# Solutions for Practice Test 9, The Official SAT Study Guide, Section 5

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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 5 of practice test #9 in The Official SAT Study Guide , second edition, found on pages 903–908. 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. The fifth term of the sequence is (30 + 1) × 2 = 62. The sixth term of the sequence is (62 + 1) × 2 = 126. Select (E) 126.
2. Expand the first equation and substitute ax = 15:
a(x + y) = 45
ax + ay = 45
15 + ay = 45
ay = 30
Select (E) 30.
• Estimate the answer: The needle appears to be somewhat beyond 30 but not significantly so; perhaps the answer is around 40 or somewhat less.
• Look at the answer choices: Answers (B) and (C) are within range of our estimate; we can eliminate answers (A), (D), and (E).
• Including only one of the endpoints, there are four tick marks between 30 and 60, so each tick mark represents 7.5 miles per hour. The needle is on the first tick mark past 30, so the speed is 30 + 7.5 = 37.5 miles per hour. Select (B) 37.5.
3. List out all of the possibilities:
456, 465, 546, 564, 645, 654
There are six possibilities. Select (C) Six.
4. There are three faces, each with area r. There are two triangular faces, each with area t. Therefore, the total surface area is 3r + 2t. Select (B) 3r + 2t.
5. Guess and check: Try each answer until you find one that works:
• For (A), (1 + 1)/21 = 1, not ½. Eliminate that answer.
• For (B), (2 + 1)/2² = ¾, not ½. Eliminate that answer.
• For (C), (3 + 1)/2³ = ½. Select (C) 3.
6. Try a special case: Say that each of the 14 books weighs p pounds. Then, the average weight will also be p pounds. The total weight of the books will be 14p pounds. Select (E) 14p.
7. As point B is halfway between points C and A, its y-coordinate (t) will be halfway between the y-coordinate of C (5) and the y-coordinate of A (−1). The number halfway between 5 and −1 is (5 + (−1))/2 = 2. Select (C) 2.
8. If you multiply a bunch of terms together and the result is 0, then one (or more) of the terms must be 0. So, either k = 0 or 2x + 3 = 0 or x − 1 = 0. However, 2x + 3 cannot equal 0 because then x would equal −3/2, which violates the condition that x > 1. Similarly, x − 1 cannot equal 0 becuse x > 1. So, k must equal 0. Select (B) 0.
• Solution 1: Look at the answer choices: Read through each answer choice and see which makes sense:
• Does (A) make sense? It seems to.
• Does (B) make sense? No, because there could be men over six feet tall who are members of another family.
• Does (C) make sense? No, because there could be men over six feet tall who are members of another family.
• Does (D) make sense? No, because the statement doesn't say anything about only the men being over six feet tall.
• Does (E) make sense? No, because it contradicts the statement given.
Select (A) No man under six feet tall is a member of the Williams family.
• Solution 2:
• It can often help to find the contrapositive of the given statement. The contrapositive of "All men in the Williams family are over six feet tall" is "No man under six feet tall is a member of the Williams family."
• Look at the answer choices: Look through the answer choices for something that appears to be the same as either the statement or the contrapositive. In this case, (A) is the exact same as the contrapositive. Select (A) No man under six feet tall is a member of the Williams family.
9. From the reference information, the circumference of a circle is 2πr, where r is the radius of a circle. Setting the circumference equal to π:
π = 2πr
1 = 2r
r = ½
Select (B) ½.
• First, find the ratio between y and x². We are told that y = 1/8 when x = ½. So, when y = 1/8, x² = ¼. So, y/x² = (1/8)/¼ = ½.
• If y = 9/2 and y/x² = ½, then x² = y ÷ (y/x²) = (9/2) ÷ ½ = 9.
• So, x = √ = √9 = ±3. We are asked for the positive value of x, so select (D) 3.
• Try a special case: Say that x = 6. Then, u = 4. So, u < x.
