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,
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 Practice Test 4: Sections
3,
6,
9.
 Practice Test 5: Sections
2,
4,
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 Practice Test 6: Sections
2,
4,
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 Practice Test 7: Sections
3,
7,
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 Practice Test 8: Sections
3,
7,
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 Practice Test 9: Sections
2,
5,
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 Practice Test 10: Sections
2,
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Here are solutions for section 2 of practice test #6 in The Official SAT Study Guide, second edition, found on pages 700–705. 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.
 If each of the packages contained 8 rolls, there would be 5 × 8 = 40 rolls. However, one of the packages contained four additional rolls. So, there are 40 + 4 = 44 rolls. Select (C) 44.

 Draw a diagram: It can help to draw a simple diagram of the three points.
 If BC is 20 more than AB, and AB is 30, then BC is 50. Label the diagram accordingly.
 If AB = 30 and BC = 50, then AC = 80. Select (D) 80.
 Multiply both sides of the given equation by 2:
2(x + 3) = 2a
2x + 6 = 2a
Select (C) 2a.

 Draw a diagram: You are given a graph, but you can draw on it to help yourself. For students whose scores on the two tests were the same, their scores would fall on a line from bottom left (0,0) to top right (100,100). Draw that line.
 Estimate the answer: The student whose change in scores was greatest must be furthest away from the line just drawn. This must be either A or E. Eliminate answer choices (B), (C), and (D).
 The difference for A was 40 − 70 = 30. The difference for E was 80 − 60 = 20. Select (A) A.

 Estimate the answer: Looking at the graph, two students had scores of 70, two had scores below 70 (i.e. 60), and one had a score above 70 (i.e. 80), so the mean must be more than 60 but less than 70. Eliminate answers (A) 60, (D) 70 and (E) 72.
 To find the arithmetic mean, add all five scores together and divide by 5 (the number of scores). Hint: You may be able to save time by doing this in your head. To add 80 + 70 + 70 + 60 + 60, just add the numbers in the tens column together (34) and add a zero to the end, giving 340. Then, to divide by 5, just multiply by 2 (giving 680) and divide by 10 (giving 68). Select (C) 68.
 Estimate the answer: u + v must be > 0, so we can eliminate t and w. u and v seem to be about −½, so u + v must be around 1. The only answer near that value is y. Select (D) y.

