# Solutions for 2014 SAT Practice Test, Section 8

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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 the section 8 of the 2014–15 SAT practice test (which was also the 2012–13 practice test); you can find the test at the College Board's web site or in the Getting Ready for the SAT booklet. The following solutions illustrate faster, less formal methods that may work better than formal methods 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 Heather ran 3 miles in 30 minutes and she ran 45 minutes at the same rate, then she would have ran 3 × (45/30) = 4.5 miles. Select (C) 4.5.
2. Divide both sides of the equation by 2:
mk = 3
. Select (A) 3.
• Estimate the answer: If 3 times a number is equal to 3/2, the number must be a fair bit less than 3/2.
• Look at the answer choices: (D) and (E) are greater than 3/2, so they can't be right. Eliminate them.
• Convert the sentence into an equation:  3 3 times × a number x is equal to = 3/2 3/2
This results in:
3x = 3/2
• Solving the equation, we get:
x = ½
Select (B) ½.
• Work backwards: We could determine the perimeter if we knew the lengths AB, BC, CD, DE, and EA. We would know CD and DE if we knew CE (since all three are sides of an equilateral triangle, they're all equal). We would know CE if we knew the area of the square (which we are given). We would also know AB, BC, and EA if we knew the area of the square (which we are given). So, we can determine the perimeter given just the area of the square.
• Work forwards: If the area of the square is 1, each side is 1, as is each side of the equilateral triangle. So, each of the five sides is 1 and the perimeter is 1 + 1 + 1 + 1 + 1 = 5. Select (B) 5.
• Estimate the answer: If −8 is at the left of the number line and 10 is at the right, then 0 must be slightly to the left of centre. Since c is in the centre of the number line, then 0 must be slightly to the left of that, so the smallest positive number must be c.
• Guess and check: Evaluate each answer, starting with our estimate c. c is mid-way between −8 and 10, so c = ½(−8 + 10) = 1. Now, because there are only five tick marks on the number line, b must be less than zero (if you aren't sure, you can evaluate it). So, the answer is (C) c.
• List the 7 numbers out in order. You are told that x is the median, so it must be the middle (fourth) number:
3, 4, 10, x, 15, 18, 21
So, x must be between 10 and 15.
• Look at the answer choices: The only choice between 10 and 15 is (D) 14. Select that number.
• Draw a diagram: Draw a diagram of two spheres. A rough sketch, like the one below that doesn't suggest much three-dimensionality, is fine: • The diagram might give you the insight that the longest distance is from one side of one sphere to the point of tangency, and from there to the other side of the other sphere. This distance is 7 + 7 + 4 + 4 = 22. Select (E) 22.
• Read the question carefully and determine what it is asking: To calculate the number of video rentals for each year, you need to calculate (members) × (video rentals / members). So, the total for the three years can be calculated by: 500 × 12 + 1,000 × 15 + 1,250 × 20
• We can make an estimate for the year 2000 by calculationg 500 × 10 = 5,000.
• For 2001, 1,000 × 15 = 15,000.
• We can make an estimate for the year 2002 by calculating 1,200 × 20 = 24,000.
For our final estimate, add the quantities up: 5,000 + 15,000 + 24,000 = 44,000.
• Look at the answer choices: The only choice anywhere near our estimate is (B) 46,000. Select that answer.
• Draw a diagram: A diagram is given, but it isn't drawn to scale. You may find it helpful to draw a diagram that is more approximately to scale than the given one is (you might find it easier to re-draw the existing diagram so that line ED is about halfway between A and BC). As well, label quantities that you know: • Estimate the answer: Draw a line from D parallel to EB until it meets BC. Let F be the point where it meets BC, as shown below: Now, BF = 4, and, from the diagram, FC appears to be about equal to BF, or 4. So, the length we're looking for is around 4 + 4 = 8.
• Look at the answer choices: (B) 8 is there; however, some of the answers contain x's, so it might be a good idea to investigate further. If we can't get any further, (B) is a good guess, though.
• Now, triangles AED and ABC share an angle, A. Also, the two sides comprising that angle (AE/AB and AD/AC) are in proportion, so the two triangles are similar. Since the other two sides of ABC are twice as long as the corresponding ones in AED, BC must be as well. So, it is 4 × 2 = 8. Select (B) 8.
