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

Math Lair Home > Test Preparation > Solutions for Practice Test 2, The Official SAT Study Guide, Section 2
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 second practice test in The Official SAT Study Guide , second edition, found on pages 452–457. 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.

• Estimate the answer: The answer should probably be slightly larger than 20.
• Look at the answers: (A), (B), and (C) are not slightly larger than 20. Eliminate them.
• If the third term is 10, then the next term is 2 more than twice 10. 2 + 2(10) = 22. Select (D) 22.
• Estimate the answer: Five minutes is less than 1/10 of an hour (fifty minutes is less than an hour, and 50 = 5 × 10). So, the machine can fill less than 24/10 = 2.4 crates.
• Look at the answers: The only answer less than 2.4 is (A) Two. Select that answer.
1. Cathy sold 48 cars in May. She sold 20 + 18 cars in the months of January and February combined. So, she sold 48 − (20 + 18) = 10 more cars in May than in January and February combined. Select (A) 10.
• Estimate the answer: There are six months. Looking at the graph, April seems to be an average month. So, April will probably take somewhere around 1/6 of the graph, or around (1/6)×360 = 60°
• Look at the choices: (B) 54° and (C) 60° are near our estimate. The other answers can be safely eliminated.
• 20 + 18 + 22 + 30 + 48 + 42 = 180 cars were sold in total. So, the degree measurement of the circle graph would be:
30/180 × 360°
= 60°
Select (C) 60°.
2. Since the figure is being rotated 90° counterclockwise, you can see what it looks like by rotating your paper 90° counterclockwise (in the direction of the arrow; the right edge of your paper will now be at the top of your desk). Now, rotate your paper back and find the shape that looks like the one you just saw (it may help to just rotate your paper back 45°, allowing you to switch your view between the figure and the answers). The answer is (D).
3. Convert the sentence into an equation. The first part of the sentence results in:
3 + 2x = 10
The second part of the sentence asks us to find 4x. So, solve the equation for 4x:
2x = 7
4x = 14
Therefore, the answer is (D) 14.
4. Solution 1: Try a special case: Say that a = − 1. Then a = −1, 2a = −2, 4a = −4, and 8a = −8. a is the largest, so we can eliminate (B), (C), and (D). If you aren't convinced that a must always be the largest, you may want to try another special case or two. You will always find that a is the largest, so select (A) a.

Solution 2: Using a graphing calculator, plot y = x, y = 2x, y = 4x, and y = 8x. Examine the area of the graph where y < 0. You will find that y = x is always the largest. Therefore, a must be the largest of the four and the answer is (A) a.
5. Solution 1: The shape can be thought of as a large rectangle and a small piece that appears square. We are not given the height of the square, but since the height of the entire figure is 6 and the height of the figure below the square is 4, the height of the square is 2 (or, if you don't see that, you can use a finger or your pencil to measure the height and the width; you'll find they're the exact same). So:
Area = area of large rectangle + area of square
= 4(6) + 2(2)
= 24 + 4
= 28
Select (B) 28.

Solution 2: This shape can be thought of as a 4×4 square plus a long rectangle. The height of the rectangle is 6 − 4 = 2, and the width of the rectangle is 2 + 4 = 6. So:
Area = area of square + area of long rectangle
= 4(4) + 2(6)
= 16 + 12
= 28
Select (B) 28.

Solution 3: This shape can be thought of as a 6 × (2 + 4) square minus a 2×4 rectangle. Similar to the above solutions, the area is 36 − 8 = 28. Select (B) 28.
6. Solution 1:
• Solve the equation by taking the square root of both sides:
(x − 2)² = 25
x − 2 = ±5
x = 5 + 2 or x = −5 + 2
x = 7 or x = −3
We're looking for a solution where x < 0, so only x = −3 is admissible (and 7 isn't one of the choices anyway). Select (D) −3.
Solution 2: Guess and check:
• Try the middle number first, (C) − 5. (−5 − 2)² = (−7)² = 49, which is not 25.
• Our first solution was too big (in absolute terms), so try something smaller in absolute terms, say (D) −3. (−3 − 2)² = (−5)² = 25. Therefore, the answer is (D) −3.
• Estimate the answer: T is much closer to S than P, so the ratio PT/PS is much larger than ½, perhaps ¾ or 4/5.
• Look at the answer choices: The only choice that is larger than ½ is (E) 4/5. Select that answer.
• Try a special case: From the graph, if W = 5, L = 50.
• Look at the answers: When W = 5, the answers evaluate to:
• (A) L = 5
• (B) L = 10
• (C) L = 15
• (D) L = 50
• (E) L = 60
The only answer that gives the correct answer for our special case is (D) L = 10W. Select that answer.
7. We are given that the only mode is 5. Therefore, since there are four 5's, there cannot be four of any other number. There are already three 6's. So, n cannot be another 6, or else there would be two modes. Select (A) 6.
8. Find where circles Y and Z overlap, and sum all of the numbers found there. The two circles overlap in two places, one with a "3" and one with a "7". The number of elements in the intersection of Y and Z is 3 + 7 = 10. Select (C) 10.
9. Solution 1:
• Try a special case: If t = 2, then m = 8, and so w = 8² + 8 = 72.
• Look at the answers: If t = 2, the answers evaluate to:
• (A) 6
• (B) 8
• (C) 10
• (D) 40
• (E) 72
The only answer that gives the correct result for our special case is (E) 72. Select (E) t6 + t³.
Solution 2: Substitute m = t³ into w = m² + m:
w = m² + m
w = (t³)² + t³
w = t6 + t³
(don't forget laws of exponents!)
Therefore, the answer is (E) t6 + t³.
• Don't try to tackle the entire problem in one step. Break the problem up and solve the first part first:
6Δ − 5Δ
= (5)(7) − (4)(6)
= 11
• Guess and check: Try the answers until you come up with the correct answer. 11 is between 3² and 2², so (B) seems like a good starting point:
3Δ + 2Δ
= (4)(2) + (3)(1)
= 11
Therefore, the answer is (B) 3Δ + 2Δ.
• Look at the answers: For (A), (C), and (E), x/y is an integer. Elminate those three answers.
• Guess and check: For (B), x²/y is not an integer. Eliminate (B). The only answer remaining is (D) x = 6, y = 4. Select that answer.
10. Solution 1: Draw a diagram: Plot y = |−2x + 6| on your graphing calculator. You should find that the graph corresponds with that of (B). Select that answer.

