sides of the triangle 29. 60. You are given that all the dimensions of a must be 9, 20, and 20. The perimeter is 9 + rectangular box are integers greater than 1. 20 + 20 = 49. Since the area of one side of this box is 12, 23. 21. In the given triangle, 10 – 5 < x < 10 + 5 or, the dimensions of this side must be either 2 equivalently, 5 < x < 15. Since the smallest by 6 or 3 by 4. The area of another side of possible integer value of x is 6, the least possi- the box is given as 15, so the dimensions of ble perimeter of the triangle is 5 + 6 + 10 = 21. this side must be 3 and 5. Since the two 24. 15. Factor 120 as 4 × 5 × 6. Since each number of sides must have at least one dimension in the set 4, 5, and 6 is less than the sum, and common, the dimensions of the box are 3 by 4 by 5, so its volume is 3 × 4 × 5 = 60. greater than the difference of the other two, a possible perimeter of the triangle is 4 + 5 + 30. 96. A cube whose volume is 8 cubic inches has an edge length of 2 inches, since 2 × 2 × 2 = 8. 6 = 15. 1 25. 9 . Since the ﬁgure is a square, x = 4x – 1 Since a cube has six square faces of equal area, the surface area of this cube is 6 × 22 or 6 × 4 1 = 3x 1 3 =x or 24. The minimum length, or L, of 1 1 1 1 To ﬁnd the area of the square, square 3 = ( 3 )2 = 9 . 4 -inch-wide tape needed to completely cover the cube must have the same surface area of 179
5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 180 – THE SAT MATH SECTION – 1 the cube. Therefore, L × 4 = 24 and L = 36. 0.01. If x + 2x + 3x + 4x = 1, then 10x = 1, so x = 1 12 1 1 24 × 4 = 96 inches. 2 10 and x = ( 10 ) = 100 . Since 100 does not 31. 22. Let x, x + 2, x + 4, and x + 6 represent four ﬁt in the grid, grid in .01 instead. 37. 21. Since 441p = 9 × 49 × p = 32 × 72 × p, let p = consecutive even integers. If their average is 3 × 7, which makes 441p = 33 × 73 = (3 × 7)3 19, then = 213. x + (x + 2) + (x + 4) + (x + 6) = 19 4 38. 82. If the average of 14 scores is 80, the sum of or the 14 scores is 14 × 80 or 1,120. If the aver- 4x + 12 = 4 × 19 = 76 age of four of these scores is 75, the sum of Then, these four scores is 300, so the sum of the 4x = 76 – 12 = 64 remaining 10 scores is 11,200 – 300 or 820. 64 x = 4 = 16 820 The average of these 10 scores is 10 or 82. Therefore, x + 6, the greatest of the four 39. 3. To ﬁnd the value of x , given that 3x – 1 = 9 y consecutive integers is 16 + 6 or 22. and 4y + 2 = 64, use the two equations to 32. 36. Since the average of x, y, and z is 12, then x ﬁnd the values of x and y. + y + z = 3 × 12 = 36. Thus, 3x + 3y + 3z = Since, 3x – 1 = 9 = 32, then x – 1 = 2, so x = 3. 3(36) or 108. And 4y + 2 = 64 = 43, then y + 2 = 3, so y = 1. The average of 3x, 3y, and 3z is their sum, x 3 Therefore, y = 1 = 3. 108, divided by 3 since three values are being 108 1 3 p+p+p 3p added: 3 or 36. 40. 2 . Since p × p × p = p × p = p2 = 12 1 1 7 p2 1 3 1 1 33. 24. Since 1 – ( 4 + 3 ) = 1 – 12 , the probability of then 3 = 12 , so p2 = 12 = 4 . Therefore, p = 2 , 5 selecting a yellow marble is 12 . If 10 of the x 11 1 1 since 4 × 4 = 2 . Grid in as 2 . 5 10 marbles in the jar are yellow, then 12 = x . Since 10 is two times 5, x must be two times 12 or 24. F inally 34. 10. Since x y = 2x + 3y, evaluate a 4 by let- ting x = a and y = 4: Don’t forget to keep track of your time during the Math a 4 = 2a + 3(4) = 2a + 12 section. Although most questions will take you about a Evaluate 1 a by letting x = 1 and y = 4: minute or so—the amount of time it takes to answer a 1 a = 2(1) + 3a = 2 + 3a particular question can vary according to difﬁculty. Since a 4 = 1 a, then 2a + 12 = 2 + 3a, Don’t hold yourself to a strict schedule, but learn to be or 12 – 2 = 3a – 2a, so 10 = a. aware of the time you are taking. Never spend too much 35. 32. In the sequence x, y, 22, 14, 10, . . . each term time on any one question. Feel free to skip around and after the ﬁrst term, x, is obtained by halving answer any questions that are easier for you, but be sure the term that comes before it and then to keep track of which questions you have skipped. adding 3 to that number. Hence, to obtain y, Remember that, in general, each set of questions begins do the opposite to 22: Subtract 3 and then with easy problems and becomes increasingly harder. double the result, getting 38. To obtain x, Finally, if you can eliminate one or more answers on a subtract 3 from 38 and then double the tough question, go ahead and make a guess, and if you result, getting 70. Thus, x – y = 70 – 38 = 32. have time at the end of the section, go back and check your answers. Good luck! 180

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– THE SAT MATH SECTION – 45-45-90 Right Triangles 45° 30-60-90 Triangles In a right triangle

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