grubers complete sat guide 2009 phần 4 docx

298 • COMPLETE SAT MATH REFRESHER 22. Choice B is correct. Let S represent the side of the large square. Then the perimeter is 4S. Let s represent the side of the smaller square. Then the perimeter is 4s. Line NQ is the diagonal of the smaller square, so the length of NQ is ͙2 ෆ s. (The diagonal of a square is ͙2 ෆ times the side.) Now, NQ is equal to DC, or S, which is the side of the larger square. So now S ϭ ͙2 ෆ s. The perimeter of the large square equals 4S ϭ 4͙2 ෆ s ϭ ͙2 ෆ (4s) ϭ ͙2 ෆ ϫ perimeter of the small square. (520) 23. Choice A is correct. Angles A and B are both greater than 0 degrees and less than 90 degrees, so their sum is between 0 degrees and 180 degrees. Then angle C must be between 0 and 180 degrees. (501, 505) 24. Choice D is correct. Let the four angles be x, 2x, 3x, and 4x. The sum, 10x, must equal 360°. Thus, x ϭ 36°, and the largest angle, 4x, is 144°. (505) 25. Choice C is correct. The diagonals of a rectangle are perpendicular only when the rec- tangle is a square. AE is part of the diagonal AC, so AE will not necessarily be perpendi- cular to BD. (518) 26. Choice D is correct. Draw the three cities as the vertices of a triangle. The length of side CB is 400 miles, the length of side AB is 200 miles, and x, the length of side AC, is unknown. The sum of any two sides of a triangle is greater than the third side, or in algebraic terms: 400 ϩ 200 Ͼ x, 400 ϩ x Ͼ 200 and 200 ϩ x Ͼ 400. These simplify to 600 Ͼ x, x ϾϪ200, and x Ͼ 200. For x to be greater than 200 and Ϫ200, it must be greater than 200. Thus, the values of x are 200 Ͻ x Ͻ600. (506, 516) 27. Choice C is correct. At 7:30, the hour hand is halfway between the 7 and the 8, and the minute hand is on the 6. Thus, there are one and one-half “hour units,” each equal to 30°, so the whole angle is 45°. (501, 526) 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 298 28. Choice E is correct. If a ship is facing north, a right turn of 90° will face it eastward. Another 90° turn will face it south, and an additional 45° turn will bring it to southwest. Thus, the total rotation is 90° ϩ 90° ϩ 45° ϭ 225°. (501) 29. Choice E is correct. Since y ϭ z ϩ 30° and x ϭ 2y, then x ϭ 2(z ϩ 30°) ϭ 2z ϩ 60°. Thus, x ϩ y ϩ z equals (2z ϩ 60°) ϩ (z ϩ 30°) ϩ z ϭ 4z ϩ 90°. This must equal 180° (the sum of the angles of a triangle). So 4z ϩ 90° ϭ 180°, and the solution is z ϭ 22 ᎏᎏ 1 2 ° ; x ϭ 2z ϩ 60° ϭ 45° ϩ 60° ϭ 105°. (505) 30. Choice D is correct. Since AB is parallel to CD, angle 2 ϭ angle 6, and angle 3 ϩ angle 7 ϭ 180°. If angle 2 ϩ angle 3 equals 180°, then angle 2 ϭ angle 7 ϭ angle 6. However, since there is no evidence that angles 6 and 7 are equal, angle 2 ϩ angle 3 does not nec- essarily equal 180°. Therefore, the answer is (D). (504) 31. Choice B is correct. Call the side of the square, s. Then, the diagonal of the square is ͙2 ෆ s and the area is s 2 . The area of an isosceles right triangle with leg r is ᎏ 1 2 ᎏ r 2 . Now, the area of the triangle is equal to the area of the square so s 2 ϭ ᎏ 1 2 ᎏ r 2 . Solving for r gives r ϭ ͙2 ෆ s. The hypotenuse of the triangle is ͙r 2 ϩ r 2 ෆ . Substituting r ϭ ͙2 ෆ s, the hypotenuse is ͙2s 2 ϩ ෆ 2s 2 ෆ ϭ ͙4s 2 ෆ ϭ 2s. Therefore, the ratio of the diagonal to the hypotenuse is ͙2 ෆ s : 2s. Since ͙2 ෆ s : 2s is ᎏ ͙ 2 2 ෆ s s ᎏ or ᎏ ͙ 2 2 ෆ ᎏ , multiply by ᎏ ͙ ͙ 2 ෆ 2 ෆ ᎏ which has a value of 