# Solutions for Practice Test 6, The Official SAT Study Guide, Section 4

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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 4 of practice test #6 in The Official SAT Study Guide , second edition, found on pages 712–717. The solutions below 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. Start with the first equation, and substitute the second and third into it. You might be able to do this in your head, but I'll show the exact steps below:
xy = 8
x − 3z = 8
x − 3(2) = 8
x = 14
Select (E) 14.
2. Look at the answer choices: The five answers relate m, t, and s using < . If Todd is older than Marta but younger than Susan, then t must be in the middle, and t < s. The only such answer choice is (A) m < t < s. Select that answer.
3. The arithmetic mean of the areas is the total area (5) divided by the number of regions (2), or 5/2. Select (B) 5/2.
• List all square numbers less than the largest answer, 50:
1, 4, 9, 16, 25, 36, 49
• Add 1 to each of the above numbers:
2, 5, 10, 17, 26, 37, 50
• The numbers above can all be written in the form n² + 1. The only one listed above that is also in the answer choices is (E) 50. Select that answer.
• The two sides of the triangle that meet at O are both radii of the circle, so they are equal. Therefore, the angles opposite those sides are equal, and the third angle of the triangle must also be y°.
• From the reference information, the sum of the measures in degrees of the angles of a triangle is 180. So:
x + y + y = 180
40 + 2y = 180
2y = 140
y = 70
Select (D) 70.
4. Guess and check: 121 = 11 × 11. Since 11 is a prime number, the only factors of 121 are 121, 11, and 1. Select (A) 121.
• Estimate the answer: We are told that XZ is slightly less than h. So, the area of the triangle (½h(XZ)) must be slightly less than ½h².
• Look at the answer choices: (A) and (B) are both slightly less than ½h². (C) doesn't even have an h² term in it, and (D) and (E) are greater than ½h², so eliminate (C), (D), and (E). You might have noticed that, out of the two remaining answers, answers (C), (D), and (E) look more similar to (B) than they do to (A), so that might suggest that (B) is more likely to be correct than (A). If you're really low on time, you might want to guess (B) and move on. Otherwise, proceed to the next step:
• If XZ = (6/7)h, then the area = ½(6/7)(h)(h) = (3/7)h². elect (B) (3/7)h².
5. This looks gross, but don't panic:
• Just remember the laws of exponents to simplify the left-hand side of the equation:
a³b² = 432
b(ab)² = 432
• Guess and check: Try each answer until you find a possible one. Can ab = 6? If ab = 6, then (ab)² = 36. Plugging that into the above equation, we get b = 12, but that means that a = ½. However, a must be a positive integer, so this answer choice doesn't work. Can ab = 12? If so, then (ab)² = 144, and b = 3. This works. Select (B) 12. If you started by checking other answers, note that, for (C), (ab)² = 324 and so b = 4/3, which is not a positive integer. Similarly, b will not be a positive integer for (D) and (E) either.
6. If a number has a factor of 10, its ones digit must be 0. The largest such three-digit number is 990.
7. If a recipe for chili for 20 people requires 4 pounds of beans, then chili for one person at that rate requires 4/20 = 1/5 pounds. So, chili for 150 people requires 1/5 150 = 30.
8. Guess and check: Say that n = 10. If 50 percent of n is 5, so if n is increased by 50 percent of itself, the result is 15, which is between 10 and 20. Enter 10. If you weren't as lucky with your guess the first time, select a larger number (if the result was too small) or a smaller number (if the result was too large) and try again.
• Draw a diagram: Draw a diagram similar to the following: • We can find the other side because we are given that the perimeter = 250:
perimeter = 40 + w + 40 + w = 250
2w = 170
w = 85
• From the reference information, the area = lw = (40)(85) = 3400.
9. There are several ways to solve this problem; here's an innovative way of doing so: Because the \$2 light bulbs cost twice as much each but only half as many were ordered, then exactly the same amount of money was spent on the \$1 bulbs as on the \$2 bulbs. Since \$600 was spent in total, \$300 was spent on the \$1 bulbs and \$300 on the \$2 bulbs. So, there were 300 \$1 bulbs and 150 \$2 bulbs, or 450 in total.
10. Substitute xy = 20 in 4(x + y)(xy) = 40:
4(x + y)(20) = 40
4(x + y) = 2
x + y = ½
Enter 1/2.
• Draw a diagram: Draw a graph such as the following: • Looking at the diagram, the radius of the circle is the same length as the y-axis between y = 12 and y = 0, or 12. Enter 12.
11. If 40% of the voting-age population voted, that would be 40%(1,200 + 1,300) = 40%(2,500) = 1,000. Now, the number of registered voters is 1,000 + 1,200 = 2,200, and the turnout is 1,000/2,200 = 10/22 = 5/11. Enter 5/11.
12. Looking at the diagram, there are three edges connecting V to another vertex. So, if 11 lines are drawn to the other vertices, 3 will lie on an edge and so 8 will not. Enter 8.
• Because the area of ABCD = 4, the area of each quadrant of the rectangle is 1.
• Because the width of each of these rectangles is ½, the height must be 2. This must mean that c = 2.
• We now have a point on y = px³, (½, 2). Substitute this into y = px³:
2 = p(½)³
2 = p(1/8)
16 = p
Enter 16.