SAT Math Practice Problems with Solutions and Explanations, Study Guides, Projects, Research of Mathematics

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Sample Section 1: Answers and Explanations

1. A

Use the Bowtie to find the value of each quantity. Quantity A =. Quantity B =.

Quantity A is greater than Quantity B.

2. C

When two lines intersect, the big angles and the small angles are equal, so s = v , r = u , and t = w. Therefore, the quantities are equal. To prove it, Plug In. Let u = 50, v = 100, and w = 30. Because the angles are all small angles, r = 100, s = 50, and t = 30. So, both r + t + v and s + u + w total 180.

3. D

First, find the value of x. Multiply both sides by 3 to get 2 x < 21. Divide both sides by 2 to find that x < 10. as well as being greater than 0. If x = 2, then Quantity B is greater because 2^2 = 4. Eliminate (A) and (C). If x = 10, then the two quantities are equal. Eliminate (B), and select (D) because different numbers have given different answers.

4. C

Solve for k: 9(3)^3 + 4 = 9(27) + 4 = 243 + 4 = 247. So, k = 247. Use trial and error to find the prime factors of 247. They are 13 and 19. The average of 13 and 19 is 16, so the quantities are equal.

5. C

First, divide $90 by 50 to find the bulk rate for a single pound of flour:. The amount saved per pound is

$.09. The quantities are equal.

6. A

Remember PEMDAS. x = ((4) + 2) × 5 = 6 × 5 = 30; y = (24 ÷ (8)) × 5 = 3 × 5 = 15. Quantity A = 15, and Quantity B = −15.

7. B

The solutions to the equation 32 = | k | are 32 and −32. Of those, only −32 works in the equation 31 = | k + 1|. Quantity A has the value of −32, so Quantity B is greater.

8. B

Plug In. First, try a = 100 and b = 2: Alice weighs 100 kilograms, Bob weighs 98 kilograms, and the sum of their weights is 198 kilograms. Quantity B is 2 × 100, or 200. Quantity B is greater, so eliminate (A) and (C). Now, try a second set of numbers: If a = 50 and b = 10, then Quantity A is 90 and Quantity B is 100. Quantity B is, again, greater—as it will be with any set of numbers that meets the restriction ( ab ≠ 0).

9. D

less than ( × × 1 = ). Therefore, (A) is correct.

17. D

Each number in the set occurs only once, and the mode is the most frequently occurring number. To create a

mode, x must be the same as one of the other numbers in the set. Eliminate (C) and (E). Plug In the

Answers. For (B), the mode is 5. The median of {1, 3, 5, 5, 7, 11} is 5. The median is equal to the mode.

You need a larger number for x. For (D), the mode is 7 and the median of {1, 3, 5, 7, 7, 11} is = 6.

The mode is now one less than the median. Note that (C) is equal to the median and would be a trap answer

if you didn’t read the question carefully enough.

18. B

Redraw this figure by adding two descending lines from the upper corners of the figure perpendicular to the

base below so that you have a rectangle and two right triangles. Each triangle has a base of 1 and a

hypotenuse of 2. Use the Pythagorean Theorem or the ratio for 30 : 60 : 90 triangles ( x : x : 2 x ) to find

the height is. The area of each triangle is (1)( ). The area of the rectangle is (2)( ). The sum of the

rectangle and two triangles is 2 + 2 = 2 + = 3.

19. A

If Mary’s wage is $2 more than Mark’s, and Andy’s wage is $6 more than Mark’s, then Andy’s wage is $ more than Mary’s. Anne’s wage is $10 more than Mary’s, so Anne’s wage is $6 more than Andy’s. Quantity A is greater than Quantity B.

20. D

Plug In, and let t = 5. M = 5 + 7 + 9 = 21—the sum of the three consecutive odd integers, of which t is the smallest. For the second part of the problem, t is the greatest integer in the series. The sum of the new series is 5 + 3 + 1 = 9. Plug M = 21 into the answers, and (D) matches the target of 9.

21. A

The bigger circle that includes the garden and the sidewalk has an area of 169p. The radius of the large circle with the garden and sidewalk is 13. The garden plot has an area 144p, so the radius of the small circle is 12. Since the circles share a center, the width of the sidewalk, T , is the difference between the radii, or 13 − 12 = 1. Therefore, (A) is correct.

22. A

Factor the quadratic expression: a^2 – b^2 becomes ( a + b )( a – b ). The average of two numbers is 6, so their

sum, ( a + b ), is 2 × 6 = 12. Substitute the value to get: 12( a – b ) = 2. Thus, a – b =.

23. B

Plug in numbers that have a sum of 19, and check to see if their product is 88. For example, the sum of 9 and 10 is 19, but their product is 90. Therefore, 9 and 10 can’t be the two numbers. Try another pair, such as 8 and 11. Their sum is 19, and their product is 88. These numbers match the information given in the problem, and their difference is 3.

24. E

Plug In. If l = 2, then. The target is. Plug 2 for l into the answer

choices. Choice (E) is the only answer choice that matches the target.

25. D , E , and F

First, figure out Pierre’s per-glass profit. His per-glass revenue is $2; his per-glass costs are 25 cents × 5 lemons per glass = 125 cents per glass for lemons, and 5 cents × 3 tablespoons of sugar per glass = 15 cents per glass for sugar, for a total of 140 cents per glass. Pierre’s profit on each glass is 60 cents. At a profit of 60 cents per glass, Pierre must sell at least 9 glasses of lemonade to make 5 dollars. Each answer choice greater than or equal to 9 is correct.

26. A

Plug In the Answers, starting with (C). If the combined total was 220 and A is 150 more than B , then A is 75 more than half of 220, and B is 75 less than half of 220. So, A is 185 and B is 35. Remove 20 from each to get A is 165 and B is 15. Is 165 four times 15? No, so (C) is not correct. It’s hard to tell whether (C) was too large or too small, so just pick a direction. For (A), if the combined total was 290, and A is 150 more than B , then A is 75 more than half of 290, and B is 75 less than half of 290. So, A is 220 and B is 70. Remove 20 from each to get A is 200 and B is 50. Is 200 four times 50? Yes, so (A) is correct.