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# Present Value Help

based on 2 ratings
By McGraw-Hill Professional
Updated on Oct 4, 2011

## Introduction to Present Value

Suppose a couple wants to give their newborn grandson a gift of \$50,000 on his 20th birthday. They can earn interest, compounded annually. How much should they deposit now so that it grows to \$50,000 in 20 years? To answer this question, we will use the formula A = P (1 + r ) t , where we know that A = 50,000 but are looking for P .

The couple should deposit \$11,770.66 now so that the investment grows to \$50,000 in 20 years.

We say that \$11,770.66 is the present value of \$50,000 due in 20 years, earning interest, compounded annually. The present value formula is P = A (1 + r ) t , for interest compounded annually, and , for interest compounded n times per year.

#### Example

• Find the present value of \$20,000 due in years, earning 6% annual interest, compounded monthly.

The present value is \$12,025.18.

## Present Value Practice Problems

#### Practice

For Problems 1-7 find the compound amount.

1. \$800, after ten years, interest, compounded annually
2. \$1200 after six years, interest, compounded annually
3. A 20-year-old college student opens a retirement account with \$2000. If her account pays interest, compounded annually, how much will be in the account when she reaches age 65?
4. \$800, after ten years, earning annual interest

(a) compounded quarterly

(b) compounded weekly

5. \$9000, after five years, earning annual interest, compounded daily (assume 365 days per year).

6. \$800, after 10 years, earning annual interest, compounded continuously.

7. \$9000, after 5 years, earning annual interest, compounded continuously.

8. The population of a city in the year 2002 is 2,000,000 and is expected to grow 1.5% per year. Estimate the city’s population for the year 2012.

9. A construction company estimates that a piece of equipment is worth \$150,000 when new. If it loses value continuously at the annual rate of 10%, what would its value be in 10 years?

10. Under certain conditions a culture of bacteria grow at the rate of about 200% per hour. If 8000 bacteria are present in a dish, how many will be in the dish after 30 minutes?

11. Find the present value of \$9000 due in five years, earning 7% annual interest, compounded annually.

12. Find the present value of \$50,000 due in 10 years, earning 4% annual interest, compounded quarterly.

13. Find the present value of \$125,000 due in years, earning annual interest, compounded weekly.

#### Solutions

1. A = 800(1 + 0.065) 10 = 800(1.065) 10 ≈ 800(1.877137) ≈ 1501.71

The compound amount is \$1501.71.

2. A = 1200(1 + 0.095) 6 = 1200(1.095) 6 ≈ 1200(1.72379) ≈ 2068.55

The compound amount is \$2068.55.

3. A = 2000(1 + 0.0825) 45 = 2000(1.0825) 45 ≈ 2000(35.420585) ≈ 70,841.17

The account will be worth \$70,841.17.

4. (a) n = 4

The compound amount is \$1487.39.

The compound amount is \$1494.04.

5. n = 365

The compound amount is \$12,612.56.

6. 6. A = 800 e 0.065(10) = 800 e 0.65 ≈ 800(1.915540829) ≈ 1532.43

The compound amount is \$1532.43.

7. A = 9000e 0.0675(5) = 9000 e 0.3375 ≈ 9000(1.401439608) ≈ 12,612.96

The compound amount is \$12,612.96.

8. n 0 = 2,000,000, r = 0.015 The growth formula is n(t) = 2,000,000 e 0.015 t and we want to find n ( t ) when t = 10.

n (10) = 2,000,000 e 0.015(10) ≈ 2,323,668

The population in the year 2012 is expected to be about 2.3 million.

9. n 0 = 150,000, r = 0.10 We will use the decay formula because value is being lost. The formula is n(t) = 150,000 e −0.10 t . We want to find n ( t ) when t = 10.

n (10) = 150,000 e −0.10(10) ≈ 55,181.92

The equipment will be worth about \$55,000 after 10 years.

10. n 0 = 8000, r = 2 The growth formula is n(t) = 8000 e 2 t . We want to find n(t) when t = 0.5.

n (0.5) = 8000 e 2(0.5) ≈ 21,746

About 21,700 bacteria will be present after 30 minutes.

11. P = 9000(1.07) −5 ≈ 6416.88

The present value is \$6416.88.

12.

The present value is \$33,582.66.

13. The present value is \$93,316.45.

Find practice problems and solutions for these concepts at: Exponents and Logarithms Practice Test.

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