sat virtual reality ii explanations phần 3 doc

So put in your grid. (And even though is the same as 1 , don’t be tempted to put 1 into the grid, since you can’t grid in mixed numbers!) 22. 9,984 If the ratio of students to teachers to parents is 11:2:3, then there might be 11 students, 2 teachers, and 3 parents, for a total of 16 people. On the other hand, there might be 10 × 11, or 110 students, 10 × 2, or 20 teachers, and 10 × 3, or 30 parents, for a total of 10 × 16, or 160 people. In fact there could be any number of people at all, as long as they are there in the ratio 11:2:3. That means that the number of people in attendance must be a multiple of 11 + 2 + 3, or 16. So what this question is really asking is – what’s the greatest multiple of 16 that fits on the grid? Since the grid only has 4 columns, the largest multiple of 16 that fits will be a 4-digit number, in other words a number greater than 999 but less than 10,000. So you need the largest multiple of 16 that is less than 10,000. The easiest way to find that number is to use your calculator and your knowledge of arithmetic. First of all, = 625, so 10,000 is a multiple of 16. Therefore, the largest multiple of 16 that the grid can accomodate must be 10,000 – 16 = 9,984. You can check that 624 × 16 = 9,984 on your calculator, but you don’t have to. 23. 75 Since we’re starting off with a group of 10 numbers and then removing 2 of them, the highest and the lowest, let’s express the sum of the 10 numbers as the sum of the highest number, the lowest number, and the other 8 numbers. The sum of the 10 numbers is equal to the highest number + the lowest number + the sum of the other 8 numbers. If we call the highest number h, the lowest number , and the sum of the other 8 numbers S, we can express the sum of the 10 numbers as h +  + S. Then the average of the 10 numbers is , which we’re given is 87. The average of the eight numbers is just , and we’re given that that’s 90. We want to know the average of the 2 grades that were removed, or . If = 90 then S = 90 × 8 = 720. And if = 87 then h +  + S = 87 × 10 = 870. But we know that S = 720, so h +  + 720 = 870, and h +  = 150. Then the average of h and  is , or 75, so grid in a 75. 150 ᎏ 2 h +  + S ᎏᎏ 10 S ᎏ 8 h +  ᎏ 2 S ᎏ 8 h +  + S ᎏᎏ 10 10,000 ᎏ 16 1 ᎏ 3 1 ᎏ 3 4 ᎏ 3 4 ᎏ 3 SAT Virtual Reality II 25 section three 24. 12 Here we have a figure that is not drawn to scale, so eyeballing won’t help much. The area of the top of the solid is much greater than the areas of the sides, so the solid probably doesn’t look much like the figure anyway. Don’t let the figure mislead you. You know that volume is length × width × height, so label the edges accordingly. Let’s say that the vertical lines represent the height of the solid and label all the vertical lines h. Let’s also say that the horizontal lines represent the width and label them w, and the diagonal lines are the length, so label them . So the area of face I is w , the area of face II is hw, and the area of face III is h. This means that we can write down 3 algebraic equations; w  = 24, hw = 2, and h = 3. Now you have 3 equations and 3 unknowns and you should be able to solve. It’s not obvious how to solve, however, so you’ll have to play around with these equations. Adding or subtracting them doesn’t get you very far, but if you multiply any 2 of them together something interesting happens. For example, if you multiply the first 2 equations together you get a new equation, w × hw = 24 × 2, or w 2 h = 48. Notice that the left side of the new equation is w 2 times h, which is also w 2 times the left side of the third equation. So, if you divide the new equation by the third equation you get = , or w 2 = 16. That means that w must be 4, so  is 6 and h is , and the volume is 4 × 6 × , or 12. 25. 25 Increasing 45 by P percent means adding P percent of 45 to 45, so increasing 45 by P percent results in 45 + (P% of 45) or 45 + × 45 which equals 45 + . Decreasing 75 by P percent means taking P percent of 75 away from 75 so decreasing 75 by P percent results in 75 – (P% of 75) or 75 – × 75 which equals 75 – . Increasing 45 by P percent gives the same result as decreasing 75 by P percent, so we have the equation 45 + = 75 – . Now let’s solve this equation for P. Subtracting 45 from both sides results in = 30 – . Adding to both sides gives us = 30. Multiplying both sides by results in P = × 30 = . Cancelling a factor of 30 from the top and bottom of gives us which equals 25. So P = 25. 