STUDY GUIDE
 
FLASHCARDS

Number and Quantity Practice Questions

  1. Simplify the following expression:
\((5\strut^{\tfrac{1}{3}}\cdot 5^2)^6\)
  1. \(5\strut^{\tfrac{2}{3}}\)
  2. \(54\)
  3. \(5\strut^{(8\tfrac{1}{3})}\)
  4. \(5^{14}\)
Show Answer
The correct answer is D!

The exponent on a base is the sum of the powers in a product (if the bases are the same), and the resulting power when an exponent is raised to a power is the product of the two exponents.

Begin by adding the exponents within the parentheses. Next, multiply that sum and the exponent outside the parentheses.

\((5\strut^{\tfrac{1}{3}} \cdot 5^2)^6=(5\strut^{\tfrac{7}{3}})^6=5^{14}\)

 

  1. Which of the following is false?
  1. \((8\strut^{\tfrac{1}{2}})^2=2^3\)
  2. \(4 \cdot 4 \cdot4 \cdot4 \sqrt{4}=4\strut^{\tfrac{9}{2}}\)
  3. \(5\strut^{\tfrac{1}{2}} \cdot 5\strut^{\tfrac{2}{3}} \cdot 5\strut^{\tfrac{1}{2}} = 5\sqrt{5}\)
  4. \(\tfrac{6\sqrt{5}}{6\sqrt{5^2}}= 5\strut^{-\tfrac{1}{2}}\)
Show Answer
The correct answer is C!

Each of the statements follows an exponent rule. In Choice A, the exponents inside and outside the parentheses are multiplied. In Choice B, the base is the same for each term in the product, so the exponents can all be added. In Choice D, the base is the same for each term in the quotient, so the exponents can be subtracted.

For Choice C, we multiply three powers of 5. Because the base is the same, we add the exponents, which gives us a product of \(5\strut^{\tfrac{5}{3}}\). Meanwhile, \(5\sqrt{5}=5^1\cdot 5\strut^{\tfrac{1}{2}}=5\strut^{\tfrac{3}{2}}\). Because \(\tfrac{5}{3}\neq \tfrac{3}{2}\), the two sides are not equal. This is the false statement.

 

  1. Rewrite the expression below using fractional exponents:
\(5\sqrt[{\LARGE 3}]{2^2} + \sqrt[{\LARGE 7}]{3^2}\)
  1. \(5\cdot2\strut^{\tfrac{2}{3}}+3\)
  2. \(5\cdot2\strut^{\tfrac{2}{3}}+3\strut^{\tfrac{2}{7}}\)
  3. \(5^3 \cdot 2\tfrac{1}{2}+7\cdot3\tfrac{1}{2}\)
  4. \(10\strut^{\tfrac{2}{3}}+3\strut^{\tfrac{2}{7}}\)
Show Answer
The correct answer is B!

Fractional exponential notation can be obtained with the following identity:

\(\sqrt[{\LARGE y}]{b^{\large x}}=b\strut^{\tfrac{x}{y}}\quad\Rightarrow\quad 5\sqrt[{\LARGE 3}]{2^2} + \sqrt[{\LARGE 7} ]{3^2}=5\cdot2\strut^{\tfrac{2}{3}} + 3\strut^{\tfrac{2}{7}}\)

This cannot be simplified further by other rules governing exponentials because the bases are not the same.

 

  1. Simplify the following expression:
\(\dfrac{5^8 \cdot \sqrt[{\LARGE 3}]{5^{{\large 2}}} \cdot 5^{-3}}{\sqrt[{\LARGE 3}]{5^{{\large 5}}}}\)
  1. \(5^5 \cdot 5\strut^{\tfrac{7}{3}}\)
  2. \(5^6\)
  3. \(5^4\)
  4. \(5\strut^{\tfrac{2}{3}}\)
Show Answer
The correct answer is C!

First, write the radicals as fractional exponents on the same base. Next, add the exponents in the numerator to write the product of those three terms. Finally, subtract the exponent in the denominator and simplify:

\(\dfrac{5^{8} \cdot \sqrt[{\LARGE 3}]{5^{\large 2}} \cdot 5^{-3}}{\sqrt[{\LARGE 3}]{5^{\large 5}}}\) \(=\dfrac{5^{8} \cdot 5^{\tfrac{2}{3}} \cdot 5^{-3}}{5\strut^{\tfrac{5}{3}}}\) \(=\dfrac{5^{8+\tfrac{2}{3}-3}}{5\strut^{\tfrac{5}{3}}}\)

\(=5^{\tfrac{17}{3}-\tfrac{5}{3}}\) \(=5^{\tfrac{12}{3}}\) \(=5^4\)

 

  1. Which of the following is a rational number?
  1. \(\sqrt{121}+7\)
  2. \(4+\sqrt[{\LARGE 3}]{150}\)
  3. \(\sqrt[{\LARGE 7}]{30}\)
  4. \(8\strut^{\tfrac{2}{3}}+8\strut^{\tfrac{1}{5}}\)
Show Answer
The correct answer is A!

