ab = 12
b2 = 16
| Quantity A | Quantity B |
| a | b |
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Solution
Plug in for c and d, in both equations, and solve for a and b. If c = 8 and d = 4, then a = 1⁄2 and b = 3, your target answer. Now plug in 1⁄2 for a in the answer choices; only choice (A) hits your target answer of 3.
b is an integer, and 0 ≤ b ≤ 2
| Quantity A | Quantity B |
| \(\left ( b^{2}+\frac{1}{2} \right )^{b}\) | 20+1⁄4 |
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Solution
Compare the quantities, looking for ways that the numbers in one quantity can be rewritten using the numbers from the other quantity. In Quantity A, rewrite 420 as 4 × 105.The 105 values cancel out, and the remaining multiples of 2 also cancel except for one 2 left in Quantity B.Choice (B) is correct.
Which expression is equivalent to \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\)?
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Solution
Try plugging in values for x and y. If x = 1 and y = 3, then Quantity A is 27 and Quantity B is 22. Quantity A is greater, so eliminate choices (B) and (C).Then try x = 1 and y = 1; now both quantities are equal, so eliminate choice (A) and select choice (D).
| Quantity A | Quantity B |
| 0.82 + 0.82 + 0.82 + 0.82 1.62 | 1.62 |
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Solution
24
Start by expressing both terms in the original equation as powers of 3: 9x + 1 = 27x – 1 becomes (32)x + 1 =(33)x – 1.To raise a power to another power, multiply the exponents, so your equation becomes 32x + 2 = 33x – 3. Now that the bases are the same, set the exponents equal to each other and solve for x: 2x + 2 = 3x – 3, so x = 5. Finally, remember to enter the correct value:The problem asks for x2 – 1, so 52 – 1 = 25 – 1 = 24.
| Quantity A | Quantity B |
| 207 | \(\left ( \frac{4^{13}}{4^{6}} \right )(5^{4}\times 5^{3})\) |
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Solution
First, square both sides of the equation to get x = 16.Then, square both sides of the equation again to get x2 = 256.The answer is choice (E).
If m > 0 and n > 0, which of the following is equivalent to \(\frac{nm}{m^{2}}\sqrt{\frac{m^{2}}{n}}\)?
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Solution
You could use the common quadratic pattern (x + y)2 = x2 + 2xy + y2. So, (√5 + √7)2 = √52 + 2 \(\sqrt{5\times 7}+\sqrt{7}^{2}=5+2\sqrt{35}+7=12+2\sqrt{35}\) .The answer is choice (E). Alternatively, you can just FOIL it. So,\(\left ( \sqrt{5}+\sqrt{7} \right )^{2}=(\sqrt{5}+\sqrt{7})(\sqrt{5}+\sqrt{7})=5+\sqrt{5\times 7}+\sqrt{5\times 7}+7=12+2\sqrt{35}\).
0 > d > e
| Quantity A | Quantity B |
| de | \(\sqrt{de}\) |
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Solution
Plug in a value for x; you’re dealing with exponents, so keep your numbers small. If x = 0, then Quantity A is greater, so eliminate choices (B) and (C). If x = –1, though, then Quantity A = 0; Quantity B is now greater, so eliminate choice (A), and you’re left with choice (D).
If x ≥ 0, then \(\sqrt{0.49x^{16}}\) must be equal to
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Solution
Simplify Quantity A:\(\sqrt{\frac{1}{4^{2}}}=\frac{\sqrt{1}}{\sqrt{4^{2}}}=\frac{1}{4}\).The answer is choice (C).
0.1 is how many times greater than (0.01)3?
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Solution
Although it may be tempting to try some fancy factoring, this problem is more easily solved by calculating the individual exponential expressions:\(\frac{9^{2}-3^{2}}{6^{2}}=\frac{81-9}{36}=\frac{72}{36}=2\).Be careful if you got choice (B):You may have incorrectly subtracted 92 – 32 in the numerator and gotten 62.
(\(\sqrt{10y}\)) = 5
z4 = 81
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Solution
First, use the exponent rules to find the values you need to add: \(\left ( \frac{1}{2} \right )^{3}+\left ( \frac{2}{3} \right )^{2}+\left ( \frac{1}{6} \right )^{1}=\frac{1^{3}}{2^{3}}+\frac{2^{2}}{3^{2}}+\frac{1^{1}}{6^{1}}=\frac{1}{8}+\frac{4}{9}+\frac{1}{6}\).Then, because you have so many types of fractions, convert them all to the common denominator of 72:\(\frac{9}{72}+\frac{32}{72}+\frac{12}{72}=\frac{53}{72}\).The answer is choice (D).