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 3 of practice test #7 in The Official SAT Study Guide, second edition, found on pages 768–773. 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.
 Estimate the answer: To make 30 rolls, 25/10 = 2.5 pounds of flour are needed. Since 12 is less than half of 30, the answer is definitely less than 2 and is probably somewhere around 1.
 Look at the answer choices: The only answer within range of our estimate is (A) 1. Select that answer.
 Simplify the expression 2 · (x/y) · y² by cancelling out the y's, giving 2xy. Since xy = 10, 2xy = 20. Select (E) 20.
 If x + y = 30, then x = 30 − y. Since x > 8, then 30 − y > 8. Rearranging this equation, we get 22 > y. Select (B) y < 22.
 Draw a diagram: Draw a diagram similar to the following:
 Estimate the answer: PQ is equal to 4 and QR is equal to 2. Because PR is the hypotenuse of a right triangle, it must be greater than 4, but because of the triangle inequality, it must be less than 6. So, the perimeter must be greater than 10 but less than 12.
 Look at the answer choices: Answers (A), (B), and (E) are greater than or equal to 12. Eliminate those three answers.
 The length of the hypotenuse is √4² + 2² = √20 (see Reference Information if you've forgotten). So, the perimeter of the triangle is 6 + √20. Select (C) 6 + √20 (approximately 10.47).

Solution 1:
 Estimate the answer: 8 + (26 − 1)9 is probably a bit less than 25 × 10, so say maybe around 240. Given that the sequence increases by 9 each time, the answer is probably somewhere between 25 and 30.
 Look at the answer choices: Answers (A) and (B) are nowhere in the range of our estimate, so eliminate them.
 Look for a pattern: The first term is 8 + (0)9, the second term is 8 + (1) 9, the third term is 8 + (2), ... the 26th term is 8 + (25)9 = 8 + (26 − 1)9. Select (D) The 26th.
 Solution 2: You may know that the general term for the n^{th} term of an arithmetic sequence is a + (n − 1)d, where a is the first term and d is the difference. In this case, a = 8 and d is 9, giving 8 + (n − 1)9. The term equal to 8 + (26 − 1)9 must be the 26^{th} term. Select (D) The 26th.
 Solution 1: Look at the answer choices: If you know t and y, you can determine x, and from there the others. If you know s and x, you can determine t, and from there the others. Similarly, if you know r and t, or r and s, you can determine the other angle measurements. However, if you know t and z, you cannot determine any other angles. Select (A) t and z.
 Solution 2: Try each of the answer choices, select specific values for the two variables given in the answer, and see which one does not allow you to find values for all 6 angle measures:
 For (A), say that t = 45. Then, because z is opposite to t, it must also be 45. However, it doesn't seem possible to determine any more angles.
 For (B), say that t = 45, and y = 45. Then, x = 90, s = 45, r = 90, and z = 45. We can eliminate answer (B).
 For (C), say that s = 45 and x = 45. Then, y = 90, r = 45, z = 90, and y = 45. We can eliminate answer (C).
 For (D), say that r = 45 and t = 45. Then, s = 90, x = 45, y = 90, and z = 45. We can eliminate answer (D).
 For (E), say that r = 45 and s = 45. Then, t = 90, x = 45, y = 45, and z = 90. We can eliminate answer (E).
Select (A) t and z.
 Estimate the answer: If two consecutive numbers sum to t, then the greater of the two numbers is probably a bit more than t/2, but less than t.
 Look at the answer choices: (C) and (D) seem plausible based on our estimate. (A) and (B) are too small, and (E) is too big. Eliminate those three answers.
 Try a special case: Say that t = 5. Then the greater of the two numbers is 3. If t = 5, then answer (C) evaluates to 3, and answer (D) evaluates to 3.5 (which is wrong). So, the answer must be (C) (t + 1)/2.
 If a thirteenth child joins the class, the median number of siblings will be the seventh greatest value (which would also be the seventh smallest value).
 Looking at the table, there are currently 3 children with 0 siblings and 6 children with 1 sibling. So, the fifth, sixth, seventh, eighth, and ninth smallest values for the number of siblings are all 1. No matter how many siblings the new student has, the seventh smallest number of siblings will be 1 after the new student joins. So, the median will be 1.
 Right now, the students have 6 + 2(2) + 3 = 13 siblings. In order for the new arithmetic mean to equal the median (1), there must be 13 siblings in total (since 13/13 = 1). Therefore, the new child can't have any new siblings. Select (A) 0.

 Convert the sentence into an equation:
When  [ignore]

twice  2 ·

a number  x (or any variable)

is decreased by  −

3  3

the result is  =

253  253

This gives us 2x − 3 = 253. Solving the equation, 2x = 256 or x = 128. Enter 128.
 Draw a diagram: We are asked to find how many black sneakers were produced. Draw a box around, or put an asterisk in, or otherwise mark, the square in the table corresponding to the number of black sneakers. This will be the rightmost square in the second row.
 We know how many black lowtops were produced. We can find the number of black hightops as follows.
Since 10,000 sneakers were manufactured in total, and 5,500 of those were lowtops, 4,500 must be hightops. Since 3,600 of those were white, 900 must be black. So, if 900 black hightops and 1,500 black lowtops were produced, then 2,400 in total were produced. Enter 2400.
 Draw a diagram: A diagram is already given, but it helps to label it. The top and bottom of the rectangle are 2 units in length (the distance between (1, 0) and (1, 0)). If the perimeter of PQRS is 10, then the left and right edges of the rectangle must be 3 units in length. Then, point R must be (1, 3).
 Since (1, 3) is on the graph of y = ax², substitute x = 1, y = 3 into the equation:
3 = a(1)²
a = 3
Enter 3.
 Substitute a = 2 and c = 3 into the equation:
2b + b = 2 2·3
3b = 8
b = 8/3
Enter 8/3.
 Work backwards: Because line l intersects parallel lines, we could find the value of x if we knew the angle that BC makes with line l, as the two values would be equal. We are told that line l bisects ∠ABC, so the angle we are looking for is ½ of ∠ABC. Because line k also intersects parallel lines, its measurement is also y°. Putting this all together, x must be half of y.
 Try a special case: Say that y = 50. Then, x = 25. Enter 25.
 First, it helps to introduce suitable notation and possibly draw a diagram. Say that the four plumbers are P1, P2, P3, and P4, and the trainees are T1, T2, T3, and T4.
 A team may consist of any one of the four experienced plumbers, so there are four possibilities as to the experienced plumber on the team.
 List out all of the possibilities for the two trainee plumbers on the team:
(P1, P2), (P1, P3), (P1, P4), (P2, P3), (P2, P4), (P3, P4)
There are six possibilities.
 To determine the total number of different teams, multiply the number of combinations of plumbers by the number of combinations of trainees: 4 × 6 = 24.
 Work backwards: We could find the radius of the larger circle if we knew its tootal area. We already know the area of the shaded region (64π square inches), so we could find the total area if we knew the area of the smaller circle. We can find the area of the smaller circle, because we know its radius: 6 inches.
 Putting all of the above together, the area of the smaller circle is π(6)² = 36&pi. The total area is 36π + 64π = 100π. Now, because Area = π(r)², and the area is 100π, then:
100π = πr²
r² = 100
r = 10
Enter 10.
 List the factors of n: 1, p, r, s, pr, ps, rs, prs. There are 8 in total. Because p, r, and s are all prime, they can't be broken down any further, so there are only 8 factors. Enter 8.
 This problem looks be a little bit timeconsuming, but there is a shortcut if you read the problem carefully: