# Solutions for Practice Test 5, The Official SAT Study Guide, Section 8

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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 8 of practice test #5 in The Official SAT Study Guide , second edition, found on pages 667–672. 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. Look at the answer choices: Look for a number that is even, positive, and less than 5. The only one is (B) 4. Select that answer.
2. Solve the equation:
8 + √k = 15
k = 7
k = 49
Select (B) 49.
3. 35 out of 35 + 14 + 1 = 50 people were in support of the library, so the fraction of those in support is 35/50 or 70/100 or 7/10. Select (A) 7/10.
• Because the angles of a triangle sum to 180°, the third angle in the upper triangle is 80°.
• The angle opposite the 80° triangle is also 80°.
• Since the angles in a triangle sum to 180°, then 80 + t + u = 180, or t + u = 100. Select (C) 100.
4. Look at the graph for adjacent bars that have very different values. Two large changes occur between 1982 and 1983, and 1984 and 1985. Looking at those two a little more closely, the difference between 1984 and 1985 is the greater difference. Select (D) 1984 and 1985.
5. Look at the answer choices: Looking at the graph, g(k) = 1 between k = −1 and k = 0. Look for an answer choice between those two numbers. The only such choice is (B) −0.5. Select that answer.
6. If 2a·2b·2c = 64, then 2abc = 64. Since 64 = 26, then abc = 6. If a, b, and c are different positive integers that multiply to 6, one must be 1, one must be 2, and one must be 3. So, 2a + 2b + 2c = 21 + 2² + 2³ = 2 + 4 + 8 = 14. Select (A) 14.
• Draw a diagram: It can be helpful to draw a diagram of the situation.
• The radius of the circle is 3 − (−2) = 5. So, the other endpoint of the diameter is 3 + 5 = 8. Select (E) (8, −7).
7. Set up an inequality:
30 < h < 50
-10 < h − 40 < 10
|h − 40| < 10
Select (D) |h − 40| < 10.
• Draw a diagram: Draw a quick sketch of the two cylinders.
• Looking at the sketch, the larger cylinder is the same as two of the smaller cylinders stacked on top of each other. So, the volume of the larger will be twice that of the smaller, which is v. Select (B) 2v.
8. By the definition in the question, the statement −2 <> (n, 0) is true if n < −2 < 0. If n < −2, the only possible value among those listed is I. −3. Select (A) I only.
• Solution 1: Convert the sentence into an equation:  20 percent 20% of · x x equals = 80 percent 80% of · y y
Putting it all together, we get:
20% · x = 80% · y
0.2x = 0.8y
y = (0.2/0.8)x
y = 0.25x
y = 25% · x
Select (B) y = 25% of x.
• Solution 2: Try a special case: Say that y = 5. Then, 80% of y equals 4. If 20% of x equals 4, then x = 20. So, y is ¼ of x (or 25% of x, or 0.25 of x). Select (B) y = 25% of x.
9. If x + y is even, then (x + y)² is even. So, if (x + y)² + x + z is odd, then x + z must be odd, because if it were even, we would be adding two even numbers, resulting in an even number. So, if x + z is odd, then one or the other of x and z, but not both, must be odd. So, if z is even, then x must be odd. Select (C) If z is even, then x is odd.
10. Try a special case: Say that x = 0.5 or ½. Then:
• For I, (0.5)² = ¼ and (0.5)³ = (1/8). ¼ > 1/8, so I is true.
• For II, x = ½ and x/2 = ¼. x is > x/2. so II is true.
• For III, x = ½ and x³ = (1/8). ½ > 1/8, so III is true.
If desired, you can try other cases if you feel that you need to verify that I, II, and III are always true. Select (E) I, II, and III.
11. Draw a diagram: Sketch the five functions in the answer section on the graph given. The graphs for (B), (C), and (D) are nowhere near the vast majority of the dots, so these answers are obviously wrong. This leaves us with (A) t(p) = 44 and (E) t(p) = p + 44. Now, (A) lies above 5 dots and below 7, and (E) lies above 10 dots, on one dot, and above one dot. Because (A) appears to be more in the middle of the data, it is the better choice. Select (A) t(p) = 44.
• Estimate the answer: If we were asked how many rectangles of L by W were needed to cover an area 12L units long and 10W units wide, the answer would be 120. However, the area in question is 12L units long and 10L units wide. From the diagram, L appears to be about 1½ times as long as W, so our estimate for the answer is 1½ times 120, or 180.
• Look at the answer choices: (E) 180 looks like a good choice. We can definitely eliminate (A), (B), and (C), since they're all less than 120. It may be worthwhile doing some extra calculation to ensure that the answer is not (D) 150, though.
• Next, determine W in terms of L. Because 2L and 3W are opposite sides of a rectangle, they must be equal, so 2L = 3W, or W = ⅔L.
• Finally, the number of tiles of dimension L by ⅔L required to cover an area of 12L by 10L is (12 · 10)/(1 · ⅔) = 120 ÷ ⅔ = 180. Select (E) 180.