25. c. You have to factor this expression accord- ingly. Notice that there are only two terms and there is a subtraction sign between them. Sometimes, that is a clue to try to factor using the difference of two perfect squares technique. However, in this case, 3x 2 and 27 are not perfect squares. Therefore, you have to try a different method. First, notice that there is a common factor of 3 in both terms. Factor this term out of both terms. Once you do, the expres- sion is 3(x 2 – 9). The job is not done. You have to factor COMPLETELY! Look at the expression (x 2 – 9). This is a binomial with two perfect squares separated by a subtraction sign. Thus, this binomial can be factored according the difference of two perfect squares. The expres- sion now becomes: 3(x + 3)(x – 3). The answer is choice c. If you are not careful, you may select one of the alternate choices. Remember, factor completely and do not stop factoring until each term is sim- plified to lowest terms. 26. d. This is a word problem involving geometry and figures. The best way of solving a prob- lem like this is to read it carefully and then try to draw a diagram that best illustrates what is being described. You should draw a diagram similar to the one below. You are trying to find out the height of the ladder as it rests against the house. This height is represented as x. The ground and the house meet at a right angle because you are told that it is level ground. This makes the diagram a right tri- angle. Thus, in order to solve for x,you have to use the Pythagorean theorem. Remember, the Pythagorean theorem is a 2 + b 2 = c 2 , where c is the hypotenuse, or longest side, of a right triangle. It can also be written as (leg 1) 2 + (leg 2) 2 = (hypotenuse) 2 . In this case, the ladder is 5 feet from the house. This distance is leg 1 or a. The ladder is across from the right angle. This makes it the hypotenuse. The hypotenuse, or c, is 13 feet. Thus, you have to solve for leg 2, or b, the following way. 5 2 + b 2 = 13 2 25 + b 2 = 169 b 2 = 144 b = 12 feet The answer is choice d. Note: You could easily solve this equa- tion if you recognize that this right triangle is a Pythagorean triplet. It is a 5-12-13 right tri- angle and 12 feet had to be the length of leg 2 once you saw that 5 feet was leg 1’s length and 13 feet was the length of the hypotenuse. 27. b. You can outline all the possibilities that can occur. First, you have either boys or girls at the party. You also know that they are either wearing a mask or not wearing a mask. Therefore, you can start outlining the possi- ble events. You are told that 20 students did not wear masks. In addition, you know that 9 boys did not wear masks. Therefore, calculations tell you that 11 girls did not wear masks. Now, if 11 girls did not wear masks and 7 girls did wear masks, then 18 girls attended the party. 5 feet x feet 13 feet –THE SAT MATH SECTION– 167 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 167 –THE SAT MATH SECTION– 168 If 18 girls attended the party and 15 boys were at the party, then 33 students attended the school costume party overall. The answer is choice b. 28. c. You have to be able to read and interpret the wording in this problem in order to develop an equation to solve. Let x = the number. Now, “One-half of a number” is ᎏ 1 2 ᎏ x. The word “is” means equals. So, you have written ᎏ 1 2 ᎏ x = . The last phrase is “8 less than two-thirds of the number.” The phrase “less than” means to subtract and switch the order of the num- bers. The reason for reversing the order of the terms is that 8 is deducted from ᎏ 2 3 ᎏ of the number. Thus, the last part is ᎏ 2 3 ᎏ x – 8. The equation to solve is ᎏ 1 2 ᎏ x = ( ᎏ 2 3 ᎏ )x – 8. Finally, you have to solve the equation. ᎏ 1 2 ᎏ x = ( ᎏ 2 3 ᎏ )x – 8 – ᎏ 1 2 ᎏ x – ᎏ 1 2 ᎏ x 0 = ᎏ 6 4 x ᎏ – 8 → – ᎏ 3 6 ᎏ x 0 = ᎏ 1 6 ᎏ x – 8 + 8 + 8 8 = ᎏ 1 6 ᎏ x → 48 = x The answer is choice c. 29. b. This problem can be difficult if you simply look at it and try to guess. It becomes easier if you try each answer by substituting into the expression. Here is a way of doing it. Choice a: a(bc) You are told that a and c are odd and b is even. Following order of operations, you multiply bc first. Remember, that an even × odd = even number. This is always true. Thus, the product of bc is even. Since a is odd, a × (even #) = even number. Therefore, this expression is even and not the answer you are searching for. Another way you could try this prob- lem (if you do not remember the even × odd = even number rule) is to substitute num- bers for a, b, c.Let’s say a = 5; b = 6; c = 9. Then a(bc) = 5(6 × 9) = 5(54) = 270. This is an even number and not the answer that you are looking for. Choice b: acb 0 This expression requires that you eval- uate b 0 first. This is an important rule to remember. Any term raised to the zero power is 1. Well, a × c is an odd number times an odd number. The product of any two odd numbers is an odd number. Thus, an odd number times 1 is an odd number. Choice b is an odd number. There is no need to try the other expressions. 