34. C Since you’ve been dealing with paragraph 4 for the past two questions, you should have a good idea of the author’s point there. Quickly go to the quoted lines to check the specifics. (A) is too narrow; the author’s big point is not about “tribal boundaries.” (D) makes a larger point, but it overstates the case. The author doesn’t say that Cro-Magnon and aboriginal art serve the same functions, but that because we know that aboriginal artists paint for many reasons, we shouldn’t oversimplify the motivations of Cro-Magnon artists (C). The author says nothing about depictions of people in early rock art (B). And (E) infers too much—stick to what the author says. 35. E A big picture question. Unlike the questions you’ve been dealing with, this one doesn’t give you a line reference. You need an answer that represents the whole passage. Does (A) do this? No, Cro-Magnon contemporaries are never discussed. Does (B)? No, agriculture isn’t discussed. As for (C), the author’s point isn’t that Cro-Magnons weren’t cavemen, but that they differed from our image of cavemen. Anyway, this isn’t the author’s main point. Neither is (D)—the author never claims Cro-Magnons were artistically more sophisticated than later civilizations. (E) is best: at points throughout the passage (paragraphs 2, 3, and 4), the author likens Cro-Magnons to modern people. section three SAT Virtual Reality III 23 Section 4 (Math) 1. C There are 60 minutes in 1 hour, so there are 120 minutes in 2 hours. Likewise, there are 60 seconds in 1 minute, so there are 120 seconds in 2 minutes. Both columns equal 120, so the answer is (C). 2. D If 2 balls are yellow out of a total of 10 balls, then the probability of choosing a yellow ball is just 2 out of 10, so write in Column A. We don’t know exactly how many of the remaining 8 balls are blue. It’s possible that 7 are blue and 1 is green, in which case the probability of choosing a blue ball is , which is greater than . On the other hand, perhaps there’s 1 blue ball and 7 green balls. In that case the probability of choosing a blue ball is , which is less than . Since the relationship depends on the exact number of blue balls in the bag, which is unknown, the correct answer is (D). 3. A If the areas of the 2 triangles are equal, then times the base times the height of triangle A is equal to times the base times the height of triangle B. You can make this into an equation: b A h A = b B h B If you multiply both sides of this equation by 2, you get: b A h A = b B h B Yo u ’re given that the the base of triangle B is greater than the base of triangle A, in other words, b B > b A . This means that the height of triangle B must be less than the height of triangle A in order for the equation above to be true. If this doesn’t make sense to you, try plugging in numbers for the 2 bases. Suppose b B = 3 and b A = 2. Then you have: 1 ᎏ 2 1 ᎏ 2 1 ᎏ 2 1 ᎏ 2 2 ᎏ 10 1 ᎏ 10 2 ᎏ 10 7 ᎏ 10 2 ᎏ 10 section four SAT Virtual Reality III 24 section four SAT Virtual Reality III 25 2h A = 3h B What values for the heights would make this equation true? Some possibilities are h A = 6 and h B = 4, or h A = 9 and h B = 6. There are an infinite number of other possibilities. In all cases, however, h A > h B , so the correct answer is (A). 4. C The angle labeled a° and the angle labeled 45° are vertical angles. Vertical angles are the angles opposite each other that are formed when two lines cross. Vertical angles are always equal, so a must equal 45. The 45° angle and the angle labeled b° lie on a straight line, so their sum must be 180°. So, b + 45 = 180, and b = 135. Since 3a = 3 × 45 = 135, the 2 columns are equal and the correct answer is (C). 