1. Thus ᎏ ͙ 2 2 ෆ ᎏ и ᎏ ͙ ͙ 2 ෆ 2 ෆ ᎏϭ ᎏ 2͙ 2 2 ෆ ᎏ ϭ ᎏ ͙ 1 2 ෆ ᎏ or 1 : ͙2 ෆ , which is the final result. (507, 509, 520) 32. Choice D is correct. The formula for the number of degrees in the angles of a polygon is 180(n Ϫ 2), where n is the number of sides. For a ten-sided figure this is 10(180°) Ϫ 360° ϭ (1800 Ϫ 360)° ϭ 1440°. Since the ten angles are equal, they must each equal 144°. (521, 522) 33. Choice C is correct. If three numbers represent the lengths of the sides of a right trian- gle, they must satisfy the Pythagorean Theorem: The squares of the smaller two must equal the square of the largest one. This condition is met in all the sets given except the set 9,28,35. There, 9 2 ϩ 28 2 ϭ 81 ϩ 784 ϭ 865, but 35 2 ϭ 1,225. (509) 34. Choice D is correct. Let the angle be x. Since x is its own supplement, then x ϩ x ϭ 180°, or, since 2x ϭ 180°, x ϭ 90°. (502) 35. Choice A is correct. The length of the arc intersected by a central angle of a circle is pro- portional to the number of degrees in the angle. Thus, if a 45° angle cuts off a 6-inch arc, a 360° angle intersects an arc eight times as long, or 48 inches. This is equal to the circle’s circumference, or 2␲ times the radius. Thus, to obtain the radius, divide 48 inches by 2␲. 48 inches Ϭ 2␲ϭ ᎏ 2 ␲ 4 ᎏ inches. (524, 526) COMPLETE SAT MATH REFRESHER • 299 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 299 300 • COMPLETE SAT MATH REFRESHER 36. Choice C is correct. Refer to the diagram pictured below. Calculate the distance from ver- tex 1 to vertex 2. This is simply the diagonal of a 1-inch square and equal to ͙2 ෆ inches. Now, vertices 1, 2, and 3 form a right triangle, with legs of 1 and ͙2 ෆ . By the Pythagorean Theorem, the hypotenuse is ͙3 ෆ . This is the distance from vertex 1 to vertex 3, the two most distant vertices. (509, 520) 37. Choice A is correct. In one hour, the hour hand of a clock moves through an angle of 30° (one “hour unit”). 70 minutes equals ᎏ 7 6 ᎏ hours, so during that time the hour hand will move through ᎏ 7 6 ᎏ ϫ 30°, or 35°. (501, 526) 38. Choice C is correct. In order to be similar, two triangles must have corresponding angles equal. This is true of triangles ODC and OBA, since angle O equals itself, and angles OCD and OAB are both right angles. (The third angles of these triangles must be equal, as the sum of the angles of a triangle is always 180°.) Since the triangles are similar, OD : DC ϭ OB : AB. But, OD and OA are radii of the same circle and are equal. Therefore, substitute OA for OD in the above proportion. Hence, OA : DC ϭ OB : AB. There is, however, no information given on the relative sizes of any of the line segments, so statement III may or may not be true. (509, 510, 524) 39. Choice C is correct. Let the three angles equal x, 2x, and 6x. Then, x ϩ 2x ϩ 6x ϭ 9x ϭ 180°. Therefore, x ϭ 20° and 6x ϭ 120°. (505) 40. Choice A is correct. Since AB ϭ AC, angle ABC must equal angle ACB. (Base angles of an isosceles triangle are equal). As the sum of angles BAC, ABC, and ACB is 180°, and angle BAC equals 40°, angle ABC and angle ACB must each equal 70°. Now, DBC is a right triangle, with angle BDC ϭ 90° and angle DCB ϭ 70°. (The three angles must add up to 180°.) Angle DBC must equal 20°. (507, 514) 41. Choice C is correct. Є AEB and ЄCED are both straight angles, and are equal; similarly, Є DEC and Є BEA are both straight angles. Є AEC and ЄBED are vertical