• Try a special case: Say that u = 5. Then v = 6. So, u < v.
• Try a special case: Say that v = 7. Then w = 5. So, w < v.
• Look at the answer choices: The only answer where u < x, u < v and w < v is (D) w < u < v < x. Select that answer.
• Solve the equation when h(t) = −60:
−60 = 2(t³ − 3)
−30 = t³ −3
−27 = t³
t = −3
• So, 2 − 3t = 2 − 3(−3) = 2 + 9 = 11. Select (B) 11.
10. Try a special case: Say that x = 3 and y = 5. In this case:
1. xy = 15, which is divisible by 15.
2. 3x + 5y = 9 + 25 = 34, which is not divisible by 15.
3. 5x + 3y = 15 + 15 = 30, which is divisible by 15.
If desired, you can try additional cases or reason about the values of x and y to convince yourself that I and III will always be divisible by 15. Select (D) I and III only.
• Estimate the answer: Both y and z are obtuse, so the answer is definitely greater than 180°. z appears to be around 135° or so, and y appears to be around 105° or so, so the answer is probably around 240° or so.
• Look at the answer choices: Answers (D) and (E) seem plausible, based on our estimate. Eliminate the other three answers.
• Looking at the triangle in the middle of the figure, the top angle must be 65° because it and the angle labelled 115° are supplementary (add to 180°). Because the angles of a triangle sum to 180°, the other two angles in the triangle must add to 115°.
• Try a special case: Say that the other two angles in the triangle are 70° (the angle that is the supplement of y) and 45° (the angle that is the supplement of z). Then, y = 110° and z = 135°. y + z = 245. Select (E) 245.
11. If n is the smallest of the three integers, then:
n + (n + 2) + (n + 4) = 111
3n + 6 = 111
Select (D) 3n + 6 = 111.
• The diagram depicts three arcs each of lengths 2 and b. They appear to take around 70° in total. So, each arc of length 2 + b takes around 20 or 25° in total. The arc of length b appears to be about 1/5 of the arc of length (2 + b), so the degree measure of each of the arcs of length b is 1/5 of 20 or 25°, so around 4° or 5°.
• Estimate the answer: (A) 4° or (B) 6° seem reasonable. Eliminate answers (C) through (E).
• If there are 18 arcs of length 2 and 18 arcs of length b, and the total circumference is 45, then:
18(2) + 18b = 45
18b = 9
b = ½
• So, each arc of length b takes up ½ ÷ 45 = 1/90 of the circle. If the number of degrees in a circle is 360°, then each arc of length b takes up 360° × (1/90) = 4°. Select (A) 4.
• If the cost of maintenance on an automobile increases 10% each year, then next year Andrew will pay \$330 for automobile maintenance. Therefore, c(1) = 330.
• Setting n = 1 in the function definition and substituting c(1) = 330:
c(1) = 300x1
330 = 300x
x = 33/30 = 11/10 = 1.1
Select (C) 1.1.
• Draw a diagram: Draw AC on the diagram.
• Estimate the answer: The figure isn't drawn to scale, but it doesn't seem to be too inaccurate, so we can make a rough estimate. AC appears to be larger than BD, but not twice as large.
• Look at the answer choices: Answer choices (A), (B), and (E) are between 1 and 2, so they seem reasonable. Answer choices (C) and (D) are less than 1, so they are not reasonable. Eliminate those two answers.
• If the five line segments are all congruent, then triangles ABD and triangles BCD are equilateral triangles. Line AC divides these two equilateral triangles into two 30°-60°-90° triangles.
• From the information about 30°-60°-90° triangles given in the Reference information, each half of line AC is equal to √3/2 of the side length of the equilateral triangle, so line AC is equal to √3 of the side length of the equilateral triangle. Since BD is the side length of the equilateral triangle, then the ratio of AC to BD is √3 to 1. Select (B) √3 to 1.