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
- Practice Test 2: Sections
- Practice Test 3: Sections
- Practice Test 4: Sections
- Practice Test 5: Sections
- Practice Test 6: Sections
- Practice Test 7: Sections
- Practice Test 8: Sections
- Practice Test 9: Sections
- Practice Test 10: Sections
Here are solutions for section 7 of practice test #7 in The Official SAT Study Guide, second edition, found on pages 785–790. 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.
- 60 is divisible by 3, so 60 − 3 = 57 is divisible by 3. Select (C) 57.
- Try drawing lines through the figures and see if you can find one through which two different lines can be drawn. You will find that the answer is (D) X.
- Try a special case: Say that Bobby does three chores (n = 3). He will make $16 for the week.
- Look at the answer choices: See what each answer evaluates to when n = 3:
The only answer evaluating to the expected value (16) is (D). Select (D) 10 + 2n.
- (A) evaluates to 13
- (B) evaluates to 36
- (C) evaluates to 32
- (D) evaluates to 16
- (E) evaluates to 26
- Estimate the answer: It doesn't hurt to make a rough estimate of the area of figure B. It appears to definitely be less than the area of figure A, but it is more than half as large. Perhaps its area is somewhere around 16–18 sq cm or so.
- Look at the answer choices: We can eliminate answer (A), since it is less than half of the area of figure A. The other answers are too close to our estimate for us to be able to eliminate any.
- There are 13 little squares in figure A, so each little square has an area of 2 sq cm. There are 8 little squares in figure B, so its area must be 2 × 8 = 16. Select (C) 16 sq cm.
- Looking at the graph, the only person showing a significant increase (i.e. the shaded bar being much higher than the unshaded bar) is Goldberg. Select (B) Goldberg.
- The average of 6, 6, 12, and 16 is ¼(6 + 6 + 12 + 16) = 10. If x were 10, the average of the 5 numbers would still be 10. Select (D) 10.
- If the program selected 13, it would have printed 26, as 13 is odd.
- If the program selected 26, it would have printed 26, as 26 is even.
- If the program selected 52, it would have printed 52, as 52 is even. Since 26 could be printed only if 13 or 26 were selected, select (C) I and II only.
- Try a special case: Say that m = 3 and s = 2. There are 60 × 3 + 2 = 182 seconds in 3 minutes and 2 seconds.
- Look at the answer choices: See which answer evaluates to 182 when m = 3 and s = 2:
Select (A) m + 60s.
- (A) evaluates to 182
- (B) evaluates to 123
- (C) evaluates to 300
- (D) evaluates to 1/12
- (E) evaluates to 2 + 1/20
- If two expressions multiply to zero, either one expression must be zero, or the other must be zero. So, either 2x − 2 = 0 (Equation 1) or 2 − x = 0 (Equation 2). Solving equation 1, x = 1. Solving equation 2, x = 2. Select (D) 1 and 2 only.
- Take the cube root of both sides of the equation:
x = y³Select (C) y³
- Solution 1: A line segment with a slope of −1 would run from the upper left to the lower right, and would be at about a 45° angle to both the x- and y-axes. The only such line segment is DC. Select (E) DC.
- Solution 2:
- Estimate the answer: A line segment with a slope of −1 will run down as it runs to the right.
- Look at the answer choices: OC and OD run up as they run to the right. Their slopes are positive. Eliminate answers (C) and (D).
- Draw a diagram: On the diagram, write the values of O (0,0), D (1,3), C (3,1), B (3,−1), and A (1,−3).
- Guess and check: Find the slopes of lines OA, OB and DC (slope = (y2 − y1)/(x2 − x1). For DC, the slope is (1 − 3)/(3 − 1) = −1. Select (E) DC.
- Look at the answer choices: Look at each answer, and eliminate any answers where any number does not satisfy exactly one criterion:
Select (A) 14-20-13.
- In (B), 25 is both odd and a multiple of 5. Eliminate that answer.
- In (C), 15 is both odd and a multiple of 5. Eliminate that answer.
- In (D), 15 is both odd and a multiple of 5. Eliminate that answer.
- In (E), 34 is neither odd, a multiple of 5, nor Kyle's birth date. Eliminate that answer.
- If x > 3, then both sides of the equation are positive. We can then square both sides:
x + 9 = (x − 3)²Select (C) x = x² − 6x.
x + 9 = x² − 6x + 9
x = x² − 6x
- To find the number of integers between 1 and 100 that are not squares, we can find the number of squares and subtract that from 100. Now, 1² = 1 and 10² = 100, so there are 10 squares in total between 1 and 100. Subtracting 10 from 100, 100 − 10 - 90. Select (E) 90.
- Draw a diagram: It can help to draw the direct route from A to D, as well as the total distance travelled in each direction. See the diagram below, with these lines drawn in thicker:
- You can see that the direct distance is the hypotenuse of a right triangle with legs 15 and 20. So, by the Pythagorean theorem, the distance travelled by the direct route is √15² + 20² = 25. The distance from A to B to C to D is 16 + 15 + 4 = 35. So, she would save 35 − 25 = 10 miles by travelling there directly. Select (C) 10.
- The larger circle's area is π(1)² = π. The smaller circle's area is π(½)² = ¼π. The ratio of the area of the larger to the area of the smaller is 1:¼ = 4:1. Select (D) 4:1.
- This question is a lot easier if you realize that the sum of the integers from −22 to 22 is 0. Now, summing the next few integers after 22, we get 23 + 24 + 25 = 72. So, the sum of the integers between −22 and 25 is 72. Select (B) 25.
- The second graph is the same as the first graph of y = f(x), shifted 2 units down and 3 units to the left. So, if the first graph is the graph of y = f(x), the second graph would be the graph of y = f(x − 2) −3. So, h = −2 and k = −3, and hk = 6. Select (E) 6.