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.
Here are solutions for section 9 of practice test #8 in The Official SAT Study Guide, second edition, found on pages 856–861. The solutions below demonstrate faster, more informal methods that might work better for you on a fastpaced test such as the SAT. To learn more about these methods, see my ebook Succeeding in SAT Math or the SAT math tips page.
 Guess and check: Try each answer choice, starting with the largest, to see for which 3m − 1 is greater than 10. For (A), 3(4) − 1 = 11 > 10. Select (A) 4.
 Divide both sides of the equation by a:
k = 1
Select (D) 1.
 Work backwards: We could determine the value of z if we knew the measures of the other two angles in the triangle. The first angle is opposite x°, so it must be equal to x°, or 80°. The second angle is opposite y°, so it must be equal to y° or 70°. Since the angles of a triangle add to 180°, z = 180 − 80 − 70 = 30. Select (A) 30.
 If Mia takes 35 kilometers to go by the scenic route and return by the direct route, and the scenic route is 5 kilometres longer, then she would take 30 kilometres to go by the direct route and return by the direct route. So, each way along the direct route must be 15 kilometres. Select (C) 15.
 The light is red for 30/80 of the time, so it is not red for 50/80 of the time. So, the probability that the light will not be red at a randomly chosen time is 50/80, or 5/8. Select (B) 5/8.
 This is a fairly straightforward proportion problem.
 Estimate the answer: x and y average to 45, and u, v, and w average to 60. So, the average of all 5 must be somewhere between 45 and 60, probably somewhere in the 50s.
 Look at the answer choices: We can eliminate answers (A) and (B). There are still 3 answers that seem within range of our estimate.
 The average of the five numbers is (u + v + w + x + y)/5. Since the angles of a triangle sum to 180°, u + v + w = 180 and x + y = 90. So, the answer is (180 + 90)/5 = 54. Select (E) 54.
 Guess and check: Try each answer and see which produces the expected result:
 For (A), if x = −2, then x² = 4, which is > x. However, x² has to be less than x, so eliminate that choice. Similarly, we can eliminate (B).
 For (C), if x = ¾, then x² = 9/16, which is < x, and x³ = 27/64, which is < x². This is what we're looking for. Select (C) ¾
 Since the question doesn't tell us where (h, k) is located, it could be located anywhere on the line. One possibility is at (1, 3), so k/h = 3/1 = 3. Select (A) 3.


 Solution 1: According to the table, oil with a 5W rating flows twice as fast as 10W oil. In turn, 10W oil flows twice as fast as 15W oil, so 5W oil flows 2 × 2 = 4 times as fast ast 15W oil. 15W oil flows twice as fast as 20W oil, so 5W oil flows 4 × 2 = 8 times as fast as 20W oil. Select (C) 8.
 Solution 2: Try a special case: Say that 5W oil flows at a rate of 32 (the units you choose here don't really matter). Since 10W oil flows half as fast as 5W oil, 10W oil flows at a rate of 16. Since 15W oil flows half as fast as 10W oil, 15W oil flows at a rate of 8. Since 20W oil flows half as fast as 15W oil, 20W oil flows at a rate of 4. So, the ratio between the speed of 5W oil and 20W oil is 32/4 = 8. Select (C) 8.
 Draw a diagram: Label the lengths of PB, PR, and AQ on the diagram. As well, since points P, A, and B are equally spaced, then PA and AB are both half of PB, or 2. Label those as well. Finally, because points P, Q, and R are equally spaced, PQ = 3 and PR = 3. Label those as well.
 In order to determine the perimeter of QABR, we still need to find the side length of BR. To do so, note that triangles PAQ and PBR are similar, because they share angle P, and the two sides that meet at that vertex are in proportion in the two triangles. Now, the sides of PBR are twice the length of sides PAQ, so if AQ = 4, then BR = 8.
 The perimeter of quadrilateral QABR is 2 + 8 + 3 + 4 = 17. Select (E) 17.
 Evaluate the functions:
g(5) − h(4)
= 5² + 5 − (4² − 4)
= 30  12
= 18
Select (D) 18.
 Expand and simplify:
h(m + 1)
= (m + 1)² − (m + 1)
= m² + 2m + 1 − m − 1
= m² + m
= g(m)
Select (A) g(m).

 Draw a diagram: You probably won't be able to solve the problem correctly without a diagram. It should look like this:
 Estimate the answer: If ABED has area ⅔, ABCD has area greater than ⅔ but probably less than 1.
 Look at the answer choices: Only (B) and (C) have area > ⅔ and < 1. Eliminate the other three answers.
 Draw a diagram: On the diagram you drew, draw lines EA, and a vertical line from E to the midpoint of AD:
This breaks the diagram up into four triangles. Because E is the midpoint of BC, the three side lengths of any triangle equal the three side lengths of any other triangle, so all four triangles are congruent. Figure ABED is made up of 3 of these triangles, so the area of each triangle is #8531;(⅔) = 2/9. Figure ABCD is made up of four of these triangles, so its area is 4(2/9) = 8/9. Select (C) 8/9.