5. A In Column A, you can factor and reduce: = = a + 1 The expression in Column B is equal to 1, since any number divided by itself is equal to 1 and 1 2 = 1. So, you’re comparing a + 1 with 1. If you subtract 1 from both columns you’re left with a in Column A and 0 in Column B, and since the centered information tells you that a > 0, Column A must be greater and the correct answer is (A). 6. A This is an exercise in graph-reading, a topic you might want to do some review on if you had trouble with this question. The chart shows 6 small computer symbols next to Company A and 5 small computer symbols next to Company B. These symbols represent the number of computers that each company sold. Since you’re told that Company A has sold 50 more computers than Company B, and the chart shows one more symbol for Company A than for Company B, the symbol must represent 50 computers. Therefore 5 small computer symbols must represent 5 × 50 = 250, which is greater than 125, so the correct choice is (A). 7. B The sum of all the odd integers from –11 to 29 is the sum of all the negative odd integers from –11 to –1, plus all the positive odd integers from 1 to 29. But adding up the first 12 terms in this sum gives you –11 – 9 – 7 – 5 – 3 – 1 + 1 + 3 + 5 + 7 + 9 + 11, which adds up to 0. So, the sum of all the odd integers from –11 to 29 is the same as the sum of all the odd integers from 13 to 29. The sum of all the odd integers from 11 to 29 is the same as 11 plus the sum of all the odd integers from 13 to 29, so Column B is 11 more than Column A and the answer is (B). 8. C Here you have 2 equations and 2 unknowns, so you should be able to solve for both x and y. Start by multiplying both sides of the equation = by y to give you x = . Then substitute for x in the 2nd equation. Then instead of y – x = 12 you have y – = 12, or = 12. Multiply both sides by to get y = 16. 4 ᎏ 3 3y ᎏ 4 y ᎏ 4 y ᎏ 4 y ᎏ 4 1 ᎏ 4 x ᎏ y a(a + 1) ᎏ a a 2 + a ᎏ a section four SAT Virtual Reality III 26 Since x = and is 4, x = 4. Then Column A, which is 2x, must be 8, and Column B, which is y, must also be 8 and the correct answer choice is (C). 9. D If the average of x, y, and 9 is 7 then = 7. That means that x + y + 9 = 21 and so x + y = 12. Then the average of x and y is = = 6. The average of x, y, and y is = . We don’t know what the value of y is, only what the value of x + y is, so we can’t tell exactly what this average is. The average of x, y, and y could be less than 6, equal to 6, or greater than 6, so the answer is (D). 10. A The area of a circle is πr 2 , so the area of a circle with radius ͙5 ෆ is π(͙5ෆ) 2 = π × ͙5 ෆ × ͙5 ෆ = π × 5 = 5π. If the diameter of a circle is ͙1 ෆ 0 ෆ , then the radius of the circle is . Then the area of the circle is π = π × × = π × = π × = . 5π is greater than , so the correct answer is (A). If you had any trouble multiplying out those square roots you might want to do some review in your Math Reference Book. 11. B An isosceles triangle has 2 equal sides, and here the sum of those 2 sides plus the third side is equal to 7. Ordinarily that wouldn’t tell you much, since there are infinitely many combinations of numbers that add up to 7, but you’re also told that the sides have integer values. That narrows things down quite a bit. How many combinations of 3 integers add up to 7? Remember, 2 of those integers have to be the same since the triangle is isosceles. 