30. a. This question fortunately, or unfortunately, requires simple memorization. You must remember the properties of a parallelogram in order to get this question correct. There are six basic properties of every parallelogram. They are: 1. The opposite sides of a parallelogram are congruent. 2. The opposite sides of a parallelogram are parallel. 3. The opposite angles of a parallelogram are congruent. 4. The consecutive angles of a parallelo- gram are supplementary. 5. The diagonals of a parallelogram bisect each other. 6. The diagonal of a parallelogram divides the parallelogram into two congruent triangles. Find a common denominator in order to subtract the like terms. Multiply both sides by ᎏ 6 1 ᎏ in order to solve for x. 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 168 Every parallelogram has these six proper- ties. However, specific types of parallelograms, such as rectangles, rhombus, and squares, have additional properties. One of the properties shared by both rectangles and squares happens to be that the diagonals are congruent. So, the answer is choice a. Not every parallelogram has this property, only specific parallelograms such as rectangles or squares. 31. d. 52% is the same as .52 (drop the % sign and move the decimal point two places to the left); ᎏ 1 2 3 5 ᎏ = ᎏ 2 5 6 0 ᎏ = ᎏ 1 5 0 2 0 ᎏ ; 52 × 100 = .52; And 52 × 10 –2 = 52 × .01 = .52. Obviously, .052 does not equal .52, so your answer is d. 32. c. The mean is the average. First, you add 80 + 85 + 90 + 90 + 95 + 95 + 95 + 100 + 100 = 830. Divide by the number of tests: 830 ÷ 9 = 92.22, which shows that statement I is false. The median is the middle number, which is 95. And the mode is the number that appears most frequently, which is also 95; therefore, statement II is correct. 33. d. An obtuse angle measures greater than 90°. A square has four angles that are 90° each, as does a rectangle and cube. The angles inside a triangle add up to 180°, and one angle in a right triangle is 90°, so the other two add up to 90°, so there cannot be one angle that alone has more than 90 degrees. Therefore, the answer is d. 34. c. Set up a proportion: ᎏ 1 5 00 ᎏ = ᎏ 2 x 0 ᎏ . Cross multi- ply: 5x = 2,000. Then divide both sides by 5 to get x = 400. This is only the first part of the problem. If you chose answer d, you for- got to do the next step, which is to find what number is 50% of 400; ᎏ 1 5 0 0 0 ᎏ = ᎏ 40 x 0 ᎏ , or reduce to ᎏ 1 2 ᎏ = ᎏ 40 x 0 ᎏ . Then again, cross multiply: 400 = 2x. Divide both sides by 2 to get x = 200. 35. d. If you look at the pattern, you will see it is 3x – 1. Plug in some numbers, like 3(1) – 1 = 2, 3(2) – 1 = 5, 3(3) – 1 = 8, etc. You can see that since every other number is even, of the first 100 terms, half will be even. 36. c. The total amount of profit according to the graph is 9% of the year’s income. Therefore, 225,198 × .09 = 20,267.2. 37. b. First, solve for x: x 2 – 1 = 36 Add 1 to both sides. +1 +1 x 2 = 37 x 2 = 37 Take the square root of both sides. ͙x 2 ෆ = ͙37 ෆ x = ͙37 ෆ ͙37 ෆ is an irrational number. Irrational numbers cannot be expressed as a ratio of two integers. (Simply put, irrational num- bers have decimal extensions that never ter- minate or extensions that never repeat.) A prime number has only two positive factors, itself and 1. Rational numbers can be expressed as a ratio of two integers. The set ofintegers is:{ –3,–2,–1,0,1,2,3, }. ͙37 ෆ is not prime, rational, or an integer. You can use your calculator to see that it is 6 with a decimal extension that neither termi- nates nor repeats. 38. c. An effective way figure out this question is to plug in some low, easy numbers to see what will happen. Below we picked (1,7) as our point A and (5,15) as our point B. (Note that the x-coordinate of our point B is 4 greater than the x-coordinate of our point A.) xy 05 1 7 pick as A 29 311 413 5 15 pick as B 617 As you can see, the y-coordinate of B is 8 greater than the y-coordinate of A. –THE SAT MATH SECTION– 169 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 169 39. c. Converting mixed numbers into improper fractions is a two-step process. First, multiply the whole number by the denominator (bot- tom number) of the fraction. Then add that number to the numerator of the fraction. So 1 ᎏ 2 7 ᎏ becomes ᎏ 9 7 ᎏ and 1 ᎏ 4 5 ᎏ becomes ᎏ 9 5 ᎏ . Since Area = length × width, ᎏ 9 7 ᎏ × ᎏ 9 5 ᎏ = ᎏ 8 3 1 5 ᎏ = 2 ᎏ 1 3 1 5 ᎏ .Remem- ber, to convert the improper fraction ( ᎏ 8 3 1 5 ᎏ ) back into a mixed number, you divide the denominator (35) into the numerator (81). Any remainder becomes part of the mixed number (35 goes into 81 twice with a remainder of 11, hence 2 ᎏ 1 3 1 5 ᎏ ). 