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 2 of practice test #5 in The Official SAT Study Guide, second edition, found on pages 638–643. 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.
 Solution 1: Solve the equation for x:
3x + 9 = 5x + 1
8 = 2x
4 = x
Select (D) 4.
Solution 2: Guess and check: Try each of the ansswer choices to see which value of x makes the equation true. You'll find, that for x = 4, 3(4) + 9 = 5(4) + 1, so the answer is (D) 4.
 Look at the answer choices:
 If m = 1, then the difference between each term would be the same (p), but this isn't the case.
 If m = 2, then this describes the sequence if p = 1.
Select (B) 2.
 Solution 1: To find the total number of combinations, multiply the number of colors by the number of sizes. There are 3 colors and 4 sizes. 3 × 4 = 12. Select (B) 12.
Solution 2: Write each of the possibilities out, as shown below (where R = Red, W = White, B = blue, S = Small, M = Medium, L = Large, E = Extralarge):
RS, RM, RL, RX, WS, WM, WL, WX, BS, BM, BL, BX
There are 12 possibilities. Select (B) 12.
 Solution 1: Guess and check: Check each answer at −3 and 3 until you find one where f(−3) is the graeter. Start with the answers where the value of x appears to decrease as x increases:
 For f(x) = 4/x, f(−3) = −4/3, and f(3) = 4/3, so that doesn't work.
 For f(x) = 4 − x³, f(−3) = 68, and f(3) = −60, so that works.
Select (D) f(x) = 4 − x³.
Solution 2: Using your knowledge of graphs and transformations, quickly sketch the five functions. It should be apparent that (D) f(x) = 4 − x³ is the only one where f(−3) > f(3). Select that answer.
Solution 3: Plot each function on your graphing calculator until you find one where f(−3) > f(3). You will find that (D) f(x) = 4 − x³ is the only such function. Select that answer.

 Read the question and determine what it is asking: This is a proportion question. The rate given is (15 pounds)/(8 centimetres) and the value given is 20 centimeters. If 15 pounds corresponds to 8 centimetres, what number of pounds corresponds to 20 centimetres?
 Estimate the answer: 15/8 is a bit less than 2, so, for 20 centimeters, the number of pounds should be a bit less than 2 × 20 = 40.
 Look at the answer choices: (D) 35 and (E) 37.5 look possible. Eliminate the other three choices.
 To calculate the answer, evaluate (15/8) × 20 = 37.5. Select (E) 37.5.


Draw a diagram: Draw a diagram of the scenario, such as that shown at right.
 Look at the answer choices: See which answer choices correspond to the diagram:
 Does YZ = ½XZ? Yes.
 Does ½XZ = 2XY? No, 2XY = XZ, not ½XZ.
 Does 2XY = XZ? Yes.
Select (E) I and III.
 If 2r = 5s and 5s = 6t, then 2r = 6t. Solving that equation for r, we get r = 3t. Select (C) 3t.

 Try a special case: Say that there were 2 buses (n = 2), each seating 10 people (x = 10). One bus would have all 10 seats full and the other would have 7 seats full, so there were 17 passengers in total (k = 17).
 Look at the answers: See which answers result in k = 17 when n = 2 and x = 10:
 For (A), (2)(10) − 3 = 17, so (A) is possible.
 For (B), (2)(10) + 3 = 23, which is incorrect.
 For (C), 2 + 10 + 3 = 15, which is incorrect.
 For (D), 2k = 10 + 3 or k = 6.5, which is incorrect.
 For (E), 2k = 10 − 3 or k = 3.5, which is incorrect.
The only possibility is (A) nx − 3 = k. Select that answer.

 Estimate the answer: x appears to be a fairly large angle, perhaps around 140 or 150 degrees.
 Look at the answer choices: We can safely eliminate (D) 110 and (E) 100. This leaves us with (A) 150, (B) 140, and (C) 130.
 The angle between the angles labelled 50° and x° must be 30°, because l is parallel to m, so the sum of 50° and that angle must be 80°.
 That 30° angle and the angle labelled x° form a straight (180°) angle, so the measure of x must be 180 − 30 = 150. Select (A) 150.
 Solution 1:
 Start by simplifying the equation, as shown below:
3x² < (3x)²
3x² < 9x²
If x ≠ 0, we can divide both sides by x²
3 < 9 if x ≠ 0
This equation is always true, so the original statement can never be false if x ≠ 0. Important note: It's important to remember that you can only divide by x if x ≠ 0; otherwise you'll get the wrong answer.
 Look at the answer choices: The only possibilities are (B) 0 and (E) For no value of x. Eliminate the other two choices.
 Guess and check: Test whether the original statement is false for x = 0. If so, select (B) 0. Otherwise, (E) For no value of x must be true. For x = 0, we get:
0 < 0
which is false. So, select (B) 0.
Solution 2: Guess and check: Try each of the first four answers. If none of those work, then the answer must be (E) For no value of x.