1 + 1 + 5 is a possibility, and so is 2 + 2 + 3 and 3 + 3 + 1. Those are the only 3 possibilities since any other combinations of positive integers add up to more than 7 or don’t have 2 numbers the same. (Remember that since we’re dealing with the sides of a triangle here, you can’t include negative numbers or 0.) Now we have 3 combinations of numbers, so it seems like the correct answer should be (C). However, that’s a little too easy for a QC number 11. There’s another consideration here — could all 3 of those combinations really be the lengths of the sides of a triangle? The key point to figuring that out is to remember that the sum of the lengths of any 2 sides of a triangle has to be greater than the length of the 3rd. That means that 1, 1, and 5 5π ᎏ 2 5π ᎏ 2 5 ᎏ 2 10 ᎏ 4 ͙1 ෆ 0 ෆ ᎏ 2 ͙1 ෆ 0 ෆ ᎏ 2 10 2       2 ͙1 ෆ 0 ෆ ᎏ 2 12 + y 3 x + y + y ᎏᎏ 3 12 ᎏ 2 x + y ᎏ 2 x + y + 9 ᎏᎏ 3 1 ᎏ 2 16 ᎏ 4 y ᎏ 4 could not possibly represent the lengths of the 3 sides of a triangle. If you try to draw a triangle with sides that are 1 inch, 1 inch, and 5 inches long, you’ll see that it’s impossible. It turns out that the other 2 combinations are OK, so there are 2 possibilities. But even just knowing that 1 of the 3 possibilities is not possible tells you that there are less than 3 different-sized triangles that meet the given criteria, so the correct answer is (B). 12. A Since the centered information has a 3a and a b in it and the columns have a 15a and a 5b in them, you can try multiplying the centered information equation by 5 to make everything more alike. If you multiply both sides of 3a – 5 = b by 5 you get 5(3a – 5) = 5b and multiplying out gives you 15a – 25 = 5b. This means that Column B is the same as 15a – 25. Which is greater 15a – 24 or 15a – 25? If you’re not sure, subtract 15a from both columns. That leaves you with –24 and –25, and since –24 is greater, the correct answer is choice (A). 13. A In the figure on the left, the square has been divided into 2 equal triangles. Each of the legs of either one of those triangles is a side of the square, and the angle formed by the legs is a right angle. Let’s call the length of a side s. Then we have a right triangle with 2 equal legs, which is an isosceles right triangle. You should remember that the sides of an isosceles right triangle are in the ratio 1:1:͙2 ෆ . (If you didn’t remember this, you could always figure it out by using the Pythagorean theorem.) So if the legs have length s then the hypotenuse must have length s͙2 ෆ . So the perimeter of the triangle is s + s + s͙2 ෆ = 2s + s͙2 ෆ = s(2 + ͙2 ෆ ). The figure on the right shows an identical square divided into 2 rectangles. The length of each rectangle is s, and since the rectangles are equal in area, the width of the rectangles must be . So the perimeter of either rectangle is s + s + + = 3s. You’re comparing s(2 + ͙2 ෆ ) with 3s, and since s > 0, you can just compare 2 + ͙2 ෆ with 3. Since ͙2 ෆ is greater than 1, 2 + ͙2 ෆ must be greater than 3 and the correct answer is (A). 14. C Each of Callie's children are exactly 2 years apart. That means that in a given year, all of her children's ages will be all odd or all even. For example, during the year when her youngest child turns 1, the ages of her children will be 1, 3, 5, 7, and 9. The following year, the ages will be 2, 4, 6, 8, and 10. Therefore, the probability that that the sum of the ages of any two of her children will be even is 1. In other words, there is a 100% chance that the sum of the ages of any two of s ᎏ 2 s ᎏ 2 s ᎏ 2 section four SAT Virtual Reality III 27 her children will be 1 because the sum of any two even or any two odd numbers will always be even. 15. D You should consider the possibility that choice (D) is correct because there are so many possibilities for x, y, and z. So let’s pick some values for x, y, and z. Suppose that x = 10, y = 4, and z = 3. Then Column A is xyz = 10 × 4 × 3 = 120 while Column B is xy + yz + xz = 10 × 4 + 4 × 3 + 10 × 3 = 40 + 12 + 30 = 82. In this case Column A is greater. Now let’s let each of x, y, and z be negative. Let x = –1, y = –2, and z = –3. Then in Column A, xyz = (–1) × (–2) × (–3) = 2 × (–3) = –6. In Column B, xy + yz +xz = (–1) × (–2) + (–2) × (–3) + (–1) × (–3) = 2 + 6 + 3 = 11. In this case Column B is greater. Since different relationships between the columns are possible, choice (D) is correct. 