Exponential Growth Applications

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GRE Quantitative Reasoning › Exponential Growth Applications

Questions 1 - 10

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?

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 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:

$2.50=e^{75k}$

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

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?

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 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:

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

$1.23=e^{4k}$

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

The rate of decrease of the gluten-eating demographic of the US is proportional to the population. The population decreased by 8 percent between 2014 and 2015. What is the constant of proportionality?

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

Since the population decreased by 8 percent between 2014 and 2015, we can solve for this constant of proportionality:

$(1-0.08)y_0=y_0e^{k(2015-2014)}$

$0.92=e^{k}$

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?

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 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:

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

$1.21=e^{7k}$

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

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?

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 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:

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

$1.15=e^{7k}$

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

Suppose a population of bacteria increases from to in . What is the constant of growth?

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

$\ln(10)$

$e^{30}$

None of these

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 and solve for .

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

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

$1000=100e^{30k}$

$e^{30k}=10$

$30k=\ln(10)$

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

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

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

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

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

$\ln(8)$

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 and solve for .

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

Now that we have we can solve for at .

$1600=200e^{6k}$

$e^{6k}=8$

$6k=\ln(8)$

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

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

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

$\ln(4)$

$\ln(7)$

$\frac{4}{7}$

None of these

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 and solve for .

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

Now that we have solved for we can solve for at

$200=50e^{7k}$

$e^{7k}=4$

$7k=\ln(4)$

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

The rate of growth of the Land of Battlecraft players is proportional to the population. The population increased by 72 percent between February and October of 2015. What is the constant of proportionality?

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

Since the population increased by 72 percent between February and October, we can solve for this constant of proportionality. It'll help to represent the months by their number in the year:

$(1+0.72)y_0=y_0e^{k(10-2)}$

$1.72=e^{8k}$

$k=\frac{ln(1.72)}{8}=0.068$

Bob invests in a bank that compounds interest continuously at a rate of . How much money will Bob have in his account after years? (Round answer to decimal places.)