Because \(\sqrt{121}=11\), the sum is \(11+7=18\), which is rational.

 

  1. Derek is purchasing lunch meat for the class picnic. There are 300 people attending the picnic. Derek expects that 11 sandwiches will be eaten for every 10 people and that 20 pounds of meat will make 90 sandwiches. How much lunch meat should Derek buy?
  1. \(73\frac{1}{3}\) \(\text{pounds}\)
  2. \(67\frac{2}{3}\) \(\text{pounds}\)
  3. \(110\) \(\text{pounds}\)
  4. \(330\) \(\text{sandwiches}\)
Show Answer
The correct answer is A!

First, figure out how many sandwiches are needed. Then, convert that to pounds of meat.

For every 10 people, 11 sandwiches are eaten:

\(\text{sandwiches} = 300 \times \frac{11}{10} = 330\)

Twenty pounds makes 90 sandwiches, so set a proportion:

\(\frac{20\ \text{lb}}{90\ \text{sandwiches}} = \frac{m\ \text{lb}}{330\ \text{sandwiches}}\)

\(m = 330 \times \frac{20}{90} = 330 \times \frac{2}{9} = \frac{660}{9} = \frac{220}{3} = 73\frac{1}{3}\ \text{lb}\)

 

  1. Choose the graph that applies to the problem below:
Meghan bought cookies and milk for her classmates. She spent $0.45 per cookie, one cookie each per student, and $2.65 on one gallon of milk to share. How much does she spend altogether?
Graph 1

Graph 2

Graph 3

Graph 4

  1. Graph 1
  2. Graph 2
  3. Graph 3
  4. Graph 4
Show Answer
The correct answer is D!

Since she buys one cookie for every student, the amount she spends depends on how many students there are. The cost is the dependent variable, dependent on the number of students, and the students are the independent variable.

Graphs 3 and 4 show this relationship with the independent variable on the \(x\)-axis and the dependent on the \(y\)-axis. The scale of the cost axis in Graph 3 is too large, though. For 30 students, Meghan only spends about $16. Since the scale of the cost axis in Graph 4 is more appropriate, Choice D is the best answer.

 

  1. A certain radioactive material has a half-life of seven days. A test sample of this material has an initial mass of 400 kg. Which of the following would be the most appropriate unit to use when labeling the time axis of a graph of remaining material vs. time elapsed until 6.1 grams remain?
  1. Hours
  2. Days
  3. Weeks
  4. Years
Show Answer
The correct answer is C!

The time period of one half-life is one week. It will take many half-lives for the mass to decay from 400 kilograms to 6.1 grams. At 10 weeks, for example, the mass is \(400\frac{1}{2}^{10}=0.3906 \text{ kg}=390.6 \text{ grams}\). Expressing time in units smaller than weeks will result in an unwieldy and unnecessarily finely-divided time axis.

On the other hand, labeling the time axis in years does not provide enough divisions to accurately resolve individual half-lives. Labeling the axis in weeks is therefore the best choice.

 

  1. Usain Bolt set a world record of 9.58 seconds for the 100-meter dash. White-tailed deer can run at 48 kilometers per hour for short periods. If Bolt and a white-tailed deer run a 100-meter dash at these speeds, who would win and by how much?
  1. Bolt wins by 3.75 seconds
  2. Bolt wins by 2.08 seconds
  3. Deer wins by 3.75 seconds
  4. Deer wins by 2.08 seconds
Show Answer
The correct answer is D!

First, convert the deer’s speed to the time for 100 m:

\(t=\dfrac{100\text{ m}}{48\ \text{km/h}}=\dfrac{100}{48,000}\text{ h}\times 3,600\text{ s/h}=7.5\text{ s}\)

Then, comparing the deer’s speed with Bolt’s, we can see that the deer’s time is faster. To see the difference, subtract the lower time from the higher time:

\(9.58-7.50=2.08 \text{ s}\)

 

  1. After getting his wisdom teeth pulled, Matt was prescribed 60 mg of codeine to be taken every four hours for five days. The pharmacy has 30 mg pills. How many pills should Matt receive when filling the prescription?
  1. 72
  2. 60
  3. 30
  4. 24
Show Answer
The correct answer is B!

First, we need to figure out how many doses Matt has to take. He takes a dose every 4 hours and there are 24 hours in a day, so that’s \(\frac{24}{4}=6\) doses per day. Over 5 days, he’ll need \(6\times5=30\) doses.

Next, we translate each dose into pills. Each dose is 60 mg, but each pill is 30 mg, so he needs \(\frac{60}{30}=2\) pills per dose.

Finally, multiply the number of doses by the pills per dose to get \(30\times2=60\).