100 × 1 ᎏ 4 100 × 30 ᎏᎏ 120 100 × 30 ᎏᎏ 120 100 ᎏ 120 100 ᎏ 120 120P ᎏ 100 75P ᎏ 100 75P ᎏ 100 45P ᎏ 100 75P ᎏ 100 45P ᎏ 100 75P ᎏ 100 P ᎏ 100 45P ᎏ 100 P ᎏ 100 1 ᎏ 2 1 ᎏ 2 48 ᎏ 3 w 2 h ᎏ h SAT Virtual Reality II 26 section three Section 4 (Math) 1. B Instead of trying to multiply the fractions in Column A, try canceling as much as you can: Now compare the columns. Since is greater than 1 and is less than 1, is greater than , and Column B is greater than Column A. 2. B This problem is relatively easy to calculate. Since 4 quarters make up one dollar and 10 dimes make up one dollar, simply multiply each of these numbers by 5 and compare them. 20 quarters can be changed for $5.00, while 50 dimes can be changed for the same amount. Column B is greater. An even faster way to do this problem is to simply compare and skip the calculations altogether. Since a quarter is worth more than a dime, it takes fewer quarters than dimes to make up $5.00. Therefore, the number of dimes that are exchanged for $5.00 is greater than the number of quarters exchanged for $5.00 and the answer is (B). 3. A Solve the centered equation for x : Now plug 3 in for the x in Column A. 3 + 2 = 5, which is greater than 4, so Column A is greater than Column B. 4. C The easiest way to do this problem is to list all the possible values of a + b : Now count up how many DIFFERENT sums there are. There are 6 combinations total, but 4 + 6 and 7 + 3 both give you 10, so there are only 5 different sums for a + b. The columns are equal and the answer is (C). 4 + 3 = 7 4 + 6 = 10 7 + 3 = 10 7 + 6 = 13 8 + 3 = 11 8 + 6 = 14 3x − 12 = 3 − 2x 5x = 15 x = 3 4 5 5 4 4 5 5 4 4 9 × 18 6 × 12 20 = 1 1 × 2 1 × 2 5 = 4 5 section four SAT Virtual Reality II 27 5. A The easiest way to deal with this problem is to get .2 out of the denominator of Column B. To do this, multiply each column by .2, yielding .9p in Column A and p in Column B. Multiplying p by .9 brings its value closer to 0, and since p is negative, it increases in value. Therefore, Column A is greater than Column B. If this is too abstract for you, pick numbers. Choose a number for p that’s easy to work with and meets the criteria of the centered information, like –.2. Multiplying –.2 by 4.5 gives you –.9, while dividing –.2 by .2 gives you –1. Since –.9 is greater than –1, Column A is greater than Column B. 6. C First, try to make the expressions in each column look more like the centered information. You can do this by breaking each expression into its factors, rewriting Column A as (6) 2 (x – y ) 2 and Column B as (6) 2 (y – x ) 2 . Now you can compare the columns piece by piece. Since both columns contain (6) 2 , ignore it. Now compare the second term of each expression. You know that x – y = 6, so (x – y ) 2 = 36. What does y – x equal? Well, y – x = (–1)(x – y), which equals –6. Since (–6) 2 = 36, the two columns are equal. 7. B Compare—don’t calculate! Use common sense to find the relationship between the two columns. In Column A, the average price of 2 dozen grade A eggs and 2 dozen grade B eggs is the same as the average of 1 dozen of each. In Column B, however, the average of 2 dozen grade A eggs and only 1 dozen grade B eggs is closer to the price of grade A eggs than it is to the price of grade B eggs. Since grade A eggs are more expensive than grade B eggs, Column B is greater than Column A. 8. B You don’t have to know the values of a and b to make the comparison. Since SRU and TRV share TRU, the only difference between the two angles is the difference between a and b. TRV is larger than SRU, which means that b must be larger than a and Column B must be greater than Column A. 9. D First solve the equation in the centered information for x 2 : Your first instinct may be to jump to the conclusion that x = 3, which makes Column A greater than Column B. However, don’t forget that x could also equal –3, which is less than 2. In that case, Column B would be greater than Column A. Since there are two possible relationships between the columns, the answer is (D). 5x 2 + 1= 46 5x 2 = 45 x 2 = 9 section four SAT Virtual Reality II 28 10. C At first you may look at this QC and decide that there’s not enough information to make a comparison. This problem is doable, however, if you realize that the two triangles share the unmarked angle on the right. We will call the measure of that shared angle c for purposes of discussion. The sum of the angles of the larger triangle can be written as x + y + c = 180, while the sum of the angles of the smaller triangle can be written as a + b + c = 180. Because both sums equal 180, set them equal to each other: x + y + c = a + b + c Subtract c from both sides: x + y = a + b Now that we have the relationship between the measures of the angles in this form, we can rearrange the variables to match the expressions in the columns: x – b = a – y The answer is (C). 