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GRE Subject Test: Math : Algebra

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2 Diagnostic Tests 148 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Constant Of Proportionality

Suppose a blood cell increases proportionally to the present amount. If there were $\displaystyle 30$ blood cells to begin with, and $\displaystyle 45$ blood cells are present after $\displaystyle 2$ hours, what is the growth constant?

Possible Answers:

$\frac{1}{2}ln\left(\frac{3}{2}\right)$ $\displaystyle \frac{1}{2}ln\left(\frac{3}{2}\right)$

$ln\left(\frac{2}{3}\right)$ $\displaystyle ln\left(\frac{2}{3}\right)$

$ln\left(\frac{3}{2}\right)$ $\displaystyle ln\left(\frac{3}{2}\right)$

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

Correct answer:

$\frac{1}{2}ln\left(\frac{3}{2}\right)$ $\displaystyle \frac{1}{2}ln\left(\frac{3}{2}\right)$

Explanation:

The population size P(t) $\displaystyle P(t)$ after some time $\displaystyle t$ is given by:

$\displaystyle P(t)= P_ie^k^t$

where P_i $\displaystyle P_i$ is the initial population.

At the start, there were 30 blood cells.

$\displaystyle P_i=P(0)=30$

Substitute this value into the given formula.

$\displaystyle P(t)= 30e^k^t$

After 2 hours, 45 blood cells were present. Write this in mathematical form.

P(2)=45 $\displaystyle P(2)=45$

Substitute this into $\displaystyle P(t)= 30e^k^t$, and solve for $\displaystyle k$.

$\displaystyle 45= 30e^k^(^2^)$

$\frac{3}{2}= e^2^k$ $\displaystyle \frac{3}{2}= e^2^k$

$ln\left(\frac{3}{2}\right)=ln(e^2^k)$ $\displaystyle ln\left(\frac{3}{2}\right)=ln(e^2^k)$

$2k=ln\left(\frac{3}{2}\right)$ $\displaystyle 2k=ln\left(\frac{3}{2}\right)$

$k=\frac{1}{2}ln\left(\frac{3}{2}\right)$ $\displaystyle k=\frac{1}{2}ln\left(\frac{3}{2}\right)$

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Example Question #1 : Exponential Growth Applications

Suppose a population of bacteria increases from 100 $\displaystyle 100$ to 1000 $\displaystyle 1000$ in $30\: min$ $\displaystyle 30\: min$. What is the constant of growth?

Possible Answers:

$\frac{\ln(10)}{30}$ $\displaystyle \frac{\ln(10)}{30}$

$e^{30}$ $\displaystyle e^{30}$

$\displaystyle 10$

$\ln(10)$ $\displaystyle \ln(10)$

None of these

Correct answer:

$\frac{\ln(10)}{30}$ $\displaystyle \frac{\ln(10)}{30}$

Explanation:

The equation for population growth is given by $P=Ae^{kt}$ $\displaystyle P=Ae^{kt}$. $\displaystyle P$ is the population, $\displaystyle A$ is the intial value, $\displaystyle t$ is time, and $\displaystyle k$ is the growth constant. We can plug in the values we know at time t=0 $\displaystyle t=0$ and solve for $\displaystyle A$ .

$100=Ae^{0k}=A$ $\displaystyle 100=Ae^{0k}=A$

Now that we solved for $\displaystyle A$, we can plug in what we know for time t=30 $\displaystyle t=30$ and solve for $\displaystyle k$.

$1000=100e^{30k}$ $\displaystyle 1000=100e^{30k}$

$e^{30k}=10$ $\displaystyle e^{30k}=10$

$30k=\ln(10)$ $\displaystyle 30k=\ln(10)$

$k=\frac{\ln(10)}{30}$ $\displaystyle k=\frac{\ln(10)}{30}$

Report an Error

Example Question #1 : How To Find Constant Of Proportionality Of Rate

A population of deer grew from 50 to 200 in 7 years. What is the growth constant for this population?

Possible Answers:

$\frac{\ln(4)}{7}$ $\displaystyle \frac{\ln(4)}{7}$

$\ln(7)$ $\displaystyle \ln(7)$

$\frac{4}{7}$ $\displaystyle \frac{4}{7}$

None of these

$\ln(4)$ $\displaystyle \ln(4)$

Correct answer:

$\frac{\ln(4)}{7}$ $\displaystyle \frac{\ln(4)}{7}$

Explanation:

The equation for population growth is given by $P=Ae^{kt}$ $\displaystyle P=Ae^{kt}$. P is the population, $\displaystyle A$ is the intial value, $\displaystyle t$ is time, and $\displaystyle k$ is the growth constant. We can plug in the values we know at time t=0 $\displaystyle t=0$ and solve for $\displaystyle A$ .