 Solution 1:
 Estimate the answer: We may assume that the diagram is drawn to scale. By eyeballing the figure or by using your pencil or finger, the distance between S and the yaxis appears to be somewhere around three times the distance between the origin O (0,0) and (1,0). Because S is to the left of the yaxis, the xcoordinate must be negative. So, the xcoordinate must be around −3.
 Look at the answer choices: The only answer near −3 is (A) −3. Select that answer.
 Solution 2: If the coordinates of S are (k, 3), then the coordinates of R are (k, 0). So, the length of RS is 3. Therefore, the length of ST must also be 3. Since T is located at (0, 3), S must be located at (−3, 3). Therefore k = −3. Select (A) −3.
 Try a special case: We are given that, when x = 0, f(x) = 1. Look at each answer choice and see which evaluate to 1 when x = 0. As it turns out, only A and E do. Elminate the other three choices.
 Try a special case: We are given that, when x = 1, f(x) = 2. Look at the remaining answer choices and see which evaluates to 2 when x = 1. Only A does so. Select (A) f(x) = x^{2} + 1.
 Try a special case: It may be easier to see the answer if we use numbers instead of letters. Say that y = 5 and x = 2. If exactly 2 years ago someone was 5 years old, they are now 7 years old, and exactly 1 year ago they were 6 years old.
 Look at the answer choices: See which of the answers evaluate to 6 when y = 5 and x = 2:
 (A) evaluates to 5 − 1 = 4
 (B) evaluates to 5 − 2 − 1 = 2
 (C) evaluates to 2 − 5 − 1 = 4
 (D) evaluates to 5 + 2 + 1 = 8
 (E) evaluates to 5 + 2 − 1 = 6
The only matching answer is (E) y + x − 1. Select that answer.
 One way to get W X Y Z is to:
 Interchange W and Y
 Interchange W and X
 Reverse the entire sequence
So, the answer must be no more than 3. Eliminate (C) 4, (D) 5, and (E) 6.
 If you're low on time, it's probably best to just select (B) 3 now and move on. Otherwise, spend a bit of time ensuring that the answer cannot be (A) 2. There isn't any easy formal way to prove this (not that you'd want to do that), but just play around with the letters a bit until you see that the sequence can never be W X Y Z after two changes. Then select (B) 3.
 Draw a diagram: Draw a diagram similar to the following:
 There will be 20 ÷ 4 = 5 cubes along the one edge of the box, 24 ÷ 4 = 6 cubes along the other edge of the box, and 32 ÷ 4 = 8 cubes along the third edge of the box. So, there will be 5 × 6 × 8 = 240 cubes in totl in the box. Select (D) 240.
 Try a special case: Say n = ¼. Then, √n = ½, and n^{2} = 1/16. So, n^{2} < n < √n. Select (E).
 The median of the slopes will be the middle value. From the graph, this is the slope of OC. The slope of OC is rise/run = 3/4. Select (C) 3/4.
 From the given information, we can conclude that the plane took off from New York City at 9:00 am PST and arrived at San Francisco at 4:00 pm PST, so it took 7 hours. If a plane left San Francisco at noon PST, it would arrive at New York at 7:00 pm PST. To convert from PST to EST, we need to add three hours, so 7:00 pm PST is 10:00 pm EST. Select (A) 10:00 pm EST.
 Estimate the answer: If the radius of each quarter circle is 1, then QP = 1 and PS = 2, so the area of the entire rectangle is 2. Looking at the diagram, the shaded region appears to be about ¼ or 1/5 of the rectangle, so the answer should be somewhere around ½ or 2/5.
 Look at the answer choices: We can quickly get a rough estimate of the value of each answer by substituting 3 for π. (A) is around 1  ¾, or about ¼. (B) is around 2 − 3/2 = ½. (C) is around 5/4. (D) is around 3/4. (E) is ⅔. The only one anywhere near our estimate is (B) 2 − π/2. Select that answer.
 Solution 1:
 Take the equation y = f(x). From this equation and the graph, we know that f(1) = 0 and f(0) = 1. Now, given these and looking at y = f(x + 2), when x = −2, y = 1, and when x = −1, y = 0.
 Look at the answer choices: Look for a graph that appears to contain (−2, 1) and (−1, 0). The only such graph is (C). Select that answer.
 Solution 2: If y = f(x) is the graph of a function, then y = f(x + 2) is the graph of that function, shifted (and be careful here) left two units. This would correspond to graph (C). Select (C).
 Draw a diagram: If you have time, you may want to draw a diagram that is more accurate (i.e. reflects the facts that AB = BC and that DE = EF = DF). If not, fill in what you know on the diagram. You are told that ∠ABC = 30°, and ∠BDE = 50°. Since DE = EF = DF, you know that all three angles of triangle DEF = 60°. Fill those in as well.
 Work backwards: How can we find ∠DFA? There are a few possibilities. If we knew ∠ADF and ∠DAF, we could find ∠DFA.
 We can find both angles: ∠ADF = 180 − 50 − 60 = 70°, and, because the triangle ABC is isosceles, ∠DAF = ∠ECF, and both are equal to ½(180 − 30) = 75. So, ∠DFA = 180 − 70 − 75 = 35°. Select (B) 35°.
 Look at the answer choices: It may be difficult to see which equations are equivalent as they are, but it's easier if you crossmultiply the equations. Four of the equations result in af = bc, while (A) is the odd one out, resulting in ac = bf. Select (A) a/f = b/c.
 Look at I, II, and III individually:
 From the question, a [] b = ab − b. Factoring b out of both terms, the expression becomes b(a − 1). Now, if a and b are positive integers, then this expression can only be equal to zero if a == 1. So, we can eliminate answer choices (B) and (C).
(a + b)[] b
= b(a + b) − b
= b(a + b − 1)
Because a and b must be at least 1, this expression can never be zero.
a[](a + b)
=a(a + b) − (a + b)
= (a + b)(a  1)
This expression can equal zero only if a == 1. Select (E) I and III.