16. 13 It’s always a good idea to start working with the simplest equation when you’re given more than one. The 3rd equation only has one variable in it, so let’s start with that one. If 2c = 16, then c = 8. Now that you have a value for c, the next logical step is to plug that value into any other equation that has a c in it. The second equation is c – b = 5, so that becomes 8 – b = 5. That means that b is 3. Plugging b = 3 into the first equation gives you 3a = 6 which means that a = 2. Now you know that a is 2, b is 3, and c is 8, so a + b + c = 2 + 3 + 8 = 13, so grid that in. 17. 87.5 If 50 out of 400 seniors are majoring in economics then 400 – 50 or 350 seniors are NOT majoring in economics. Since PERCENT × WHOLE = PART, then PERCENT = . 350 is the part and 400 is the whole, so the percent we’re looking for is just . Dividing both numbers by 4 gives you , which is 87.5%, so just grid in 87.5. 18. 1.25 Another way of saying that 5 pens cost as much as 2 notebooks is the equation 5P = 2N. That’s just translating from words to math, by using N to represent the cost of a notebook and P to represent the cost of a pen. If one notebook plus one pen cost $1.75, then N + P = $1.75. If you take the first equation and solve for P you get P = . You can plug that into the second equation to get N + = $1.75, or = $1.75. Solving for N gives you N = ($1.75) = $1.25, so put 1.25 in the grid. 5 7 7N ᎏ 5 2N ᎏ 5 2 N ᎏ 5 87.5 ᎏ 100 350 ᎏ 400 PART ᎏ WHOLE section four SAT Virtual Reality III 28 section four SAT Virtual Reality III 29 19. 44 The 4 scores you’re given here average to 86. That means that the sum of the 4 scores divided by 4 is 86, and so the sum of the scores is 4 × 86, or 344. That’s just another way of stating the average formula — if average = , then sum of the terms = average × number of terms. If you have 4 scores that add up to a particular number, and you want 1 score to be as low as possible, you have to make the other 3 scores as high as possible. The highest possible value for any one score is 100, so the lowest score occurs when the other 3 scores are each 100, and 100 + 100 + 100 + the lowest score = 344. A tiny little bit of arithmetic will tell you that the lowest score must be 44, which is what you should put into the grid. 20. 122, 182, or 242 Here you’re told that v leaves a remainder of 2 when it’s divided by 3, 4, or 5. This means that v – 2 is a multiple of 3, 4, and 5. Since 3, 4, and 5 have no common factors greater than 1, v – 2 must be a multiple of 3 × 4 × 5 = 60. So v is 2 more than a multiple of 60. Let’s keep in mind that v is a three-digit number less than 250. So v can’t be 2 + 60 or 62. But v could be 2 + 2(60) = 122, v could be 2 + 3(60) = 182, and v could be 2 + 4(60) = 242. Notice that 2 + 5(60) = 302 is greater than 250. So v can’t be 302, or 2 more than any larger multiple of 60. So v could be 122, 182, or 242. 21. 81 Since you’re given that angle ADF is 36°, the first thing to do is to label that angle 36° in the diagram. Do you see any other angles that you know the measure of? Since ABCD is a square, the corner angles are all 90°, and since AC is a diagonal, it cuts those corner angles in half. Therefore angles DAF, FAB, BCF, and DCF are all 45° angles. If you write all that on the diagram, you should notice that you now know the measure of 2 out of the 3 angles of triangle ADF. Since angle ADF is 36° and angle DAF is 45°, angle AFD must be 180° – 36° – 45° = 99°. Since AFD and the angle measuring x ° lie on a straight line, x must be 180 – 99 = 81. Grid in an 81. 