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 5 of practice test #10 in The Official SAT Study Guide, second edition, found on pages 965–970. 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.
 By inspection, x = 39 and x − 2 = 37. Select (B) 39.
 Read the question and understand what it is asking: z is equal to the total number of students in the class, so we need to determine which of the expressions in the answers is equal to the number of students in the class.
 Look at the answer choices: : Determine which of the answers is equal to the total number of students in the class:
 (A) is equal to the number of junior boys plus the number of senior girls, not the total number of students in the class.
 (B) is equal to the number of boys plus the number of seniors, which is not correct.
 (C) is equal to the number of junior girls plus the number of senior girls, which is not correct.
 (D) is equal to the number of junior girls plus the number of senior girls plus the number of girls, which is not correct.
 (E) is equal to the number of junior boys pluse the number of senior boys plus the number of junior girls plus the number of senior girls, which is the total number of students. Select (E) k + n + r + s.
 Estimate the answer: The angle labelled x° appears larger than the 25° angle, but not significantly so. Perhaps the angle measures 30° or 35° or so.
 Look at the answer choices: Since the answer is greater than 25°, we can eliminate (A) 25. The angle is definitely less than 60°, so we can eliminate (E) 60. The other three answers seem plausible.
 Because ∠ACB is supplemental to a 60° angle, its measure is 120°. Since the angles of a triangle add to 180°, x = 180 − 120 − 25 = 35. Select (C) 35.
 The difference between the cost of the new refrigerator and the cost of fixing the new one is 900 − 300 = $600. If it saves $15 per month, it will save an amount of $600 in $600/($15/month) = 40 months. Select (D) 40.
 If the perimeter of equilateral triangle DEF is 10, the perimeter of equilateral triangle ABC is 30. So, each side of triangle ABC = 30/3 = 10. Select (B) 10.
 If the machine mints coins at the rate of one coin per second, and there are 60 seconds in a minute and 60 minutes in a second, it mints 3,600 coins per hour. If it works 10 hours a day, it mints 36,000 coins per day. So, it will take 360,000/36,000 = 10 days to mint 360,000 coins. Select (A) 10.
 The arithmetic mean of x and 3x is (x + 3x)/2 = 4x/2 = 2x. Since the arithmetic mean is 12, 2x = 12, or x = 6. Select (C) 6.
 Draw a diagram: It may be helpful to draw a Venn diagram of the situation.
 If some members of the chess team are on the swim team and no members of the swim team are tenth graders, then some members of the chess team are not tenth graders. Select (C) Some members of the chess club are tenth graders.
 Solve for n:
3x + n = x + 1
n = −3x + x + 1
n = −2x + 1
n = 1 − 2x
Select (D) 1 − 2x.
 Look at the answer choices: List the first few members of each of the sets in the answers:
 [5] = {5, 10, 15, ...} 5 is not in [2] or [3], so eliminate (A).
 [6] = {6, 12, 18, ...} 6 is not in [5], so eliminate (B).
 [10] = {10, 20, 30, ...} 10 is not in [3], so eliminate (C).
 [21] = {21, 42, 63, ...} 21 is not in [5], so eliminate (D).
 [60] = {60, 120, 180, ...} All of these numbers are in [2], [3], and [5]. Select (E) [60].
 Label ∠AOB as having measure 80°.
 Since ∠AD is a straight angle, it has measure 180°. So, ∠BOD has measure 180 − 80 = 100°.
 Since CF bisects ∠BOD, and ∠BOD has measure 100°, then ∠BOC and ∠COD both have measure 50°.
 The measure of ∠EOF is equal to that of ∠BOC, since they are opposite (or vertical) angles. So, its measure is also 50°. Select (B) 50°.
 Guess and check: Try each value, starting with the smallest, to see if it makes √5k/3 an integer:
 √5 · 3 / 3 = √5, which is not an integer. Eliminate (A).
 √5 · 5 / 3 = √25/3, which is not an integer. Eliminate (B).
 √5 · 15 / 3 = √25 = 5, which is an integer. Select (C) 15.
 Draw a diagram: You may find it helpful to draw grid lines through all six shapes, to divide them into 1 × 1 blocks.