$50=Ae^{0k}=A$ $\displaystyle 50=Ae^{0k}=A$

Now that we have solved for $\displaystyle A$ we can solve for $\displaystyle k$ at t=7 $\displaystyle t=7$

$200=50e^{7k}$ $\displaystyle 200=50e^{7k}$

$e^{7k}=4$ $\displaystyle e^{7k}=4$

$7k=\ln(4)$ $\displaystyle 7k=\ln(4)$

$k=\frac{\ln(4)}{7}$ $\displaystyle k=\frac{\ln(4)}{7}$

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Example Question #22 : Exponential Functions

A population of mice has 200 mice. After 6 weeks, there are 1600 mice in the population. What is the constant of growth?

Possible Answers:

$\frac{\ln(6)}{8}$ $\displaystyle \frac{\ln(6)}{8}$

$\ln(8)$ $\displaystyle \ln(8)$

$\frac{\ln(6)}{6}$ $\displaystyle \frac{\ln(6)}{6}$

$\frac{\ln(8)}{6}$ $\displaystyle \frac{\ln(8)}{6}$

Correct answer:

$\frac{\ln(8)}{6}$ $\displaystyle \frac{\ln(8)}{6}$

Explanation:

$200=Ae^{0k}=A$ $\displaystyle 200=Ae^{0k}=A$

Now that we have $\displaystyle A$ we can solve for $\displaystyle k$ at t=6 $\displaystyle t=6$.

$1600=200e^{6k}$ $\displaystyle 1600=200e^{6k}$

$e^{6k}=8$ $\displaystyle e^{6k}=8$

$6k=\ln(8)$ $\displaystyle 6k=\ln(8)$

$k=\frac{\ln(8)}{6}$ $\displaystyle k=\frac{\ln(8)}{6}$

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Example Question #1 : Exponential Growth Applications

The rate of decrease of the dwindling wolf population of Zion National Park is proportional to the population. The population decreased by 7 percent between 2009 and 2011. What is the constant of proportionality?

Possible Answers:

-0.009 $\displaystyle -0.009$

-0.036 $\displaystyle -0.036$

-0.044 $\displaystyle -0.044$

0.130 $\displaystyle 0.130$

-0.011 $\displaystyle -0.011$

Correct answer:

-0.036 $\displaystyle -0.036$

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population decreased by 7 percent between 2009 and 2011, we can solve for this constant of proportionality:

$(1-0.07)y_0=y_0e^{k(2011-2009)}$ $\displaystyle (1-0.07)y_0=y_0e^{k(2011-2009)}$

$0.93=e^{2k}$ $\displaystyle 0.93=e^{2k}$

$\displaystyle 2k=ln(0.93)$

$k=\frac{ln(0.93)}{2}=-0.036$ $\displaystyle k=\frac{ln(0.93)}{2}=-0.036$

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Example Question #2 : Exponential Growth Applications

The rate of growth of the Martian Transgalactic Constituency is proportional to the population. The population increased by 23 percent between 2530 and 2534 AD. What is the constant of proportionality?

Possible Answers:

-0.37 $\displaystyle -0.37$

1.04 $\displaystyle 1.04$

0.07 $\displaystyle 0.07$

0.05 $\displaystyle 0.05$

0.12 $\displaystyle 0.12$

Correct answer:

0.05 $\displaystyle 0.05$

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased by 23 percent between 2530 and 2534 AD, we can solve for this constant of proportionality:

$(1+0.23)y_0=y_0e^{k(2534-2530)}$ $\displaystyle (1+0.23)y_0=y_0e^{k(2534-2530)}$

$1.23=e^{4k}$ $\displaystyle 1.23=e^{4k}$

$\displaystyle 4k=ln(1.23)$

$k=\frac{ln(1.23)}{4}=0.05$ $\displaystyle k=\frac{ln(1.23)}{4}=0.05$

Report an Error

Example Question #2701 : Functions

The rate of growth of the bacteria in an agar dish is proportional to the population. The population increased by 150 percent between 1:15 and 2:30. What is the constant of proportionality?