22. or .208 You probably want to start this question by drawing a number line. Put and on your number line, with to the left of since it is smaller: 1 ᎏ 4 1 ᎏ 6 1 ᎏ 4 1 ᎏ 6 5 ᎏ 24 sum of the terms number of terms 1 6 1 4 X What point is the same distance from as it is from ? Certainly it’s the point that is halfway in between and , so that must be where point X is. The distance from point X to either or is half the distance from to . The distance from to is just the difference of the 2 numbers, or – . However, you can’t subtract from because they are fractions with different denominators. You can find a common denominator, like 12. = and = , so the difference between and (and therefore the distance between them on the number line) is – = . The distance from X to is half of , or , so let’s put everything in terms of 24ths. = = and = = . If you put this on the number line, it becomes clear what X is since X is halfway between and : The coordinate of X must therefore be , so put that into the grid. 23. 54 The polygon ABCDE is clearly a five-sided polygon, which is called a pentagon. OE and OD are radii of the circle, and if you draw lines from O to A, B, and C, those lines will also be radii of the circle. If you do that you’ll see that since the sides of the pentagon are equal, all those radii divide the pentagon into 5 identical triangles. Therefore those radii divide the circle into 5 equal pieces. That means that each of the central angles formed by those radii measures of the whole circle, or × 360° = 72°. So, angle DOE measures 72°. Since 2 of the sides of triangle OED are radii of the circle and therefore equal in length, the triangle is 1 ᎏ 5 1 ᎏ 5 5 ᎏ 24 6 ᎏ 24 4 ᎏ 24 6 ᎏ 24 3 ᎏ 12 1 ᎏ 4 4 ᎏ 24 2 ᎏ 12 1 6 1 2 × 1 12 = 1 24 1 12 1 ᎏ 6 1 ᎏ 12 2 ᎏ 12 3 ᎏ 12 1 ᎏ 4 1 ᎏ 6 2 ᎏ 12 1 ᎏ 6 3 ᎏ 12 1 ᎏ 4 1 ᎏ 4 1 ᎏ 6 1 ᎏ 6 1 ᎏ 4 1 ᎏ 4 1 ᎏ 6 1 ᎏ 4 1 ᎏ 6 1 ᎏ 4 1 ᎏ 6 1 ᎏ 4 1 ᎏ 6 1 ᎏ 4 1 ᎏ 6 section four SAT Virtual Reality III 30 X 4 24 5 24 6 24 isosceles. That means that angles OED and ODE are equal and so they both measure r °. So, r ° + r ° + 72° = 180° which means that r = 54. 24. or 19.5 The figure shows 2 triangles that are formed by 2 lines that cross each other plus 2 additional lines. Each triangle has an unlabeled angle, both of which are formed where the two longer lines cross. That means that the 2 unlabeled angles are vertical angles, so they must be equal to each other. Let’s label each of those 2 angles y °. Then the triangle on the left has angles measuring y °, x °, and (3x + 1)°, and the triangle on the right has angles measuring y °, (2x)°, and 40°. The sum of the angles of any triangle is 180°, so y + x + (3x + 1) = 180 and y + 2x + 40 = 180. That gives you 2 equations and 2 unknowns so you should be able to solve for both x and y. There are a number of different ways to do so, but the easiest is to just forget the 180 and set the sums of the angles equal to each other: y + x + (3x + 1) = y + 2x + 40 y + 4x + 1 = y + 2x + 40 4x + 1 = 2x + 40 2x + 1 = 40 2x = 39 x = 19.5 So just put 19.5 into the grid. 25. If you draw yourself an 8 by 8 checkerboard, you’ll see that the only unit squares which are not completely surrounded by other unit squares are the unit squares on the border. Also, the unit squares which are completely surrounded by other unit squares make up a bigger square which is 6 by 6, that is, there are 6 unit squares along each dimension of this larger 6 by 6 square. So the number of unit squares which are completely surrounded by other squares is 6 × 6 or 36. There are 8 × 8 or 64 squares on the checkerboard. So the number of unit squares which are not completely surrounded by other squares which is also the number of unit squares on the border is 64 – 36 = 28. The ratio of the number of unit squares which are completely surrounded by other squares to the number of unit squares on the border is . Dividing the numerator and denominator by 4, we find that the ratio is . 9 ᎏ 7 36 ᎏ 28 9 ᎏ 7 39 ᎏ 2 section four SAT Virtual Reality III 31 Section 5 (Math) 1. C If = 3, then x = 9. Plug x = 9 into the expression x 2 – 1, which gives you 81 – 1 = 80. 