WORD PROBLEMS ON EXPONENTIAL GROWTH AND DECAY

Problem 1 :

Bacteria can multiply at an alarming rate when each bacteria splits into two new cells, thus doubling. If we start with only one bacteria which can double every hour. How many bacteria will we have by the end of one day ?

Solution :

Number of bacteria's initially = 1

Growing rate = 2 (doubles)

y = ab^x

a = 1, b = 2

y = 1(2)^x

So, the required function is y = 2^x

Problem 2 :

Find a bank account balance if the account starts with $100, has an annual rate of 4% and the money left in the account for 12 years.

Solution :

Initial balance = $100

Balance after 1 year = 100 + 4% of 100

= 100(1 + 4%)

Balance after 2 years

= 100(1 + 4%) + 4% of (100(1 + 4%))

= 100(1 + 4%) (1 + 4%)

= 100(1 + 4%)²

Continuing like this, amount left after 12 years

= 100(1 + 4%)¹²

= 100(1 + 0.04)¹²

= 100(1.04)¹²

= $160.10

Problem 3 :

In 1985, there were 285 cell phone subscribers in the small town at center ville. The number of subscribers increased by 75% per year after 1985. How many cell phone subscribers  were in the Center ville in 1994 ?

Solution :

The required function will be exponential growth function,

y = a(1 + r%)^x

a = number of subscribers initially

x-period of investing = 1994 - 1985

x = 9

r = rate of growth = 75%

y = 285(1 + 75%)⁹

y = 285(1 + 0.75)⁹

y = 285(1.75)⁹

y = 43871.9

Approximately 43872 subscribers will be at 1994.

Problem 4 :

The population of Winnemucca, Nevada can be modeled by

P = 6191 (1.04)^t

where t is the number of years since 1990.

a)  What was the population in 1990?

b)  By what percent did the population increase by each year ?

Solution :

P = 6191 (1.04)^t

a) To find the population in the year 1990, we apply t = 0

P = 6191 (1.04)⁰

P = 6191 (1)

P = 6191

b)  Expressing the given function P = 6191 (1.04)^t in the form of y = a (1 + r%)^x

P = 6191 (1 + 0.04)^t

P = 6191 (1 + 4%)^t

Percentage increase is 4%.

Problem 5 :

You have inherited land that was purchased for $30000 in 1960. The value of the land increased by approximate 5% per year. What is the approximate value of the land in the year 2011 ?

Solution :

y = a(1 + r%)^x

a = 30000, x = 2011 - 1960 ==> 51 and r = 5%

y = 30000(1 + 5%)⁵¹

y = 30000(1 + 0.05)⁵¹

y = 30000(1.05)⁵¹

y = 30000(12.04)

y = 361200

So, approximate value of the land at 2011 is $361200.

Problem 6 :

During normal breathing, about 12% of the air in the lungs is replaced after one breath. Write an exponential decay model for the amount of the original air left in the lungs if the initial amount of air in the lungs is 500 mL. How much of the original air is present after 240 breaths ?

Solution :

The given is an exponential decay function.

Quantity of air after every breaths :

Q(t) = a(1 - r%)^x

a = 500 mL, r = 12%, x = 240

Q(240) = 500(1 - 12%)²⁴⁰

= 500(1 - 0.12)²⁴⁰

= 500(0.88)²⁴⁰

= 500(4.74 x 10^-14)

= 2.37 x 10^-11 ml

Problem 7 :

An adult takes 400 mg of ibuprofen. Each hour, the amount of ibuprofen in the person's system decreases by about 29%. How much ibuprofen is left after 6 hours ?

Solution :

The given function is exponential decay function.

y = a(1 - r%)^x

a = 400, r = 29% and x = 6

y = 400(1 - 29%)⁶

= 400(1 - 0.29)⁶

= 400(0.71)⁶

= 51.24 mg

Problem 8 :

You deposit $1600 in a bank account. Find the balance after 3 years if the account pays 4% annual interest yearly.

Solution :

Deposit initially (a) = 1600

x = 3, r = 4%

The given function is exponential growth function.

y = a(1 + r%)^x

= 1600(1 + 4%)³

= 1600(1 + 0.04)³

= 1600(1.04)³

= 1799.78

Problem 9 :

You buy a new computer for $2100. The computer decreases by 50% annually. when will the computer have a value of $600 ?

Solution :

The given function is exponential decay function.

y = a(1 - r%)^x

a = 2100, r = 50%, y = 600 and x = ?

600 = 2100 (1 - 50%)^x

600 / 2100 = (1 - 0.50)^x

0.29 = (0.50)^x

Taking log on both sides.

log (0.29) = x log(0.50)

x = log (0.29) / log (0.50)

x = 1.78

So, approximately 2 years.

Problem 10 :

You drink a beverage with 120 mg of caffeine. Each hour the caffeine in your system decreases by about 12%. How long until you have 10 mg of caffeine ?

Solution :

The given function is exponential decay function.

y = a(1 - r%)^x

a = 120, r = 12%, y = 10 and x = ?

10 = 120(1 - 12%)^x

10/120 = (1 - 0.12)^x

1/12 = (0.88)^x

0.083 = (0.88)^x

taking log on both sides.

log (0.083) = x log(0.88)

x = log (0.083) / log(0.88)

x = 19.4 hours

x = 20 hours.

After 20 hours.