Possible Answers:

0.039 $\displaystyle 0.039$

0.102 $\displaystyle 0.102$

0.008 $\displaystyle 0.008$

0.012 $\displaystyle 0.012$

0.005 $\displaystyle 0.005$

Correct answer:

0.012 $\displaystyle 0.012$

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased by 150 percent between 1:15 and 2:30, we can solve for this constant of proportionality:

$(1+1.50)y_0=y_0e^{k(2:30 - 1:15)}$ $\displaystyle (1+1.50)y_0=y_0e^{k(2:30 - 1:15)}$

Dealing in minutes:

$2.50=e^{75k}$ $\displaystyle 2.50=e^{75k}$

$\displaystyle 75k=ln(2.50)$

$k=\frac{ln(2.50)}{75}=0.012$ $\displaystyle k=\frac{ln(2.50)}{75}=0.012$

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Example Question #2 : Exponential Growth Applications

The rate of growth of the duck population in Wingfield is proportional to the population. The population increased by 15 percent between 2001 and 2008. What is the constant of proportionality?

Possible Answers:

0.31056 $\displaystyle 0.31056$

0.01997 $\displaystyle 0.01997$

0.00341 $\displaystyle 0.00341$

0.02961 $\displaystyle 0.02961$

$\displaystyle .000287$

Correct answer:

0.01997 $\displaystyle 0.01997$

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased by 15 percent between 2001 and 2008, we can solve for this constant of proportionality:

$(1+.15)y_0=y_0e^{k(2008-2001)}$ $\displaystyle (1+.15)y_0=y_0e^{k(2008-2001)}$

$1.15=e^{7k}$ $\displaystyle 1.15=e^{7k}$

$\displaystyle 7k=ln(1.15)$

$k=\frac{ln(1.15)}{7}=0.01997$ $\displaystyle k=\frac{ln(1.15)}{7}=0.01997$

Report an Error

Example Question #3 : Exponential Growth Applications

The rate of decrease of the panda population is proportional to the population. The population decreased by 12 percent between 1990 and 2001. What is the constant of proportionality?

Possible Answers:

-0.012 $\displaystyle -0.012$

-2.526 $\displaystyle -2.526$

-0.034 $\displaystyle -0.034$

-0.191 $\displaystyle -0.191$

-0.891 $\displaystyle -0.891$

Correct answer:

-0.012 $\displaystyle -0.012$

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population decreased by 12 percent between 1990 and 2001, we can solve for this constant of proportionality:

$(1-0.12)y_0=y_0e^{k(2001-1990)}$ $\displaystyle (1-0.12)y_0=y_0e^{k(2001-1990)}$

$0.88=e^{11k}$ $\displaystyle 0.88=e^{11k}$

$\displaystyle 11k=ln(0.88)$

$k=\frac{ln(0.88)}{11}=-0.012$ $\displaystyle k=\frac{ln(0.88)}{11}=-0.012$

Report an Error

Example Question #4 : Exponential Growth Applications

The rate of growth of the salmon population of Yuba is proportional to the population. The population increased by 21 percent over the course of seven years. What is the constant of proportionality?

Possible Answers:

0.030 $\displaystyle 0.030$

0.289 $\displaystyle 0.289$

0.133 $\displaystyle 0.133$

-1.755 $\displaystyle -1.755$

0.027 $\displaystyle 0.027$

Correct answer:

0.027 $\displaystyle 0.027$

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased by 21 percent over the course of seven years, we can solve for this constant of proportionality:

$(1+0.21)y_0=y_0e^{7k}$ $\displaystyle (1+0.21)y_0=y_0e^{7k}$

$1.21=e^{7k}$ $\displaystyle 1.21=e^{7k}$

$\displaystyle 7k=ln(1.21)$

$k=\frac{ln(1.21)}{7}=0.027$ $\displaystyle k=\frac{ln(1.21)}{7}=0.027$

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GRE Subject Test: Math : Algebra

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Constant Of Proportionality

Example Question #1 : Exponential Growth Applications

Example Question #1 : How To Find Constant Of Proportionality Of Rate

Example Question #22 : Exponential Functions

Example Question #1 : Exponential Growth Applications

Example Question #2 : Exponential Growth Applications

Example Question #2701 : Functions

Example Question #2 : Exponential Growth Applications

Example Question #3 : Exponential Growth Applications

Example Question #4 : Exponential Growth Applications

All GRE Subject Test: Math Resources

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