angles, as are Є BEC and Є DEA, and are equal. ЄAED and ЄCEA are supplementary and need not be equal. (501, 502, 503) 42. Choice A is correct. All right isosceles triangles have angles of 45°, 45°, and 90°. Since all triangles with the same angles are similar, all right isosceles triangles are similar. (507, 509, 510) 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 300 43. Choice C is correct. As the diagram shows, the altitude to the base of the isosceles triangle divides it into two congruent right triangles, each with 5Ϫ12Ϫ13 sides. Thus, the base is 10, height is 12 and the area is ᎏ 1 2 ᎏ (10)(12) ϭ 60. (505, 507, 509) 44. Choice C is correct. The altitude to any side divides the triangle into two congruent 30°– 60°–90° right triangles, each with a hypotenuse of 2 inches and a leg of 1 inch. The other leg equals the altitude. By the Pythagorean Theorem the altitude is equal to ͙3 ෆ inches. (The sides of a 30°Ϫ60°Ϫ90° right triangle are always in the proportion 1 : ͙3 ෆ : 2.) (509, 514) 45. Choice E is correct. As the diagram illustrates, angles AED and BEC are vertical and, therefore, equal. AE ϭ EC, because the diagonals of a parallelogram bisect each other. Angles BDC and DBA are equal because they are alternate interior angles of parallel lines (AB ʈ CD). (503, 517) 46. Choice E is correct. There are eight isosceles right triangles: ABE, BCE, CDE, ADE, ABC, BCD, CDA, and ABD. (520) 47. Choice D is correct. Recall that a regular hexagon may be broken up into six equilateral triangles. Since the angles of each triangle are 60°, and two of these angles make up each angle of the hexagon, an angle of the hexagon must be 120°. (523) COMPLETE SAT MATH REFRESHER • 301 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 301 302 • COMPLETE SAT MATH REFRESHER 48. Choice E is correct. Since the radius equals 1Љ, AD, the diameter, must be 2Љ. Now, since AD is a diameter, ACD must be a right triangle, because an angle inscribed in a semicircle is a right angle. Thus, because ЄDAC ϭ 30°, it must be a 30°Ϫ60°Ϫ90° right triangle. The sides will be in the proportion 1 : ͙3 ෆ : 2. As AD : AC ϭ 2 : ͙3 ෆ , so AC, one of the sides of the equilat- eral triangle, must be ͙3 ෆ inches long. (508, 524) 49. Choice D is correct. Let the angles be 2x, 3x, 4x. Their sum, 9x ϭ 180° and x ϭ 20°. Thus, the largest angle, 4x, is 80°. (505) 50. Choice B is correct. The sides of a right triangle must obey the Pythagorean Theorem. The only group of choices that does so is the second: 12, 16, and 20 are in the 3 : 4 : 5 ratio, and the relationship 12 2 ϩ 16 2 ϭ 20 2 is satisfied. (509) 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 302 MATH REFRESHER SESSION 6 COMPLETE SAT MATH REFRESHER • 303 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 303 304 • COMPLETE SAT MATH REFRESHER Miscellaneous Problems Including Averages, Series, Properties of Integers, Approximations, Combinations, Probability, the Absolute Value Sign, and Functions Averages, Medians, and Modes 601. Averages. The average of n numbers is merely their sum, divided by n. Example: Find the average of: 20, 0, 80, and 12. Solution: The average is the sum divided by the number of entries, or: ᎏ 20 ϩ 0 ϩ 4 80 ϩ 12 ᎏ ϭ ᎏ 11 4 2 ᎏ ϭ 28 A quick way of obtaining an average of a set of numbers that are close together is the following: STEP 1. Choose any number that will approximately equal the average. STEP 