11. D Start this problem by multiplying out the expressions using FOIL. This gives you x 2 – 2xy + y 2 = x 2 + 2xy + y 2 . You can subtract x 2 + y 2 from each side of the equation, leaving –2xy = 2xy. Getting all the variables over to one side leaves –4xy = 0, which can be simplified to xy = 0. Because you aren’t given any information about the value of either variable, x or y must equal 0, but either one could. In fact, if x = 0, y could be any value, positive, negative or zero. Therefore, the answer is (D). 12. A If the 1990 tuition was 20 percent greater than the 1980 tuition, then it was equal to the 1980 tuition times 120 percent. Knowing this allows you to set up and solve an equation for the 1980 tuition: Column A is greater than Column B. 13. A This problem is most easily solved by making each side look more like the other. To do this, break each quantity into components. In Column A, 25 100 can be broken into (25 50 )(25 50 ). The value in Column B can be rewritten as (2 × 25) 50 , or (2 50 )(25 50 ). Since both columns have one factor of 25 50 , it can be canceled out. What’s left is 25 50 in Column A and 2 50 in Column B. Since 25 50 is greater than 2 50 , Column A is greater than Column B. 3,000 = 120%(1980 tuition) 3,000 = 1.2(1980 tuition) 3,000 ÷ 1. 2 = 1980 tuition 1980 tuition = $2,500 section four SAT Virtual Reality II 29 14. D First solve for x by subtracting 1 from both sides of the equation in the centered information. This gives you x = 2. Now you must find the range of values for y, but be careful. If y 2 is greater than 9, it may be true that y > 3, but it may also be true that y < –3. Substitute these values into Column A and see how the results compare to 5. If x = 2 and y > 3, then x + y > 5. However, if y < –3, then x + y < –1, which is less than 5. Since more than one relationship exists between the two columns, the answer is (D). 15. B There are no numbers provided, so you might think that there’s not enough information to compare the areas and that the answer is (D). That, however, would be too easy for the last QC. Manipulate the diagram a bit and you can see how the area of the triangle compares to the area of the rest of the square. First, imagine how they’d compare if AB were coincident with DE: If this were the case, the area of the triangle would be exactly half that of the square, and therefore equal to the area of the shaded regions. Now, if you move A and B a bit to the right, the area of the triangle decreases: The more you move A and B to the right, the smaller the area of the triangle: So you can see that the area of the triangle is less than half that of the square. Therefore the sum of the areas of the shaded regions is greater than the area of the triangle, and Column B is greater. A B C A B C A B C section four SAT Virtual Reality II 30 16. 12 First multiply both sides of the equation by x + 2: Now solve for x: The question asks for the value of 3x, not x, so multiply 4 by 3. Your answer is 12. 17. 192, 480, or 960 The best way to do this problem is to use trial and error. Just choose three consecutive even integers and see if their product is a 3-digit number. Start with 2, 4, and 6 and go from there: 2 × 4 × 6 = 48 That doesn’t work, so now try 4, 6, and 8: 4 × 6 × 8 = 192 That combination works, so grid in 192. Other possible combinations are 6 × 8 × 10 = 480 and 8 × 10 × 12 = 960. 18. 6.5 or 13/2 An equilateral triangle with side x has 3 sides of length x, which means it has a perimeter of 3x. The second triangle has sides of length 5, 8, and x, which means its perimeter is equal to 5 + 8 + x, or 13 + x. Since the two perimeters have the same value, you can set 3x equal to 13 + x and solve for x: Remember that 6 is not a griddable response and the answer should either be gridded as an improper fraction or a decimal. 1 2 3x = 13 + x 2x = 13 x = 13 2 = 6.5 30 = 5x + 10 20 = 5x 4 = x 30 x + 2 = 5 30 = 5(x + 2) 30 = 5x + 10 section four SAT Virtual Reality II 31 19. 3 To find how many peaches are left over, divide 675 by 48. The result of this calculation is 14 remainder 3. The remainder is the number of peaches that do not fit into a crate, which in this case is 3. 20. 1 < AE ≤ 1.41 Although you have no way of finding AE, you can find a range of values that AE must be within. Since the perimeter of the square is 4, each side of the square must have a length of 1. AE forms the hypotenuse of right triangle ABE, which means that it is longer than side AB. Therefore, AE must be greater than 1. Since AE is not a diagonal of the square, it must be shorter than the length of AC, one of the square’s diagonals. Draw in AC: Look at triangle ADC. It is an isosceles right triangle, so you can use the ratio of side lengths to find the length of AC. The ratio of the sides of an isosceles right triangle is 1:1: , and since the legs of this triangle equal 1, the length of AC is , which is approximately 1.414. Therefore, AE is somewhere between 1 and 1.414, so choose a value between them. 21. 