Quantity A | Quantity B |
\(\frac{15}{16}+\frac{1}{256}\) | \(1-\frac{1}{64}\) |
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Solution
In Quantity A,\(\frac{15}{16}+\frac{1}{256}=\frac{241}{256}\), and in Quantity B,1 - \(\frac{1}{64}=\frac{63}{64}\). If you multiply the numerator and denominator of \(\frac{63}{64}\) by 4, you obtain a common denominator:\(\frac{63}{64}=\frac{252}{256}\) .Clearly \(\frac{252}{256}>\frac{241}{256}\) , so Quantity B is greater.
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Solution
Use the Bowtie when adding or subtracting fractions:\(\frac{\frac{1}{5}-\frac{1}{2}}{\frac{1}{5}+\frac{1}{2}}=\frac{\frac{2-5}{10}}{\frac{2+5}{10}}=\frac{-\frac{3}{10}}{\frac{7}{10}}\).Next, divide the fractions by flipping the numerator and denominator and multiplying: \(\frac{-\frac{3}{10}}{\frac{7}{10}}=(-\frac{3}{10})\times (\frac{10}{7})=-\frac{3}{7}\).
3 ÷ 6⁄7 =
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Solution
When dividing by a fraction, flip the fraction and multiply:\(3+\frac{6}{7}=3\times \frac{7}{6}=\frac{21}{6}=3\frac{1}{2}\) . Alternatively, you may estimate and realize that 3 divided by something slightly smaller than 1 must be slightly larger than 3
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Solution
6
Break down the left side of the equation into prime factors to make it easier to simplify.You should get
\(\frac{(3^{2}\times 7^{2}\times 2^{2}\times 5^{2})\times (3^{3}\times 7^{3})\times 7}{(2^{4}\times 5^{4}\times 3^{4})\times 3}=7^{*}\)
Then group all the like terms:
\(\frac{2^{4}\times 3^{5}\times 5^{4}\times 7^{6}}{2^{4}\times 3^{5}\times 5^{4}}=7^{*}\)
Everything cancels out on the left side except for 76, which makes 6 your answer.
If x is divisible by 78, which of the following must be divisible by x?
Indicate all possible values.
A.\(\frac{x}{78}\) |
B.x |
C.x + 78 |
D.\(\frac{78}{x}\) |
E.78 − x |
F.78x |
G.78x + 78 |
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Solution
B and F
There are variables in the answer choices, so Plug In.Try x = 78.You can eliminate choices (A) and (D). Now try a weird number: 0. Eliminate choice (G).Try one more number: 156, which is double 78.This time you can eliminate choices (C) and (E). A number is divisible by itself and a multiple of a number is divisible by that number, so the correct answers are choices (B) and (F).
If \(\frac{2}{3^{*}} < 0.02 \) , what is the least integer value of x?
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Solution
First, rewrite 0.02 as a fraction,\(\frac{2}{100}\). For \(\frac{2}{3^{*}}\) to be less than \(\frac{2}{100}\), 3x must be greater than 100. Plugging In the Answers is the easiest way to get this right.Choice (E) is 34 = 81 and the fraction is greater than 0.02; eliminate it.Choice (D) is 35 = 243 and this makes the fraction less than 0.02.Therefore, the least value for x is 5.Be sure to answer what is asked.The inequality would be true if the denominator of \(\frac{2}{3^{*}}\) were 101, which is choice (B); however, the question is asking for the least value of x, not of 3x, so the correct answer is choice (D).
If p and x are non-negative numbers and y is a non-positive number, then which of the following must be true?
Indicate all possible values.
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Solution
D and F
As soon as you see variables in the answer choices, set up your scratch paper to Plug In. Start with easy numbers like p = 2, x = 3, and y = −4. Of the answer choices, all work except choice (C), which can be eliminated. Now plug in different numbers: Since the variables are described as non-negative and non-positive, try making p, x, and y all 0. Now choices (A), (B), and (D) all yield false statements and can be eliminated.The correct answers are choices (D) and (F).
If |x2y| = |(–w)z|, then which of the following could be true?
Indicate all possible values.
A.x2 = wz |
B.|x2y| = |(–w)z| |
C.−(wz) = wz |
D.|–x2| = |(–w)z| |
E.−w = z |
F.x2 = –|y| |
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Solution
A,C, D, and E
The best way to approach a could be question is to consider many different kinds of numbers to plug in that could work in the problem. We will have to plug in a few times here, so let’s start with easy numbers. For instance, let’s try making every variable in the problem equal to 1. Immediately, choices (A) and (D) work. If we made w = 1 and z = −1, then choice (E) works as well.Try plugging in 0 for either w or z and choice (C) can also work. In the end, choices (B) and (F) are always going to have a positive value on the left side of the equation and a negative value on the right, and therefore will not be correct. An absolute value is always positive, so it can never equal something negative.
Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8. In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y?
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Solution
You should Plug In on this question. Variables a and b can be consecutive numbers such as 6 and 12. Plug in consecutive multiples of 8 for x and y, such as 8 and 16. Finding the ratio of the averages is easy now; the averages of each pair of numbers will be the halfway point between the two numbers. For 6 and 12, the average is 9. For 8 and 16, the average is 12.The ratio of these averages is 3⁄4. Now, plug your original numbers into the answer choices and look for the one that equals 3⁄4.Choice (A) gives you , which is the reciprocal of what you want.Cross it off.Choice (B) gives you 3⁄4 which is what you’re looking for.But don’t stop yet—keep checking all of the answers.Choice (D) also works with these numbers. No harm done, just try a different set of numbers and check the two remaining choices. Plug in some really unusual numbers, like 24 and 30 for a and b, and 80 and 88 for x and y. Now you are looking for an answer to give you \(\frac{9}{28}\).Choice (B) no longer works, so choice (D) is your answer.
If w is a non-positive integer, which of the following must be positive?
Indicate all possible values.
A.−3w |
B.2w + 10 |
C.w4 |
D.w0 |
E.−w + 0.5 |
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Solution
D and E
Remember to Plug In multiple times for must be questions. First, use an easy number, such as −1, and try it in each choice:
All of the answers are positive, so don’t eliminate anything.Can we eliminate anything by making w smaller? A number such as −10 will allow us to eliminate choice (B), but everything else is still positive. But can w = 0? Non-positive just means the number can’t be positive—it doesn’t mean it can’t be zero.