Solution 2: Draw a diagram: The graph of y = −2x + 6 is given. The graph of y = |−2x + 6| would be the same, except that the negative portion of the graph would be reflected in the x-axis. Draw that in the given graph and compare with the answer choices. The only answer choice that looks similar is (B). Select that answer.

Solution 3:
• Look at the answers: Since y is equal to the absolute value of something, then y must always be greater than or equal to 0. Therefore, answers (A), (C), and (E) can be eliminated.
• Try a special case: When x = 3, y = 0. Graph (D) does not show this. Therefore, the correct answer must be (B).
11. This question is not difficult computationally, but it does require a fair bit of insight.

Solution 1: Draw a diagram: Draw a box that just surrounds the cylinder. The width of the box is d, the height of the box is h, and the depth of the box is also d (since the depth would be equal to the circle's diameter). So, the volume of the box is d²h. Select (B) d²h.

Solution 2:

• Look at the answer choices: At first glance, it appears to be impossible to eliminate any answers; however, we can look at the units that each answer would have to be expressed in. Say that h and d are measured in inches (it doesn't matter; it could be centimetres, miles, or simply "units"). Then the volume of the cylinder would be measured in inches³. Now:
• The units for (A) dh are inches × inches = inches², so that can't be correct.
• The units for (D) d²h² are inches² × inches² = inches4, so that can't be correct either.
• Similarly, (E) gives a unit of inches², which can't be correct either.
All three answers can be eliminated.
• Now, no matter what the values of d and h are, the correct answer must always be greater than the volume of the cylinder. The volume of the cylinder, from the first page of the section, is:
V = πr²h
V = π(d/2)²h
V = (&pi/4)d²h
Now, (B) d²h will always be greater than (&pi/4)d²h. However, if d is large and h small, then dh² might be less than the volume of the cylinder, so that can't be the correct answer. Having eliminated every other answer, the correct answer is (B) d²h.
12. The sentences in the question can be converted into equations:
x² = 4y² (sentence 1)
x = 1 + 2y (sentence 2)
We can rewrite sentence 1 as
x² = (2y
and sentence 2 as
x − 1 = 2y

From sentence 2, substitute x − 1 for 2y in sentence 1:

x² = (x − 1)²

Now, take the square root of both sides of the equation. Important: Don't forget about the negative square root, or else the equation will have no solution.

x = ±(x − 1)
x = x − 1 or x = 1 − x
The first equation has no solution. Looking at the second equation:
2x = 1
x = ½
Select (E) ½.

This was a complex question, so if you have time it's a good idea to check your work. From sentence 2:

½ = 1 + 2y
−½ = 2y
y = −¼
and substitute that value for y into the first sentence:
x² = 4(¼)²
x² = ¼
x = ±½
Of course, the negative solution doesn't satisfy the second sentence, so the answer x = ½ is correct.
• Draw a diagram: Draw a graph of the two lines (you might have space to do this at answer choices 17 (C) or 17 (D)). Your graph might look something like the following: • Estimate the answer: From the diagram, t is positive and quite a bit larger than 1. So, (A), (B), and (C) are definitely wrong. Depending on how well you've drawn the diagram, you may also be able to see that (D) 3 is wrong, in which case you would select (E) 5. If not, or if you want to be sure, continue on to the next step.
• Since lines l and q are perpendicular, their slopes are the negative reciprocals of each other. Now, l goes up by 1 unit as it goes right 2 units, so its slope is ½. So, the slope of q is −2. From (2, 1), if q goes left 2 units (or right −2 units), it will increase by 2 × 2 = 4. So, at this point, its value will be (2 − 2, 1 + 4), or (0, 5). Select (E) 5.