996 It would take too much time to write out the entire series until you reach 1,000. Notice, however, that each term in the series is 2 plus a multiple of 7. So start by finding the highest 3-digit number that is divisible by 7 and add 2 to it. The best way to find the biggest 3-digit number divisible by 7 is to divide 999 by 7. This yields 142 remainder 5. To find a number that is evenly divisible by 7, subtract 5 from 999, leaving you 994. This is the largest 3-digit number divisible by 7, but it is not the answer. Remember that the terms in the series equaled 2 plus a multiple of 7 and you must add 2 to 994, giving you 996. 22. 7 The formula for finding the slope given two points is: Slope = Plug the given x’s and y’s, most of which are in terms of a, into the formula and set the fraction equal to the value you were given, 9. Then solve for a: rise run = difference in y ’s difference in x ’s 2 2 AB E C D section four SAT Virtual Reality II 32 23. 4.80 You only have to find the amount of money saved by purchasing 2 dozen roses, so don’t spend your time finding out how much 2 dozen roses cost. Instead, since you are given that 1 rose is free with every 5 purchased, and since 1 rose costs $1.20, you only need to know how many roses are free of charge when you buy 2 dozen. One out of every 6 roses is free, and there are 24 roses in 2 dozen, so you would receive 24 ÷ 6 = 4 free roses in 2 dozen. 4 × $1.20 = $4.80, which is the answer. 24. 27.5 or 55/2 If you enlarge something by 150 percent, the new dimensions are actually 250 percent of the old ones. Therefore, when trying to determine the photograph’s new dimensions, multiply the original ones by 2.5, not 1.5: The perimeter of a rectangle equals 2 × (length + width), so the perimeter of the enlarged picture equals 2(8.75 + 5), which is 2(13.75), or 27.5. 25. 21 To find the area of a triangle, you need an altitude. Drop an altitude from vertex E. (For purposes of discussion, let’s call the point where this altitude hits the base point F.) The area of ∆ACE is equal to (base × height), or (AC)(EF). You’re not given enough information to find the numerical values of AC and EF, but you can find their values in terms of BC and BD. First, it’s given that BC = AC, so you can say that AC = 3BC. Second, because ∆EFC is similar to ∆BDC (they’re both right triangles 1 3 1 2 1 2 A E D C B F 3.5 × 2.5 = 8.75 2 × 2.5 = 5.0 Slope = Difference in y's Difference in x's 9 = 2a − 5 (a + 1) − a 9 = 2a − 5 1 = 2a − 5 14 = 2a a = 7 section four SAT Virtual Reality II 33 and they both include angle C) and DC = EC, you know that BD = EF, or, in other words, EF = 2BD. Now you can express the area of ∆ACE this way: Because it’s given that the area of ∆BDC is , you know that (BC)(BD) = , and therefore that (BC)(BD) = 7. Plug that into the above expression for the area of ∆ACE and you get: Area of ∆ ACE = 3(BC)(BD) = 3(7) = 21 7 2 1 2 7 2 Area of ∆ ACE = 1 2 (AC)(EF ) = 1 2 (3BC)(2BD) = 3(BC)(BD) 1 2 1 2 section four SAT Virtual Reality II 34 [...]... “using a minimum of words,” has the opposite meaning of the predicted word There is no indication in the sentence that Lovecraft’s correspondence is “unknown” (B) or “popular” (E); and 35 section five 1 SAT Virtual Reality II ... whether the city was “seedy” (D) or not is irrelevant “Seedy” means “disreputable or squalid.” This has nothing to do with population density “Deserted” (E) is the opposite of what you need in the blank 3 A Predictions can be made for both of the blanks based on the information that the rains caused landslides and washed away people’s houses The first blank has to be something like “overwhelming,” because . fraction or a decimal. 1 2 3x = 13 + x 2x = 13 x = 13 2 = 6.5 30 = 5x + 10 20 = 5x 4 = x 30 x + 2 = 5 30 = 5(x + 2) 30 = 5x + 10 section four SAT Virtual Reality II 31 19. 3 To find how many peaches. ∆ ACE = 3( BC)(BD) = 3( 7) = 21 7 2 1 2 7 2 Area of ∆ ACE = 1 2 (AC)(EF ) = 1 2 (3BC)(2BD) = 3( BC)(BD) 1 2 1 2 section four SAT Virtual Reality II 34 section five SAT Virtual Reality II 35 Section. 75. 150 ᎏ 2 h +  + S ᎏᎏ 10 S ᎏ 8 h +  ᎏ 2 S ᎏ 8 h +  + S ᎏᎏ 10 10,000 ᎏ 16 1 ᎏ 3 1 ᎏ 3 4 ᎏ 3 4 ᎏ 3 SAT Virtual Reality II 25 section three 24. 12 Here we have a figure that is not drawn to scale,