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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