 Estimate the answer: If the front wheel is smaller, it must make more revolutions than the back wheel to cover the same distance. Perhaps it must make 2 or 4 times as many revolutions.
 Look at the answer choices: We can eliminate (D) ½ and (E) ¼. The other three answers are still possible.
 Try a special case: Say that the diameter of the front wheel is 1 unit. Then the diameter of the back wheel is 2 units. Now, the circumference of the front wheel is π units and the circumference of the back wheel is 2π units. Say that the back wheel makes one revolution; then, the bicycle will travel 2π units. Since the front wheel has a circumference of π units, it will make 2 revolutions. Select (C) 2.
 Try a special case: Select a bunch of positive and negative numbers such that the probability of selecting a positive number is 3/5. If you select the following numbers, this is the case:
1, 2, 3, −1, −2
Now, there are 3 positive numbers (p) and 2 negative numbers (n). So, n/p = ⅔. Select (C) ⅔.
 Substitute c(x) = 640 and x = 20 into the given equation:
640 = 600 · 20 − 200⁄20 + k
640 = 600 − 10 + k
50 = k
Select (B) 50.

Guess and check: Try each pair of positive integers, starting with the smallest possible pairs:
 Does (1,1) satisfy the equation? 2(1) + 3(1) = 5 < 6. It does.
 Does (2,1) satisfy the equation? 2(2) + 3(1) = 7, which is not < 6.
 Does (1,2) satisfy the equation? 2(1) + 3(2) = 8, which is not < 6.
 You should be able to see that no larger values will satisfy the equation either.
We found one ordered pair that satisfies the equation. Select (A) One.

 Draw a diagram: The diagram given is not drawn to scale. Given that you know that y = 60, you may find it useful to redraw the second triangle such that the bottom two angles look more like 60° angles, and also label them as 60° angles. You should probably get something like this:
 Now, because the angles of a triangle sum to 180°, the third angle in the triangle must be 60°. This means that the triangle must be equilateral. So, each side must be 5 units in length, and the perimeter is 15.
 Going back to the first triangle, we are given that side AB, which is opposite an angle labelled as x° is 8. Therefore, side BC, which is opposite the other angle labelled as x° must also be 8. The perimeter is 8 + 8 + 5 = 21.
 The difference between the perimeter of triangle ABC and triangle DEF is 21 − 15 = 6. Select (C) 6.

Solution 1:
 Try a special case: Say that x = 5, so y = 7. y² − x² = 49 − 25 = 24.
 Look at the answer choices: See which answer choice(s) evaluates to 24 when x = 5:
 (A) 2x evaluates to 10. Try an answer choice that looks a lot bigger.
 The largest answer choice is (E) 4x + 4, which evaluates to 24, so this might be our answer.
 Glancing quickly at (B) through (D), you'll notice they must be smaller than (E), so they can't be correct.
As (E) is the only choice evaluating to 24, select (E) 4x + 4.
Solution 2:


Try a special case: Assume that x = k and y = k. This is consistent with what we're given (that the average of the two is k). The average of k, k, and z is (2k + z)/3. Select (A) (2k + z)/3.

 Work backwards:
 We are asked to find the area of the circle.
 Looking at the reference information, we can find the area of the circle if we know the radius (A = πr²).
 Since the radius is half the diameter, we would know the radius if we knew the diameter.
 Since the diameter is one side of a right triangle, we could find its length if we knew the other two sides.
 The hypotenuse of that triangle is 2 (it is the side of the given equilateral triangle). The other leg of that triangle is 1 (it is half the side of the equilateral triangle). So, we have enough information to solve the problem.
 Estimate the answer: The diameter must be a bit less than 2, since it is a leg of a right triangle with hypotenuse 2. So, the radius must be a bit less than 1. So, the area must be a bit less than π.
 Look at the answer choices: (A) is quite a lot less than π, so eliminate that. (D) and (E) are not less than π, so eliminate those. We are left with (B) and (C).
 Work forwards:
Using either the Pythagorean theorem or the properties of a 30°60°90° triangle (see Reference Information), the diameter is √3. So, the radius is √3/2. The area is π × 3/4, or 3π/4. Select (C) 3π/4.

Solution 1:

Consider another problem: If you think about it, the problem you are asked to solve is similar to "When 12 is divided by the positive integer k, the remainder is 0. For how many different values of k is this true?"
 12 is divisible by 1, 2, 3, 4, 6, and 12. So, the answer to our new problem is 6. Since that isn't one of the answer choices, that should tip you off that we aren't done yet.
 Check each of the six answers to ensure they leave remainder 3 when divided by 15. When 15 is divided by 1, the remainder is 0. When it is divided by 2, the remainder is 1. When it is divided by 3, the remainder is 0. However, when it is divided by 4, 6, and 12, the remainder is 3. Since there are three valid values, select (C) Three.
Solution 2: Guess and check: If you have time remaining, find the remainder when 15 is divided by each number from 1 to 15. You'll find three numbers (4, 6, and 12) that leave remainder 3. Select (C) Three.