2. Subtract this approximate average from each of the numbers (this sum will give some positive and negative results). Add the results. STEP 3. Divide this sum by the number of entries. STEP 4. Add the result of Step 3 to the approximate average chosen in Step 1. This will be the true average. Example: Find the average of 92, 93, 93, 96 and 97. Solution: Choose 95 as an approximate average. Subtracting 95 from 92, 93, 93, 96, and 97 gives Ϫ3, Ϫ2, Ϫ2, 1, and 2. The sum is Ϫ4. Divide Ϫ4 by 5 (the number of entries) to obtain Ϫ0.8. Add Ϫ0.8 to the original approximation of 95 to get the true average, 95 Ϫ 0.8 or 94.2. 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 304 601a. Medians. The median of a set of numbers is that number which is in the middle of all the numbers. Example: Find the median of 20, 0, 80, 12, and 30. Solution: Arrange the numbers in increasing order: 0 12 20 30 80 The middle number is 20, so 20 is the median. Note: If there is an even number of items, such as 0 12 20 24 30 80 there is no middle number. So in this case we take the average of the two middle numbers, 20 and 24, to get 22, which is the median. If there are numbers like 20 and 22, the median would be 21 ( just the average of 20 and 22). 601b. Modes. The mode of a set of numbers is the number that occurs most frequently. If we have numbers 0, 12, 20, 30, and 80 there is no mode, since no one number appears with the greatest frequency. But consider this: Example: Find the mode of 0, 12, 12, 20, 30, 80. Solution: 12 appears most frequently, so it is the mode. Example: Find the mode of 0, 12, 12, 20, 30, 30, 80. Solution: Here both 12 and 30 are modes. Series 602. Number series or sequences are progressions of numbers arranged according to some design. By recognizing the type of series from the first four terms, it is possible to know all the terms in the series. Following are given a few different types of number series that appear frequently. 1. Arithmetic progressions are very common. In an arithmetic progression, each term exceeds the previous one by some fixed number. Example: In the series 3, 5, 7, 9, . . . find the next term. Solution: Each term in the series is 2 more than the preceding one, so the next term is 9 ϩ 2 or 11. If the difference in successive terms is negative, then the series decreases. Example: Find the next term: 100, 93, 86, 79 . . . . Solution: Each term is 7 less than the previous one, so the next term is 72. COMPLETE SAT MATH REFRESHER • 305 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 305 306 • COMPLETE SAT MATH REFRESHER 2. In a geometric progression each term equals the previous term multiplied by a fixed number. Example: What is the term of the series 2, 6, 18, 54 . . . ? Solution: Each term is 3 times the previous term, so the fifth term is 3 times 54 or 162. If the multiplying factor is negative, the series will alternate between positive and negative terms. Example: Find the next term of Ϫ2, 4, Ϫ8, 16 . . . Solution: Each term is Ϫ2 times the previous term, so the next term is Ϫ32. Example: Find the next term in the series 64, Ϫ32, 16, Ϫ8 . . . Solution: Each term is Ϫ ᎏ 1 2 ᎏ times the previous term, so the next term is 4. 