 First, look at the areas of the three given shapes and figures I., II., and III. The three given shapes have area 1 + 5 + 3 = 9 square units. Figure I. has area 8. Figure II. has area 9. Figure III. has area 8. So, Figures I. and III. cannot be made using the three shapes.
 To verify that figure II. can be made from the three shapes, try drawing a possible fitting in the figure. As it turns out, it is possible (put the 1 × 1 figure in the bottom left corner, the cross just above that, and the other figure above that). Select (C) II only.
 Examine each integer greater than 20 but less than 30:
 21 = 3 × 7, both of which are prime.
 22 = 2 × 11, both of which are prime.
 23 is prime.
 24 = 4 × 6 (for example), but neither 4 nor 6 are prime.
 25 = 5 × 5. 5 is prime, but both factors are the same, so 25 does not satisfy the given criteria.
 26 = 2 × 13, both of which are prime.
 27 = 3 × 9, but 9 is not prime.
 28 = 4 × 7, but 4 is not prime.
 29 is prime.
Out of those numbers, three of them, 21, 22, and 26 meet the criteria in the problem. Select (D) 3.


 Solution 1:
 Draw a diagram: We are told that the parabola has a maximum of 2. Draw a horizontal line (x = 2) from the maximum to the xaxis.
 Look at the answer choices: First, the parabola patently does not have a value of 0 at x = 0 or x = 2, so eliminate answers (B) and (C).
 Estimate the answer: Looking at the line you just drew, the distance from x = 2 to the point where the parabola intersects the xaxis on the right is greater than the distance from x = 2 to the yaxis (x = 0). So, it must intersect the graph at a value of x greater than 4. So, we can eliminate choices (D) and (E). On the other hand, the parabola does appear to intersect the xaxis exactly at x = −1, so (A) looks good. Select (A) −1. Hint: Using your finger or your pencil to measure may help here.
 Solution 2: If the parabola has its maximum at x = 2, then it is symmetrical around x = 2. From the graph it is obvious that the parabola does not have a value of 0 at x = 0, 1, or 2, so by symmetry it can't have a value of 0 at x = 3 or 4 either. So, we can eliminate choices (B) through (E). Select (A) −1.
 Solution 1: Set x² + kx + 7 equal to (x + 1)(x + h):
x² + kx + 7 = (x + 1)(x + h)
x² + kx + 7 = x² + (h + 1)x + h
kx + 7 = (h + 1)x + h
Now, this looks problematic because we have one equation and three "letters". However, because k and h are constants, then the coefficients of x must be equal, so we have:
k = h + 1 (xcoefficient)
7 = h (constant term
So, k = 8. Select (D) 8.
 Solution 2: Try a special case: By inspection, it seems that h has to be 7; otherwise, how could you get a constant term of 7 in the expression? So, substituting h = 7 into (x + 1)(x + h) and expanding, we get:
(x + 1)(x + 7)
= x² + 8x + 7
But if this is equal to x² + kx + 7, then k must equal 8. Select (D) 8.
 Draw a diagram: Draw a vertical line from (4, 10) to the xaxis, a line from there to an origin, and a line back to (4, 10). This creates a right triangle with legs 4 and 10.
 The triangle you just drew is similar to triangle ABC; because both triangles' legs are parallel to the axes, they must make the same angles with the given line going through (4, 10). So, the legs of triangle ABC must be in the ratio 4:10 or 2:5.
 Look at the answer choices: Only answers (A) and (B) have legs in the ratio of 2:5. However, answer (B) does not satisfy either the Pythagorean theorem (a² + b² = c²) or the triangle inequality (the sum of the lengths of the two smaller sides must be greater than the length of the longest side). So, eliminate (B) and select (A) 2, 5, and √29.

 Try a special case: Say that k = 2. Then:
 (A) evaluates to 4, which is twice the value of an even integer.
 (B) evaluates to 7, which is odd.
 (C) evaluates to 8, which is twice the value of an even integer.
 (D) evaluates to 9, which is odd.
 (E) evaluates to 10, which is twice the value of an odd integer.
Only (E) is an even integer that is twice the value of an odd integer. Select (E) 4k + 2.
Note: If you tried an odd integer (e.g. 1), you might find that (A), (C), and (E) all result in an even integer twice the value of an odd integer. In that case, just try another integer, testing only (A), (C), and (E) this time.