# Solutions for Practice Test 10, The Official SAT Study Guide, Section 2

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SAT Practice Test Solutions:
2014–15 SAT Practice Test
2013–14 SAT Practice Test
The Official SAT Study Guide, second edition
• Practice Test 1: Sections 3, 7, 8.
• Practice Test 2: Sections 2, 5, 8.
• Practice Test 3: Sections 2, 5, 8.
• Practice Test 4: Sections 3, 6, 9.
• Practice Test 5: Sections 2, 4, 8.
• Practice Test 6: Sections 2, 4, 8.
• Practice Test 7: Sections 3, 7, 9.
• Practice Test 8: Sections 3, 7, 9.
• Practice Test 9: Sections 2, 5, 8.
• Practice Test 10: Sections 2, 5, 8.
SAT Math Tips

Here are solutions for section 2 of practice test #10 in The Official SAT Study Guide , second edition, found on pages 948–953. The solutions below demonstrate faster, more informal methods that might work better for you on a fast-paced test such as the SAT. To learn more about these methods, see my e-book Succeeding in SAT Math or the SAT math tips page.

1. If 3 pencils cost \$4.50, each pencil costs \$1.50. If the price is increased by \$0.50, they will cost \$2 each, so 5 pencils will cost \$10. Select (E) \$10.00.
• Try a special case: When x = 1, y = 3.
• Look at the answer choices: Look at all of the answer choices to see for which y = 3 when x = 1. The only such choices are (C) y = 3x and (E) y = 4x − 1. Eliminate the other three choices.
• Try a special case: When x = 2, y = 7.
• Look at the answer choices: When x = 2, 3x = 6, so eliminate (C). Select (E) y = 4x − 1.
2. Let x be the circumference of the circle with centre A. So, x/2 is the circumference of the circle with centre B. So, the radius of the circle with centre A is x/2π, and the radius of the circle with centre B is x/4π. The radius of the circle with centre A is twice the radius of the circle with centre B. So, if AC = 6, then AB = 4 and BC = 2. Select (B) 2.
3. Guess and check: Try each point and see which satisfies |x| − |y| = 3. For points where this is the case, the x value must be further away from x = 0 than the y value is, so try a point that looks like it meets that criterion. B is such a point. B is located at (-4, -1). |x| − |y| = |-4| − |-1| = 4 − 1 = 3. select (B) B.
• Draw a diagram: It may help to shade the portion of the graph representing people under 40 (but not so darkly you can't read the diagram!)
• Estimate the answer: The shaded portion of the graph appears to be exactly half of the graph, so the number of people is probably around 1000/2 = 500.
• Look at the answer choices: Select (D) 500.
4. When a number is divided by 3, the remainder must be either 0, 1, or 2. The only answer where all of the numbers are 0, 1, or 2 is (D) 0, 1, 2, 0. Select that answer.
5. If y is inversely proportional to x and y = 15 when x = 5, then when x is 5 times greater than 5 (i.e. 25), y will be 1/5 of 15, or 3. Select (C) 3.
6. Substitute 2x + z = 2y into the second equation:
2x + 2y + z = 20
2y + 2y = 20
4y = 20
y = 5
Select (A) 5.
7. Solve the equation:
2(x − 3) = 7
x − 3 = 3.5
x = 6.5
Enter 6.5.
8. Substitute x = 4 into the given equation:
y − 4 = 3(4 − 2)
y = 4 + 3(2)
y = 4 + 6
y = 10
Enter 10.
• First, determine how much gasoline car A used. If car A travelled 60 miles and averaged 20 miles per gallon, the amount of gasoline used is (60 miles) ÷ (20 miles/gallon) = 3 gallons.
• If car B travels 15 miles per gallon, the number of miles it will take to use 3 gallons is (3 gallons)(15 miles per gallon) = 45 miles. Enter 45.
• Estimate the answer: The value of x appears to be slightly greater than 90°, say 100° or 105° or so.
• To solve this problem, it helps to know that the sum of the angles of a quadrilateral is 360°. So, the fourth angle of the quadrilateral is 360 − 120 − 100 − 65 = 75°. Then, x = 180 - 75 = 105°. Enter 105.
9. The first term of the sequence is 20, the second is 8, the third is ½(20 + 8) = 14, the fourth is ½(8 + 14) = 11, and the fifth is ½(14 + 11) = 12.5. Enter 12.5.
• First, it can help to write the sentences out as equations:
x = (1/5)y (Equation 1)
y = (3/10)z (Equation 2)
• Multiplying Equation 1 by 5, we get 5x = y.
• Substituting 5x = y into Equation 2, we get:
5x = (3/10)z
x = (3/50)z
Enter 3/50.
• To find the value of BE, consult the Reference Information for information on the 30°-60°-90° triangle. BE is √3 times AE, or 8√3.
• The area of the square is (8√3)² = 64 × 3 = 192. Enter 192.
• Try a special case: In 7 pounds of the mixture, 2 pounds will be cashews.
• If there are only 4 pounds of mixture, there will be 4/7 of that amount of cashews, or (4/7)(2) = 8/7. Enter 8/7.
• Solution 1:
• Draw a diagram: Draw one possible value of m on the diagram.
• The slope of m must be less than the slope of OA, but greater than the slope of OB. Since the slope of OB is 0, anything slightly greater than 0 will work. Enter .001.
• Solution 2:
• Draw a diagram: Draw one possible value of m on the diagram.
• Because line l passes through (8, 3), and line m falls below line l, the y-value on line m will be less than 3. Say that it's 1. Then the slope is 1/8. Enter 1/8.
10. The median of the enrollment over the 5 years must be the third largest value, which we are told is 1351. Because the table already contains two values larger than 1351, the unknown value x must be no greater than 1351; otherwise, x would be the median, not 1351. Therefore, the greatest possible value for x is 1351. However, the question also says that no two years had the same enrolment, so x can't be 1351, so the largest possible value for it is 1350. Enter 1350.