3. In mixed step progression the successive terms can be found by repeating a pattern of add 2, add 3, add 2, add 3; or a pattern of add 1, multiply by 5, add 1, multiply 5, etc. The series is the result of a combination of operations. Example: Find the next term in the series 1, 3, 9, 11, 33, 35 . . . Solution: The pattern of successive terms is add 2, multiply by 3, add 2, multiply by 3, etc. The next step is to multiply 35 by 3 to get 105. Example: Find the next term in the series 4, 16, 8, 32, 16 . . . Solution: Here, the pattern is to multiply by 4, divide by 2, multiply by 4, divide by 2, etc. Thus, the next term is 16 times 4 or 64. 4. If no obvious solution presents itself, it may be helpful to calculate the difference between each term and the preceding one. Then if it is possible to determine the next increment (the difference between successive terms), add it to the last term to obtain the term in question. Often the series of increments is a simpler series than the series of original terms. Example: Find the next term in the series 3, 9, 19, 33, 51 . . . Solution: Write out the series of increments: 6, 10, 14, 18 . . . (each term is the difference between two terms of the original series). This series is an arithmetic progression whose next term is 22. Adding 22 to the term 51 from the original series produces the next term, 73. 5. If none of the above methods is effective, the series may be a combination of two or three different series. In this case, make a series out of every other term or out of every third term and see whether these terms form a series that can be recognized. Example: Find the next term in the series 1, 4, 4, 8, 16, 12, 64, 16 . . . Solution: Divide this series into two series by taking out every other term, yielding: 1, 4, 16, 64 . . . and 4, 8, 12, 16 . . . These series are easy to recognize as a geometric and arith- metic series, but the first series has the needed term. The next term in this series is 4 times 64 or 256. Properties of Integers 603. Even-Odd. These are problems that deal with even and odd numbers. An even number is divisible by 2, and an odd number is not divisible by 2. All even numbers end in the digits 0, 2, 4, 6, or 8; odd numbers end in the digits 1, 3, 5, 7, or 9. For example, the numbers 358, 90, 18, 9,874, and 46 are even numbers. The numbers 67, 871, 475, and 89 are odd numbers. It is important to remember the following facts: 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 306 604. The sum of two even numbers is even, and the sum of two odd numbers is even, but the sum of an odd number and an even number is odd. For example, 4 ϩ 8 ϭ 12, 5 ϩ 3 ϭ 8, and 7 ϩ 2 ϭ 9. Example: If m is any integer, is the number 6m ϩ 3 an even or odd number? Solution: 6m is even since 6 is a multiple of 2. 3 is odd. Therefore 6m ϩ 3 is odd since even ϩ odd ϭ odd. 605. The product of two odd numbers is odd, but the product of an even number and any other number is an even number. For example, 3 ϫ 5 ϭ 15 (odd); 4 ϫ 5 ϭ 20 (even); 4 ϫ 6 ϭ 24 (even). Example: If m is any integer, is the product (2m ϩ 3)(4m ϩ 1) even or odd? Solution: Since 2m is even and 3 is odd, 2m ϩ 3 is odd. Likewise, since 4m is even and 1 is odd, 4m ϩ 1 is odd. Thus (2m ϩ 3)(4m ϩ 1) is (odd ϫ odd) which is odd. 606. Even numbers are expressed in the for m 2k where k may be any integer. Odd numbers are expressed in the form of 2k ϩ 1 or 2k Ϫ 1 where k may be any integer. For example, if k ϭ 17, then 2k ϭ 34 and 2k ϩ 1 ϭ 35. If k ϭ 6, then we have 2k ϭ 12 and 2k ϩ 1 ϭ 13. Example: Prove that the product of two odd numbers is odd. Solution: Let one