12.Practice.Tests.for.the.SAT_2015-2016_1128p

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12 Practice Tests for the SAT 564 Practice Test Six Answers and Explanations 13. c Difficulty: Medium Strategic Advice: Be sure to understand the chart given before trying to tackle the problem. In multiple-choice problems like this, work through all the possibilities and eliminate choices as you go. Getting to the Answer: For Richmont, the population increases from 50 to 200, which is definitely more than l. Eliminate (A), (D), and (E). 2 For Pierdson, the population decreases, so eliminate (B). For Thomasville, the population increases from 200 to 400. Only Ill works. 14. B Difficulty: Medium Strategic Advice: To find percent increase, determine the difference between values and divide by original value. Getting to the Answer: Total population in 1986 = 950 Total population in 1999 = 1,425 15. c 1,425 - 950 = 475 = 50% 950 950 Difficulty: Medium Strategic Advice: This problem can be solved through careful reasoning, but if you're still confused, you can perform the logic-question equivalent of Picking Numbersmake up some imaginary characters to fit the situation and see which statement holds true. Getting to the Answer: Let Amanda, Ellen, Greg, and Max be in the accounting department. Some of these people, say Amanda and Greg, are on the bowling team. Since no one on the bowling team is married, Amanda and Greg are definitely unmarried. We don't know anything about whether Ellen and Max are married. Choice (A) is uncertain; it could be either true or false. Choice (B) is uncertain as well; you don't know enough about Ellen and Max. Choice (C) is necessarily true; you know Amanda and Greg are single. Choices (D) and (E) are uncertain; the question doesn't say anything about the number of employees in either group. Thus, (C) is the correct answer choice. 16. c Difficulty: Medium Strategic Advice: Use the formula for averages to find the missing pieces of an averages question. Getting to the Answer: The average is the sum of the terms divided by the number of terms. So l 0 students got l 00, leaving 20 to get y scores each (we can assume each will have the same score). We set up the problem like this: 100(10) + 20y =92 30 17. B 1,000 + 20y = 2,760 20y = 1,760 y =88 Difficulty: Medium Strategic Advice: Remember that a-2 is equal to - If you a have trouble simplifying the given expression, try picking a number for a to make the problem more concrete. Getting to the Answer: l 02 a2 x a-2 = a2 x - = - 2 = l 02 a Or, say a= 3: l l 9 32 x - =9x-=-= l 32 9 9 18. c Difficulty: High Strategic Advice: In order to visualize some geometry problems, it is necessary to draw the diagram split into its pieces. The perimeter of each semicircle is half of the perimeter (circumference) of the original circle plus the length of the straight edge.
12 Practice Tests for the SAT Practice Test Six Answers and Explanations 565 I Getting to the Answer: OD P of one semicircle = nd + 6 educated guess-you know that the slope of the line is negative, so k must be greater than 2, eliminating (A) and (B). Getting to the Answer: 2-k -l -- - 3-0 2 1_-js_ = =l_ 3 2 4-2k=-3 7=2k J_ =k 2 = 7t(6) + 6 = 37t + 6 P of two semicircles = 2(37t + 6) = 67t + 12 19. A Difficulty: High Strategic Advice: The probability that a randomly selected point will fall within the shaded area is equal to the ratio of the area that is shaded to the area of the entire figure. The area of the entire figure is equal to the area of the largest circle, and the area of the shaded area is equal to the area of the second circle minus the area of the smallest circle. To make the calculations easier, you can assume that the radius of the smallest circle equals l, so its diameter is 2. Then the radius of the second circle is 3, and the radius of the largest circle is 5. Getting to the Answer: Shaded area: 7t(32) - 7t( l 2) = 97t - 7t = 87t Total area of figure: 7t(52) = 257t b bl 87t 8 Pro a 11ty: = 257t 25 20. c Difficulty: High Y - 2 Yi Strategic Advice: Use the slope formula, m = ---, wit . h X2 - Xl the two given points to get an equation involving k. Then solve for k. If you find yourself running out of time, you can easily narrow down the answer choices and make an SECTION 4 1. c Difficulty: Low Look for the most concise version of the selection that retains the meaning of the original sentence. This sentence is awkward and unnecessarily wordy. Choice (C) solves both of these problems without adding any additional errors. Choice (B) introduces the passive voice unnecessarily. Choices (D) and (E) are grammatically incorrect in structure. 2. B Difficulty: Medium In a compound predicate, both verbs must be in parallel form. The verb phrase would increase should be to increase in order to parallel to hold; also, the pronoun they is unnecessary. Choice (B) fixes both problems. Choice (C) is awkwardly worded and fails to make the verbs parallel. Choices (D) and (E) use transition words (for and thus), which are inappropriate in context. 3. E Difficulty: Medium The gerund (-ing) verb form can never be the predicate (main) verb in an independent clause.
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