3. Look for a pattern: The first student kept ½ of the rope. The second student kept ½ × ½ = ¼ of the rope. The third student kept ¼ × ½ = 1/8 of the rope. It appears that each possible value is ½ of the previous value. So, the next value must be 1/8 × ½ = 1/16. Select (C) 1/16.
• Read the question carefully and understand what it is asking: Where f(x) = 0, y = 0 on the graph and so the graph intersects the x-axis. So, we need to find a graph that intersects the x-axis exactly twice.
• Look at the answer choices:
• (A) doesn't intersect the x-axis at all.
• (B) intersects the x-axis once.
• (C) intersects the x-axis three times.
• (D) intersects the x-axis twice. Yes!
Select (D).
4. Solution 1:
• Draw a diagram: A diagram is given, but fill in x = 20 and y = 30 on the diagram.
• y°, x°, and x° are three angles of a quadrilateral. Since the angles in a quadrilateral sum to 360°, the fourth angle must equal 360 − 30 − 20 − 20 = 290°.
• Now, looking at the diagram, the 290° angle and z form a full angle (360°). So, z = 360 − 290 = 70°. Select (B) 70°.
Solution 2:
• Draw a diagram: Drawing a line can convert our weird shape into two triangles: • Now, if the one angle in the large triangle equals 30°, the other two angles total 150° (since the angles of a triangle add to 180°). So, the two angles in the small triangle sum to 150 − 20 − 20 = 110°. So, the angle labelled z must be 180 − 110 = 70°. Select (B) 70.
Note that it's also possible to solve the problem by transforming the figure in other ways than the one illustrated here.
• Try a special case: We are told that x is an integer. Assume that x is 3, 4, 5, ... and see what results:
• If x = 3, y = 7.5 (not an integer).
• If x = 4, y = 10.
• If x = 5, y = 12.5 (not an integer).
• If x = 6, y = 15.
• If x = 7, y = 17.5 (not an integer, and also outside the range of y, so we don't need to try any more numbers).
There were two integer values of x that resulted in integer values of y. Select (B) Two.
There are some shortcuts you could take to speed this up a bit (e.g. if you realize that x must be even and y must be a multiple of 5, you wouldn't have to check as many numbers).
5. Solution 1:
• Estimate the answer: First, darken a portion of the larger circle that appears to be the same length as the darkened arc. Next, estimate what portion of the larger circle that is. To me, it appeared to be 1/10 of the large circle. This is just a rough estimate, so if you came up with 1/9 or 1/11 or something similar, that's fine. Since the large circle has a circumference of 36, then our estimate would be 1/10 × 36 = 3.6.
• Look at the answer choices: (D) 4 appears pretty close. This represents 1/9 of the large circle. (E) 2 represents 1/18 of the large circle, which is just too little. (C) 6 represents 1/6 of the large circle, which is too much. Only (D) is in the range of our estimate. Select (D) 4.
Solution 2:
• Estimate the answer: See above.
• Work backwards:
• We would know the length of the darkened arc if we knew the circumference of the small circle (it would be 80/360 of the circumference of the small circle, since there are 360 degrees in a circle).
• We would know the circumference of the small circle if we knew the radius of the small circle.
• We would know the radius of the small circle if we knew the radius of the large circle, since the small one is half the radius of the large.
• We would know the radius of the large circle if we knew the circumference of the large circle.
• We are given the circumference of the large circle, so we can use that to solve the problem.
• Work forwards:
• The radius of the large circle must be 36/2π.
• The radius of the small circle must be 18/2π.
• The circumference of the small circle must be 18.
• The length of the darkened arc must be 80/360 × 18 = 4.
Select (D) 4.
• Estimate the answer: The line should have a negative slope and a y-intercept of 120.
• Look at the answer choices: (A), (B), and (C) do not have a negative slope. Eliminate those answers.
• Guess and check: At x = 10, y = 0. Check both remaining answers to see what happens when x = 10. For (D), y = 20 when x = 10, so that one's wrong. Select (E) y = 120 − 12x.
• If x is an integer, and x + 1/x is also an integer, than 1/x must be an integer. If 1/x is also an integer, then x must equal 1 or −1.
• If x = 1, then x + 1/x = 1 + 1 = 2.
If x = −1, then x + 1/x = −1 − 1 = −2.
• Look at the answer choices: 2 is not a possible choice, but −2 is. Select (D) −2.