# Solutions for Practice Test 3, The Official SAT Study Guide, Section 2

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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.
SAT Math Tips

Here are solutions for section 2 of the third practice test in The Official SAT Study Guide , second edition, found on pages 514–519. The following solutions demonstrate faster, more informal methods that might work better for you on a fast-paced test such as the SAT. To learn more about these methods, see my e-book Succeeding in SAT Math or the SAT math tips page.

1. If 20y − 5y = 15, then:
15y = 15
y = 1
Given that y = 1 and y = x − 5, then 6 = x. Select (A) 6.
• Estimate the answer: Since ⅓ of the buttons are red and somewhat more than ⅓ of the buttons are blue, somewhat less than ⅓ of the buttons must be yellow.
• Look at the answers: (E) ⅓ is definitely out of range of our estimate. (A) 1⁄9 appears to be too small as well, but we still have several answers to work with.
• There are 9 − 4 − 3 = 2 yellow buttons. There are 9 buttons in total. The probabillity that a yellow button is selected is 2⁄9. Select (C) 2⁄9.
• Draw a diagram: On the given diagram, draw an arc to complete the circle. This should result in something similar to the following: • Look at the answers: Look at the answers for a figure that looks like the piece you just drew. This would be (B).
2. Draw a diagram: Draw two lines, from top to bottom and from left to right, dividing the circle graph into quarters. Now, looking at the diagram, the Vanilla piece occupies an entire quarter and a bit more, so it represents more than 25%. However, pecan, peach, mint, and strawberry each occupy less than one of the quarters, so there are four that represent less than 25%. Select (D) Four.
3. Solution 1:
• Estimate the answer: We can assume the diagram is drawn to scale. By looking at the diagram, you can see that the angles denoted by x, y, and z are all obtuse (> 90°). So, their sum must be greater than 270°, probably somewhere in the 300s.
• Look at the answer choices: The only answer choice greater than 270° is (E) 360. Select that answer.
Solution 2: You may know that the external angles of a polygon sum to 360°. The three angles denoted by x, y, and z are the external angles of a triangle. So, they must sum to 360°. Select (E) 360.

Solution 3:

• The Reference Information tells you that the angles of a triangle sum to 180°. So, the third angle of the triangle is 180° − 37° − 58° = 85°.
• We can now calculate x, y, and z, since each is supplemental to an angle that we know:
x = 180 − 85 = 95
y = 180 − 37 = 143
z = 180 − 58 = 122
Now, x + y + z = 95 + 143 + 122 = 360. Select (E) 360.
4. We are given that 6x + 4 = 7. Subtract 8 from both sides of the equation gives us:
6x − 4 = − 1
Select (B) −1.

Note: In a question such as the above, where you are given an equation and asked to find the value of an expression, it's usually a lot slower to solve for x than it is to try to turn one side of the equation into the expression you're looking for.

5. Solution 1: The distance between A and B is the same as the distances between B and C, C and D, and so on. You can tell this either by looking at the diagram (you can assume it's drawn to scale) and noticing that the distances appear the exact same, or by noting that the figure has rotational symmetry, so all of the distances have to be the same. Either way, ABC contains 2 equal segments, while AEC contains 3 equal segments. The ratio of their lengths is 2 to 3. Select (B) 2 to 3.

Solution 2: I wouldn't recommend a formal solution here due to the amount of time required, but if you wanted to do that, here is an outline of how you could do so: Draw a radius from the centre of the circle to each of A, B, C, D, and E, prove that each of the five triangles thus created are congruent (since two of the sides of each triangle are radii and the other side is that of the equilateral pentagon, all three sides of each triangle are equal). Since each triangle is congruent, the angle formed by the two radii must be equal in all five triangles, so each of the arcs AB, BC, ... must be equal. ABC contains two equal segments and AEC 3, so the ratio must be 2 to 3. Select (B) 2 to 3.

• Estimate the answer: When you square anything, the result is positive. When you square a number whose absolute value is greater than 0 and less than 1, the result will be greater than 0 and less than 1 but be smaller in absolute terms. So, the answer must be between 0 and ½ (but not either extreme).
• Look at the answer choices: The only answer choice between, but not on, 0 and ½ is D. Select (D) D.
6. Solution 1: You can turn the sentence into an equation as follows:
 how much greater What we're looking for (you can call it x if you like) is = the sum of [add the two following things together] s s and + t t than − ( the sum of [add the two following things together] s s and + w w )
Putting it all together, we get:
x = s + t − (s + w)
x = s + tsw
x = tw
Select (C) tw

Solution 2: Try a special case: If you find all the letters to be confusing, substitute numbers for them.

