Plug In The Answers starting with choice (C).Choice (C) yields \(\frac{1}{8}+\frac{2}{4}=\frac{1}{8}+\frac{4}{8}=\frac{5}{8}\), so it’s the correct answer.

Try plugging in values for a and c. Let a = 6 and c = 2.Then b=\(\frac{4(6)}{2}\)=3. If a is halved, equal it will 3. If c is doubled, it will equal 4. So now, b= \(\frac{4(3)}{4}\)=3.Because the question is looking for a percentage decrease, apply the percentage change formula:\(\frac{12-3}{12}\times 100=\frac{9}{12}\times 100=\frac{3}{4}\times 100=75%\).

The problem has variables in the answer choices so Plug In.Because the problem says, “in terms of a,” a is the variable that you want to plug in for. If a = 8, then b = 2, and that is now your target. When you plug in your value for a into choice (B), you get \(\frac{8}{3}-\frac{2}{3}=\frac{6}{3}=2\); thus, the correct answer.

Start by combining the 2 given equations: If a = 4 and 6 < b < 8, then the acceptable range for ^a⁄_b is ⁴⁄₆ > ^a⁄_b > ⁴⁄₈, which can be reduced to ²⁄₃ > ^a⁄_b > ¹⁄₂ or, in decimal form, 0.5 < ^a⁄_b < 0.67. Select choice (D) because ^a⁄_b can be either larger or smaller than 0.6

The given inequality is equivalent to 3x – 27 < 11x – 2, which becomes –8x < 25. Dividing both sides by –8 (and flipping the inequality sign), you get x > -3(¹⁄₈).Therefore, any number that is greater than –3-3(¹⁄₈) will satisfy the inequality, so the greatest integer that does not satisfy it is x = –4, choice (A).

20

Since Lyle bought an equal number of CDs at each price, combine the prices: 1 CD of each type would cost $8 + $12 = $20. For a total of $200, then, Lyle bought \(\frac{\$ 200}{\$ 20}\) = 10 CDs of each type, for a total of 20 CDs altogether.

Solve this “must be” problem by plugging in values for a and b. Starting with the simplest allowable values, a = 30 and b = 15, does not eliminate any answer choices. Next, try a = 100 and b = 0; now choices (A), (B), and (D) can be eliminated. Finally, try a = 30 and b = –30; now choice (E) can be eliminated, leaving only choice (C), which is the correct answer.

You know that b is greater than a, so (a – b) will always be negative, and Quantity A will always be greater. Alternatively, you can solve this one by plugging in values for a and b.Try making a= and b= :The value in Quantity B is now \(2\left ( \frac{1}{4}-\frac{1}{2} \right )=2\left ( -\frac{1}{4} \right )=-\frac{1}{2}\). Quantity A is greater, so eliminate choices (B) and (C).

You are told that \(\frac{\frac{1}{x}-1}{\frac{1}{x}}\), which you can manipulate to \((\frac{1}{x}-1)(\frac{x}{1})=\frac{x}{x}-\frac{x}{1}=1-x\), so the answer is choice (D). Alternatively, you can solve this “must be” problem by plugging in values for x: If x = 2, then \(\frac{\frac{1}{x}-1}{\frac{1}{x}}=\frac{\frac{1}{2}-1}{\frac{1}{2}}=\frac{-\frac{1}{2}}{\frac{1}{2}}=-\frac{1}{2}\times \frac{2}{1}=-1\).Eliminate choices (A), (B), and (E), because none of them hit your target of –1. Next, try x = 3: Now \(\frac{\frac{1}{x}-1}{\frac{1}{x}}=\frac{\frac{1}{2}-1}{\frac{1}{2}}=\frac{-\frac{1}{2}}{\frac{1}{2}}=-\frac{1}{2}\times \frac{2}{1}=-1\); eliminate choice (C). Choice (D) gives the target answer of –2, so it is the correct answer.

To solve this single-variable equation, you’ll just need to isolate the variable. First, add 24 to both sides to yield 7a + 32 = 8a.Then subtract 7a from both sides to yield 32 = a. Quantity A is greater.