2. E For each of 4 entrances, there are 3 exits. To find the number of combinations possible, multiply the number of entrances by the number of exits. 4 × 3 = 12. 3. B The easiest way to do this problem is to pick numbers. Choose 3 for x and 2 for y and plug them into each of the answer choices. In choice (A), 3 + 2 – 1 = 4, which is divisible by 2. In (B), (2)(3) + 3 = 9, which is not divisible by 2. That should be the answer, but look at the rest of the answer choices for the sake of discussion. (C) is (3)(2) – (2) = 4, which is even and therefore divisible by 2. (D) is 4 – 4 = 0, which is an even integer. (E) becomes 4 + 4 = 8, which is even. Therefore, (B) is indeed the correct answer. 4. D Since A is at –2 and D is at 7, AD has a length of 9 and B and C must be three units away from the endpoints. The coordinates of B and C must be –2 + 3 = 1 and 7 – 3 = 4. You aren’t told which point goes where, but only one answer choice, (D), contains one of these values. 5. B First find out how much money Tariq’s purchase will cost. 21 oranges at 30 cents each will cost 21 × $0.30 = $6.30. 12 apples at 50 cents each will cost 12 × $0.50 = $6.00. The total purchase would cost $6.30 + $6.00, or $12.30. Since Tariq only has $10, he needs $12.30 – $10.00, or $2.30 more. 6. A Because the figure in the problem is a square, you know that each angle of the quadrilateral measures 90 degrees. Therefore, when a corner is divided into 2 equal angles, the measure of each of those new angles is 90 ÷ 2, or 45 degrees, so s = 45. When a corner is divided into 3 equal angles, those new angles measure 90 ÷ 3, or 30 degrees, so r = 30. The question asks for the value of s – r, so subtract 30 from 45, leaving 15. 7. A Don’t be fooled by this question. If Sola received raffle tickets numbered 1324 to 1372 inclusive, you cannot just subtract 1324 from 1372 to find out how many tickets she was given. Instead, after subtracting you must add 1: 1372 − 1324 = 48 48 + 1= 49 x section five SAT Virtual Reality III 32 [...]... cost $1 and Sola raised $36 , it means she sold 36 tickets To find out how many she returned unsold, subtract 36 from 49 The answer is 13, choice (A) SAT Virtual Reality III 33 8 D The fastest way to do this problem is to plug the values of C and D listed in the chart into each of the answer choices Start with choice (A) Plugging in –1 and –4 works, as –4 = –1 3, but does 2 = 1 – 3? No, so move on to the... = 3( –1) – 1? 2 = 3( 1) – 1? 8 = (3) (3) – 1? 14 = 3( 5) – 1? Yes Yes Yes Yes Since all the pairs work for choice (D), that’s the answer Just for practice, look at (E) Plugging in the first pair of values gives you –4 = 3( –1) + 1 Is this true? No, because 3( –1) + 1 = –2 9 D The best way to do this problem is to find N and take 15% of it To do this, set up the following equation: 6% of N = 30 06N = 30 30 ... following equation: 6% of N = 30 06N = 30 30 N= 06 Using your calculator, you’ll get N = 500 To find 15% of N, multiply 500 by 15, and you’ll get 75 Sum of the terms Number of terms 5 + 5 + 7 + 8 + 10 = 5 35 = =7 5 Mean = Median = Middle value = 7 Mode = Value occurring most often = 5 section five 10 E This problem requires you to know how to find the mean, median, and mode of a set of data Remember that . Suppose b B = 3 and b A = 2. Then you have: 1 ᎏ 2 1 ᎏ 2 1 ᎏ 2 1 ᎏ 2 2 ᎏ 10 1 ᎏ 10 2 ᎏ 10 7 ᎏ 10 2 ᎏ 10 section four SAT Virtual Reality III 24 section four SAT Virtual Reality III 25 2h A = 3h B What. Sum of the terms Number of terms = 5 + 5 +7 + 8 +10 5 = 35 5 = 7 6% of N = 30 .06N = 30 N = 30 .06 section five SAT Virtual Reality III 33 . the grid. 5 7 7N ᎏ 5 2N ᎏ 5 2 N ᎏ 5 87.5 ᎏ 100 35 0 ᎏ 400 PART ᎏ WHOLE section four SAT Virtual Reality III 28 section four SAT Virtual Reality III 29 19. 44 The 4 scores you’re given here average