7. When t = 4:
P(4) = 3,000 · 21
=6,000
When t = 16:
P(16) = 3,000 · 24
= 48,000
To find the population increase, subtract the number of organisms at t = 16 (48,000) from the number at t = 4 (6,000). The result is 48,000 − 6,000 = 42,000. Select (D) 42,000.
• Estimate the answer: The average of 3, s, and t is 5. Say that s and t were both 5. Then s + t = 10. However, the average would be slightly less than 5. So, s + t has to be slightly more than 10.
• Look at the answer choices: The only answer greater than 10 is (E) 12. Select that answer.
8. The elements common to sets A and B will be the numbers that are both in circle A and circle B. There is a 5 and a 2 that are in both circles, so the answer is 5 + 2 = 7. Select (D) 7.
• Estimate the answer: If 60% of 800 students are female, then more than 60% of the remaining 200 students must be male to even things out.
• Look at the answer choices: (A) 100 and (B) 120 are not more than 60% of 200. Eliminate those two choices.
• When working with percentages, it's usually a good idea to convert things to absolute numbers. The number of males accepted so far is 40% × 800 = 320. State University needs 1,000 × ½ = 500 males in total. So, the number of additional males required is 500 − 320 = 180. Select (D) 180.
• This question isn't too difficult if you can think in terms of integers, which you may not have a lot of practice doing. Now, you can't add two positive integers, or multiply them together, and get something that's not a positive integer. If you subtract one from the other, you'll still be left with an integer (although perhaps not a positive one). So, any sum or product of them must be a positive integer, and any difference must be an integer.
• Read the question carefully and understand what it is asking:
We are given that t² − k² < 6. Since the left-hand side must be an integer, it must be 5 or less. The left-hand side can also be written as (t + k)(tk).
We are given that t + k > 4. Since the left-hand side is an integer, it must be 5 or more.
We are given that t > k, or tk > 0. As the left-hand side is an integer, it must be 1 or more.
Now, think about this for a moment. The only possible way that (t + k)(tk) could be no more than 5, t + k no more than 5, and tk at least 1 is if t + k = 5 and tk = 1. We can solve this system of equations the normal way, or simply by inspection: t = 3, k = 2. Select (C) 3.
9. Solution 1:
• Convert the sentence to an equation. ½ of 23 percent of 618 can be written as:
½ · 23% · 618
• Using the laws of arithmetic, we can arrange this equation as:
23% · 618 · ½
= 23% · 309
This corresponds to (A) 23% of 309. Select that answer.
Solution 2:
• Estimate the answer: We can estimate ½ of 23% of 618 as ½ of 25% of 600. This works out to about 75.
• Look at the answer choices: Look at each answer and make an estimate of its size. Answer (A) is about 75. Answer (B) is about 40. Answer (C) is about 150. Answer (D) is about 40. Answer (E) is about 6,000. The only answer around 75 is (A) 23% of 309. Select that answer.
Solution 3:
• On your calculator, evaluate ½ of 23 percent of 618. ½ × 23% × 618 = 71.07
• Guess and check: Evaluate each answer to see whether it equals 71.07. (A) 23% of 309 is equal to 71.07. Select (A) 23% of 309.
10. Break the problem up and solve each part. First:
g(x) = f(3x + 1)
g(2) = f(3 · 2 + 1)
g(2) = f(7)
Next, find the value of f(7). From the table, this is −5. Select (A) −5.
11. This problem may appear intimidating because you've probably never been taught how to find the area of little swirly things in class. However, with the right technique, it's not too difficult.
• Read the question and understand what it is asking: It turns out that the swirls are really just two semicircles that are joined to each other. We can determine the area of semicircles, so it should be possible to solve the problem.
• Estimate the answer: The diameter of the circle is 6, so the radius is 3. Therefore, the area of the entire circle is π(3)² = 9π. Looking at the diagram, it appears that the shaded region is about ⅔ of the circle, or about 6π
• Look at the answer choices: (D) and (E) are obviously too big. (A) appears too small. (C) 6π looks like the most reasonable. If you feel that (B) is definitely too small, you can select (C) at this point. Otherwise, continue, and even if you aren't able to make any further progress you have at least set yourself up to make a good educated guess.
• Draw a diagram: You are given a diagram, but it helps to draw another. It can help to draw a diagram with the two halves separated, or a diagram with the bottom half flipped right to left, producing 3 circles.
• The area of the top half of the circle can be calculated as the sum/difference of the three semicircles. This is ½(9π − 4π + π) = 3π. Similarly, the area of the bottom half is also 3π. The total is 6π. Select (C) 6π.
12. Solution 1: Draw a diagram: Draw six points, draw all lines between them, and count the number of lines very carefully. If you are sufficiently careful, you will find there are 15 lines. Select (A) 15.

Solution 2: A line extends from any point to any other point. Since there are 6 points, there are 6 × 5 = 30 lines. However, this counts each line twice (e.g. it counts the line from point 1 to point 2, and the line from point 2 to point 1). So, we have to divide this result by 2, giving 15. Select (A) 15.

• Since a = b, substitute a for b in the three possibilities. This results in:
I. f(2a) = 2f(a)
II. f(2a) = [f(a)]²
III. 2f(a) = f(2a)
• Look at the answer choices: Note that equations I. and III. are the same, so we can eliminate all choices that contain one but not the other. So, we can eliminate (B) and (D). Now, II. gives a different value for f(2a) than the other two, so we can eliminate all choices that contain II. and either I. or II. We can eliminate (D) again and (E). This leaves (A) None and (C) I and III only.
• If we substitute x = a, y = a into the given equation, we get f(2a) = 2f(a). This is the same as I. and III., so those equations must be true. Select (C) I and III only.
13. Solution 1: Since the area of the pool is 4,000 square meters, xy = 4000, or x = 4000⁄y. The total length of the rope is y + 4x = y + 4(4000⁄y = y + 16000⁄y. Select (B) y + 16000⁄y.

Solution 2:

• Try a special case: Say that y = 100. Then x = 40. The total amount of rope needed is 100 + 4 × 40 = 260.
• Look at the answer choices: If y = 100 and x = 40, then:
• (A) works out to 140.
• (B) works out to 260.
• (C) works out to 153.3
• (D) works out to 326.6
• (E) works out to 353.3
Only (B) evaluates to the expected value. Select (B) y + 16000⁄y.