of the odd numbers be represented as 2x ϩ 1. Let the other number be represented as 2y ϩ 1. Now multiply (2x ϩ 1)(2y ϩ 1). We get: 4xy ϩ 2x ϩ 2y ϩ 1. Since 4xy ϩ 2x ϩ 2y is even because it is a multiple of 2, that quantity is even. Since 1 is odd, we have 4xy ϩ 2x ϩ 2y ϩ 1 is odd, since even ϩ odd ϭ odd. 607. Divisibility. If an integer P is divided by an integer Q, and an integer is obtained as the quotient, then P is said to be divisible by Q. In other words, if P can be expr essed as an integral multiple of Q , then P is said to be divisible by Q. For example, dividing 51 by 17 gives 3, an integer. 51 is divisible by 17, or 51 equals 17 times 3. On the other hand, dividing 8 by 3 gives 2 ᎏ 2 3 ᎏ , which is not an integer. 8 is not divisible by 3, and there is no way to express 8 as an integral multiple of 3. There are various tests to see whether an integer is divisible by certain numbers. These tests are listed below: 1. Any integer is divisible by 2 if the last digit of the number is a 0, 2, 4, 6, or 8. Example: The numbers 98, 6,534, 70, and 32 are divisible by 2 because they end in 8, 4, 0, and 2, respectively. 2. Any integer is divisible by 3 if the sum of its digits is divisible by 3. Example: Is the number 34,237,023 divisible by 3? Solution: Add the digits of the number. 3 ϩ 4 ϩ 2 ϩ 3 ϩ 7 ϩ 0 ϩ 2 ϩ 3 ϭ 24. Now, 24 is divisible by 3(24 Ϭ 3 ϭ 8) so the number 34,237,023 is also divisible by 3. 3. Any integer is divisible by 4 if the last two digits of the number make a number that is divisible by 4. Example: Which of the following numbers is divisible by 4? 3,456, 6,787,612, 67,408, 7,877, 345, 98. Solution: Look at the last two digits of the numbers, 56, 12, 08, 77, 45, 98. Only 56, 12, and 08 are divisible by 4, so only the numbers, 3,456, 6,787,612, and 67,408 are divisible by 4. 4. An integer is divisible by 5 if the last digit is either a 0 or a 5. Example: The numbers 780, 675, 9,000, and 15 are divisible by 5, while the numbers 786, 5,509, and 87 are not divisible by 5. COMPLETE SAT MATH REFRESHER • 307 175-346.qxd:22678_0175-0258.qxd 5/1/08 4:02 PM Page 307 [...]... 7, 8, ? (A) (B) (C) (D) (E) 8 9 10 11 12 1 2 1 4 1 4 1 4 17 What is the average of the following numbers: 3 ᎏᎏ, 4 ᎏᎏ, 2 ᎏᎏ, 3 ᎏᎏ, 4? (A) (B) (C) (D) (E) 3.25 3.35 3 .45 3.50 3.60 18 Which of the following numbers is divisible by 24? (A) (B) (C) (D) (E) 76,300 78,132 80 ,42 4 81,2 34 83,636 175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 315 COMPLETE SAT MATH REFRESHER • 315 19 A B C D E | | | | |... 320 320 • COMPLETE SAT MATH REFRESHER Answer Key for Practice Test 6 1 2 3 4 5 6 7 8 9 10 11 12 13 E D C D B E D C B D E E C 14 15 16 17 18 19 20 21 22 23 24 25 26 D C A C C D B E E B E D B 27 28 29 30 31 32 33 34 35 36 37 38 C A B E E D A A E E A B 39 40 41 42 43 44 45 46 47 48 49 50 A E B B E A B B C C E D Answers and Solutions for Practice Test 6 1 Choice E is correct The five consecutive odd numbers... (B) (C) (D) (E) 13, 944 15, 746 15,966 16,012 None of the above 45 Which of the following numbers is a prime? (A) (B) (C) (D) (E) 147 149 153 155 161 46 What is the next number in the following series: 4, 8, 2, 4, 1, ? (A) (B) (C) (D) (E) 1 2 4 8 16 175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 319 COMPLETE SAT MATH REFRESHER • 319 47 48 A B C D E | | | | | | | | | | | | | | | | | | | | | | | |... | | | | 22 23 24 34 25 0.02 0.2 2.0 20 200 42 What is the next