MODELLING EXPONENTIAL FUNCTIONS

Problem 1 :

The amount of money A accrued at the end of n years when a certain amount P is invested at a compound annual rate r is given by

A = P(1 + r)ⁿ

If a person invests $150 at 5% interest compounded annually, find the approximate amount obtained at the end of 5 years.

a)  $191     b) $4500     c) $1139   d) $900

Solution :

A = P(1 + r)ⁿ

P = 150, r = 5%, n = 5

= 150(1 + 0.05)⁵

= 150(1.05)⁵

= 150 (1.2763)

= 191.445

Approximately $191.

So, the answer is option a.

Problem 2 :

The projected worth (in millions of dollars) of a large company is modeled by the equation

y = 246 (1.1)^x

The variable x represents the number of years since 1997. What is the projected annual percent of growth, and what should the company be worth in 2005?

a) 21%; $273.06 million         b) 21%; $629.28 million

c) 11%; $566.92 million         d) 11%; $510.74 million

Solution :

y = 246 (1.1)^x ----(1)

y = 246 (1 + 0.1)^x

y = 246 (1 + 10%)^x

Period of investment :

x = 2005 - 1997

= 8

Applying the value of x in (1), we get

y = 246 (1.1)⁸

= 246(2.14358881)

= 527.32

So, option c is correct.

Problem 3 :

You borrow $200 from a relative for six months. You agree to pay compound interest at the rate of 1% per month. How much interest will you pay your relative when you return the money at the end of the six months?

a) $11.66     b) $201.00     c) $210.00     [d)   $12.30

Solution :

p = 200, n = 6 months or 6/12 = 1/2 year = 0.5

r = 1%

Formula for compound interest :

A = P(1 + r%)ⁿ

= 200(1 + 1%)^0.5

= 200(1 + 0.01)^0.5

= 200(1.01)^0.5

= 200.99

= $201

So, option b is correct.

Problem 4 :

Which of the following accounts will yield the greatest amount of interest on an initial deposit of $500.00?

a) Account that pays 6% interest compounded annually for 3 years

b) Account that pays 4% interest compounded annually for 4 years

c) Account that pays 3% interest compounded annually for 5 years

d) Account that pays 5% interest compounded annually for 6 years

Solution :

Option a :

Account that pays 6% interest compounded annually for 3 years

A = P(1 + r%)ⁿ

= 500(1 + 6%)³

= 500(1.06)³

= 595.508

Interest = Amount - Principal

I = 595.508 - 500

I = 95.08

Option b :

Account that pays 4% interest compounded annually for 4 years

A = P(1 + r%)ⁿ

= 500(1 + 4%)⁴

= 500(1.04)⁴

A = 584.93

I = 584.93 - 500

I = 84.93

Option c :

Account that pays 3% interest compounded annually for 5 years

A = P(1 + r%)ⁿ

= 500(1 + 3%)⁵

= 500(1.03)⁵

A = 579.63

I = 579.63 - 500

I = 79.63

Option d :

Account that pays 5% interest compounded annually for 6 years

A = P(1 + r%)ⁿ

= 500(1 + 5%)⁶

= 500(1.05)⁶

= 670.04

I = 670.04 - 500

I = 170.04

So, option d will earn more interest.

Problem 5 :

The population of Mexico in mid-1994 was 91,800,000. Its annual growth rate is 2.2%. Estimate its population in mid-2000.

Solution :

Growth function will be,

A = P(1 + r%)ⁿ

Number of years = 2000 - 1994 ==> 6

Growth rate = 2.2%

= 91,800,000(1 + 2.2%)⁶

= 104603943

Problem 6 :

Use any problem solving strategy to solve the following problem. The value of a house is expected to increase from its current value of $50,000 by 3% each year. What will the value of the house be after 3 years?

If you have $55,000 in 3 years, will you have enough to buy the house?

Solution :

A = P(1 + r%)ⁿ

P = 50000, r = 3%, n = 3

A = 50000(1 + 3%)³

= 50000(1.03)³

= 54636.35

So, you have enough money to buy the house.

Problem 7 :

A position at a local company has a starting salary of $15,000. The salary is expected to increase by 5% each year. What will the salary be after 5 years?

Solution :

A = P(1 + r%)ⁿ

P = 15000, r = 5%, n = 5

A = 15000(1 + 5%)⁵

= 15000(1.05)⁵

= 19144.2