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GRE Subject Test: Math : Exponential Growth Applications

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Example Question #1 : Exponential Growth Applications

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

Possible Answers:

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

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

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

$\frac{2}{3}$

$\frac{3}{2}$

Correct answer:

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

Explanation:

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

P(t)= P_ie^k^t

where P_i is the initial population.

At the start, there were 30 blood cells.

P_i=P(0)=30

Substitute this value into the given formula.

P(t)= 30e^k^t

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

P(2)=45

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

45= 30e^k^(^2^)

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

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

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

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

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Example Question #2 : Exponential Growth Applications

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

Possible Answers:

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

$e^{30}$

$\ln(10)$

None of these

Correct answer:

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

Explanation:

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

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

Now that we solved for , we can plug in what we know for time t=30 and solve for .

$1000=100e^{30k}$

$e^{30k}=10$

$30k=\ln(10)$

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

Report an Error

Example Question #3 : Exponential Growth Applications

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

Possible Answers:

$\frac{4}{7}$

$\ln(4)$

$\ln(7)$

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

None of these

Correct answer:

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

Explanation:

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

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

Now that we have solved for we can solve for at t=7

$200=50e^{7k}$

$e^{7k}=4$

$7k=\ln(4)$

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

Report an Error

Example Question #4 : Exponential Growth Applications

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:

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

$\ln(8)$

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

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

Correct answer:

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

Explanation:

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

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

Now that we have we can solve for at t=6 .

$1600=200e^{6k}$

$e^{6k}=8$

$6k=\ln(8)$

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

Report an Error

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

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:

-0.036

-0.044

-0.009

-0.011

0.130

Correct answer:

-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:

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

Where p_0 is an initial population value, and is the constant of proportionality.

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

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

$0.93=e^{2k}$

2k=ln(0.93)

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

Report an Error

Example Question #6 : 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

1.04

0.12

0.05

0.07

Correct answer:

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:

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

Where p_0 is an initial population value, and 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:

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

$1.23=e^{4k}$

4k=ln(1.23)

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

Report an Error

Example Question #7 : Exponential Growth Applications

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

0.102

0.012

0.005

0.008

Correct answer:

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:

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

Where p_0 is an initial population value, and 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:

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

Dealing in minutes:

$2.50=e^{75k}$

75k=ln(2.50)

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

Report an Error

Example Question #8 : 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:

0.02961

0.31056

0.01997

0.00341

Correct answer:

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:

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

Where p_0 is an initial population value, and is the constant of proportionality.

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

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

$1.15=e^{7k}$

7k=ln(1.15)

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

Report an Error

Example Question #9 : 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:

-0.891

-2.526

-0.012

-0.191

-0.034

Correct answer:

-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:

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

Where p_0 is an initial population value, and is the constant of proportionality.

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

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

$0.88=e^{11k}$

11k=ln(0.88)

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

Report an Error

Example Question #10 : 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:

-1.755

0.030

0.133

0.289

0.027

Correct answer:

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:

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

Where p_0 is an initial population value, and 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:

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

$1.21=e^{7k}$

7k=ln(1.21)

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

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GRE Subject Test: Math : Exponential Growth Applications

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Exponential Growth Applications

Example Question #2 : Exponential Growth Applications

Example Question #3 : Exponential Growth Applications

Example Question #4 : Exponential Growth Applications

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

Example Question #6 : Exponential Growth Applications

Example Question #7 : Exponential Growth Applications

Example Question #8 : Exponential Growth Applications

Example Question #9 : Exponential Growth Applications

Example Question #10 : Exponential Growth Applications

All GRE Subject Test: Math Resources

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