number in the series 5, 2, 4, 2, 3, 2, ? (A) (B) (C) (D) (E) 1 2 3 4 5 43 If a, b, and c are all divisible by 8, then their average must be (A) (B) (C) (D) (E) divisible by 8 divisible by 4 divisible by 2 an integer None of the above 44 Which of the following numbers is divisible by 24? (A) (B) (C) (D) (E) 13, 944 15, 746 15,966 16,012... Mountains (2) How much does the New England area spend per pupil? $48 0 (3) How much less does the Great Plains spend per pupil than the Great Lakes? $46 4 Ϫ 44 7 ϭ $17/pupil (4) How much more does New England spend on a pupil than the Rocky Mountain area? $48 0 Ϫ 43 3 ϭ $47 /pupil 175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 329 COMPLETE SAT MATH REFRESHER • 329 Circle Graphs 705 A circle graph shows...175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 308 308 • COMPLETE SAT MATH REFRESHER 5 Any integer is divisible by 6 if it passes the divisibility tests for both 2 and 3 Example: Is the number 12 ,41 4 divisible by 6? Solution: Test whether 12 ,41 4 is divisible by 2 and 3 The last digit is a 4, so it is divisible by 2 Adding the digits yields 1 ϩ 2 ϩ 4 ϩ 1 ϩ 4 ϭ 12 12 is divisible... or 18 ᎏᎏ 3 3 175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 3 24 3 24 • COMPLETE SAT MATH REFRESHER 44 Choice A is correct To be divisible by 24, a number must be divisible by both 3 and 8 Only 13, 944 and 15,966 are divisible by 3; of these, only 13, 944 is divisible by 8 (13, 944 ϭ 24 ϫ 581) (607) 45 Choice B is correct The approximate square root of each of these numbers is 13 Merely divide each of... the next term in the following series: 8, 3, 10, 9, 12, 27, ? (A) (B) (C) (D) (E) 8 14 18 36 81 39 Which of the following numbers is divisible by 11? (A) (B) (C) (D) (E) 30,217 44 ,221 59 ,40 3 60 ,41 1 None of the above 175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 318 318 • COMPLETE SAT MATH REFRESHER 40 41 A B C D E | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |... f(Ϫ2) ϭ (Ϫ2)2 Ϫ (Ϫ2) ϭ 6 f(z) ϭ z2 Ϫ z f(2z) ϭ (2z)2 Ϫ (2z) ϭ 4z2 Ϫ 2z Let us consider still another example: Let f(x) ϭ x ϩ 2 and g(y) ϭ 2y What is f [ g(Ϫ2)]? Now 1 1 1 1 1 g(Ϫ2) ϭ 2Ϫ2 ϭ ᎏᎏ Thus f [g(Ϫ2)] ϭ f ᎏᎏ Since f (x) ϭ x ϩ 2, f ᎏᎏ ϭ ᎏᎏ ϩ 2 ϭ 2 ᎏᎏ 4 4 4 4 4 ΂΃ ΂΃ 175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 312 312 • COMPLETE SAT MATH REFRESHER Practice Test 6 Miscellaneous Problems Including... 12 12 is divisible by 3 so the number 12 ,41 4 is divisible by 3 Since it is divisible by both 2 and 3, it is divisible by 6 6 Any integer is divisible by 8 if the last three digits are divisible by 8 (Since 1,000 is divisible by 8, you can ignore all multiples of 1,000.) Example: Is the number 342 ,169 ,42 4 divisible by 8? Solution: 42 4 Ϭ 8 ϭ 53, so 342 ,169 ,42 4 is divisible by 8 7 Any integer is divisible . radius, divide 48 inches by 2␲. 48 inches Ϭ 2␲ϭ ᎏ 2 ␲ 4 ᎏ inches. (5 24, 526) COMPLETE SAT MATH REFRESHER • 299 175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 299 300 • COMPLETE SAT MATH REFRESHER 36 3 ᎏ 1 2 ᎏ , 4 ᎏ 1 4 ᎏ , 2 ᎏ 1 4 ᎏ , 3 ᎏ 1 4 ᎏ , 4? (A) 3.25 (B) 3.35 (C) 3 .45 (D) 3.50 (E) 3.60 18. Which of the following numbers is divisible by 24? (A) 76,300 (B) 78,132 (C) 80 ,42 4 (D) 81,2 34 (E). Since f(x) ϭ x ϩ 2, f ΂ ᎏ 1 4 ᎏ ΃ ϭ ᎏ 1 4 ᎏ ϩ 2 ϭ 2 ᎏ 1 4 ᎏ . COMPLETE SAT MATH REFRESHER • 311 175- 346 .qxd:22678_0175-0258.qxd 5/1/08 4: 02 PM Page 311 312 • COMPLETE SAT MATH REFRESHER Practice