40. d. We use D = RT, and rearrange for T.Divid- ing both sides by R,we get T = D ÷ R.The total distance, D = (x + y), and R = 2 mph. Thus, T = D ÷ R becomes T = (x + y) ÷ 2. –THE SAT MATH SECTION– 170 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 170  Part 2: Grid-in Questions “Grid-in”questions are also called student-response ques- tions because no answer choices are given; you, the stu- dent, generate the response. Otherwise, grid-in questions are just like five-choice questions. In responding to the grid-in questions on the SAT, there are several things you will need to know about the special four-column grid. Become familiar with the answer grid below. The above answer grid can express whole numbers from 0 to 9999, as well as some fractions and decimals. To grid an answer, write it in the top row of the column. If you need to write a decimal point or a fraction bar, skip a column and fill in the necessary oval below it. ■ Very important: No grid-in questions will have a negative answer. If you get a negative number, you have done something wrong. ■ Write the answer in the column above the oval. The answer you write will be completely disre- garded because the scoring machine will only read the ovals. It is still important to write this answer, however, because it will help you check your work at the end of the test and ensure that you marked the appropriate ovals. ■ Answers that need fewer than four columns, except 0, may be started in any of the four columns, 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. ■ Enter mixed numbers as improper fractions or decimals. This is important for you to know when working on the grid-in section. As a math stu- dent, you are used to always simplifying answers to their lowest terms and often converting improper fractions to mixed numbers. On this section of the test, however, just leave improper fractions as they are. For example, it is impossible to grid 1 ᎏ 1 2 ᎏ in the answer grid, so simply grid in ᎏ 3 2 ᎏ 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. If you get a fraction that fits into the grid, do not waste time changing it to a decimal. Changing the form of an answer can result in a miscalculation and is completely unnecessary. ■ Enter the decimal point first, followed by the first three digits of a long or repeating decimal. Do not round the answer. It won’t be marked as wrong if you do, but it is not necessary. ■ If the answer is a fraction that requires more than four digits, like ᎏ 1 2 7 5 ᎏ , write the answer as a decimal instead. The fraction ᎏ 1 2 7 5 ᎏ does not fit into the grid and it cannot be reduced; therefore, you must turn it into a decimal by dividing the numerator by the denominator. In this case, the decimal would be .68. ■ If a grid-in answer has more than one possibility, enter any of the possible answers. This can occur when the answer is an inequality or the solution to a quadratic equation. For example, if the answer is x < 5, enter a 4. If the answer is x = ±3, enter positive 3, since negative numbers cannot be entered into the grid. ■ If you are asked for a percentage, only grid the numerical value without the percentage sign. There is no way to grid the symbol, so it is simply not needed. For example, 54% should be gridded as .54. Don’t forget the decimal point! 1 2 3 4 5 6 7 8 9 • 1 2 3 4 5 6 7 8 9 0 • / 1 2 3 4 5 6 7 8 9 0 • / 1 2 3 4 5 6 7 8 9 0 • –THE SAT MATH SECTION– 171 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 171 ■ Remember these important tips:  If you write in the correct answer but do not fill in the oval(s), you will get the question marked wrong.  If you know the correct answer but fill in the wrong oval(s), you will get the question marked wrong.  If you do not fully erase an answer, it may be marked wrong.  Check your answer grid to be sure you didn’t mark more than one oval per column. Be especially careful that a fraction bar or deci- mal point is not marked in the same column as a digit. Now it is time to do some grid-in practice prob- lems. Be sure to review the strategies listed above to ensure that you fully understand the grid system. Remember: You will never be penalized for an incorrect answer on the grid-in questions—so go ahead and guess. Good luck! –THE SAT MATH SECTION– 172 1 2 3 4 5 6 7 8 9 1 2 4 5 6 7 8 9 0 • / 1 2 3 5 6 7 8 9 0 • / 1 2 3 4 5 6 8 9 0 • 1 2 4 5 6 7 8 9 • 1 2 3 5 6 7 8 9 0 • / 1 2 3 4 5 6 7 8 9 0 / 1 2 3 4 5 6 8 9 0 • 1 2 4 5 6 7 8 9 • 1 2 3 5 6 7 8 9 0 • / 1 2 3 4 5 6 7 8 9 0 • 1 2 3 4 5 6 8 9 0 • .347 34 . 7 34 / 7 5658 SAT2006[04](fin).qx 11/21/05 6:44 PM Page 172 . feet THE SAT MATH SECTION 167 565 8 SAT2 0 06[ 04](fin).qx 11/21/05 6: 44 PM Page 167 THE SAT MATH SECTION 168 If 18 girls attended the party and 15 boys were at the party, then 33 students attended the. point! 1 2 3 4 5 6 7 8 9 • 1 2 3 4 5 6 7 8 9 0 • / 1 2 3 4 5 6 7 8 9 0 • / 1 2 3 4 5 6 7 8 9 0 • THE SAT MATH SECTION 171 565 8 SAT2 0 06[ 04](fin).qx 11/21/05 6: 44 PM Page 171 ■ Remember these important tips:  If you write in the correct answer. y-coordinate of A. THE SAT MATH SECTION 169 565 8 SAT2 0 06[ 04](fin).qx 11/21/05 6: 44 PM Page 169 39. c. Converting mixed numbers into improper fractions is a two-step process. First, multiply the whole