Exam #2 Review

Math 1030Q 1. Solve for x:

Solution:

Exam #2 Review

2 + (9.2)-8x = 2.32

2 + (9.2)-8x = 2.32 (9.2)-8x = 2.32 - 2

log (9.2)-8x = log .32 -8x log 9.2 = log .32 -8x log 9.2 log .32

= -8 log 9.2 -8 log 9.2

x = .06418

Spring 2013

2. Solve for y: Solution:

3(y - 7)13 = 1540

3(y - 7)13 = 1540 (y - 7)13 = 513.33333

y - 7 = 1.61619 y = 8.61619

3. Which will be worth more in 10 years: $10,000 invested at 8.2% simple interest, or $10,000 invested at 5% interest, compounded monthly?

Solution: For simple interest:

F = (10000)(1 + (.082)(10)) = $18,200

For compound interest:

.05 12?10 F = (10000) 1 +

12 = (10000)(1.00416667)120 = (10000)(1.64701015)

= $16,470.10

... so you earn more with simple interest.

4. Suppose a friend lends you $100, and you agree to pay him back $112 in 18 months. If we assume that this is simple interest, then what is the interest rate?

Solution: Note that 18 months is t = 1.5 years. Then solving for r,

F = P (1 + rt) 112 = 100(1 + 1.5r) 1.12 = 1 + 1.5r 0.12 = 1.5r 0.08 = r = 8%

5. For an account with an annual interest rate of 6%, find the annual percentage yield (APY) if interest is compounded:

(a) quarterly?

Solution:

.06 4

APY = 1 +

-1

4

= (1.015)4 - 1

= 1.06136355 - 1

.0614 = 6.14%

(b) monthly? Solution:

.06 12

APY = 1 +

-1

12

= (1.005)12 - 1

= 1.06167781 - 1

.0617 = 6.17%

(c) daily? Solution:

.06 365

APY = 1 +

-1

365

= (1.00016438)365 - 1

= 1.06182993 - 1

.0618 = 6.18%

6. A bank advertises a Certificate of Deposit (CD) with 4.8% interest, compounded monthly. If I invest $3,500 today, how long will it take for my investment to grow to $4,200?

Solution: Using the compound interest formula and solving for t,

r nt F =P 1+

n 0.048 12t

4200 = 3500 1 + 12

0.048 12t 1.2 = 1 +

12

log(1.2) = log

0.048 12t 1+

12

0.048 log(1.2) = 12t log 1 +

12 log(1.2) = 12t log(1.004) log(1.2)

= t = 3.806 years 12 log(1.004)

7. Reba would like to make the $2,150 down payment on a new car in 6 months. If she has $2,000 in her savings account, and interest is compounded daily, what interest rate would she need to earn to have enough?

Solution: Using the compound interest formula (t must be in years, not months):

r 2150 = (2000) 1 +

365?

6 12

365

r 2150 = (2000) 1 +

365 2

365

r 1.075 = 1 +

365 2

365

2

(1.075) 365 =

2

r 1+

365 365 2

365

r 1.00039636 = 1 +

365 r 0.00039636 = 365

0.00039636 ? (365) = r

0.14466994 = r 14.47%

8. When Jed was born, his grandfather deposited $1,982 into a savings account for his grandson, under the condition that nobody touches it until Jed turns 21. If this account earns 3.9% interest compounded semi-anually (twice per year), then how much will Jed have on his 21st birthday?

Solution: Using the compound interest formula and solving for F ,

r nt F =P 1+

n 0.039 2?21

= 1982 1 + 2

= 1982(1.0195)42

= 1982(2.25042) = $4,460.33

9. Many years later, Jed's granddaughter is born, and he would like to do something similar for her. He would like her to have exactly $10,000 in the account on her 21st birthday. If the account earns 4.1% compounded annually, how much would Jed need to deposit on the day she is born?

Solution:

.041 1?21 10000 = P 1 +

1 10000 = P (1.041)21 10000 = P (2.32522680) 10000 P (2.32522680)

= 2.32522680 2.32522680

P = $4,300.66

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