> :<97 bjbjUU "27|7|%lzzzzzzzvvv84"$$$$$$qssssss$ \z$$$$$zz$$$z$z$q$qzz%$_=v%L0,`Ld`%zzzzMAT 120 Fall 2004 11/23/2004
TEST #2
Instructor Aliya Nurtaeva ANSWER KEY
1. How much money should James invest now in a money market account with an annual interest rate of 7.2% if she wants to have $2000 in the account by the end of 8 months?
Use simple interest formula with r = 0.072, t = 8/12 year and future value $2000:
S = P(1 + rt)
8
2000 = P(1 + 0.072 ----)
12
2000 = P(1 + 0.048)
2000 = P(1.048)
P = 2000/(1.048) = $1908.40 present value
2. If $2000 is invested in an account with an annual simple interest rate of 9%, how long will it take the investment to be worth $3000?
Use simple interest formula, letting S = 3000, P = 2000, r = 0.09:
S = P(1 + rt)
3000 = 2000 (1 + 0.09t)
3000 = 2000 + 2000( 0.09t
1000 = 2000(0.09t
t = 1000/(180) = 5.56 (yrs)
3. Future value of a certain amount of money deposited in a savings account for certain time period is given by the following formula:
S = 5000 ( 1 + 0.015)12
Assuming that the interest rate is compounded quarterly, find the following:
a) the number of compounding periods per year: m = 4 (quarterly)
the total number of compounding periods: n = 12 (exponent)
time period in years: mt = 12 ( t = 12/m = 12/4 = 3 years
the interest rate per compounding period: i = 0.015 (from equation)
e) annual interest rate: r = im = 0.15(4) = 6%
f) present value of the deposit: P = $5000
Indicate the symbols of the variables.
4. Karen must pay$20,000 in settlement of an obligation in 3 years. What amount can she deposit today, at 4% compounded monthly, to have enough?
The future value is known, and the present value must be found.
Interest type periodic compound interest: use formula S = P(1 + i)n
Plug into this formula the values of variables:
S = 20,000, m = 12, i = r/m = 0.04/12 = 0.00333, n = mt = 12(3) = 36
20000 = P(1 + 0.00333)36
20000 = P(1.1273)
P = 20000/1.1273 = $17,741.51 to deposit today (present value)
5. Suppose the average price of a house nationally is $137,000 in 2004. Housing prices are increasing at a rate of 3% per year (compound interest). Assuming prices continue to increase at the same rate, what will be the average price of a house in year 2015?
$137,000 is the present value, we need to find future value of this money.
S = P(1 + i)n
t = 2015 2004 = 11 years
3% - annual compound interest rate, r = 0.03 and m = 1 (one period per year)
i = r/m = 0.03/1 = 0.03
S = 137,000(1 + 0.03)11 = $189,640.04 the average price of a house in 2015
6. $200 are invested at 9% compounded continuously. Find the future value of this investment in 10 years.
Interest continuous compound, use formula S = Pert
P = 200 (present value)
r = 0.09
t = 10 years
S = 200e0.09(10) = $491.92 future value
7. Charlie is saving for a laptop computer. At the end of each month he put $60 in a money market account that pays 4% interest compounded monthly. How much is in the account after 3 years?
Annuity with regular deposits given. We need to find future value of annuity:
(1+i)n - 1
FV = PMT ------------
i
i = 0.04/12 = 0.0333
n = 12 (3) = 36 periods
(1+0.00333)36 - 1
FV = 60 -------------------- = 60 (38.1819) = $2290.91 will be in 3 yrs in his account
0.00333
8. Stereo Shack sells a stereo system for $600 down and monthly payments of $30 for the next 3 years. If the interest rate is 12% compounded monthly, find the cost of stereo system.
Stereo Shack allows to buy on loan terms, so we consider this as a loan for a buyer.
The cost of stereo system, therefore, is the present value of annuity with regular payments of $30,
n = 12(3) = 36 PMT = 30 i = 0.12/12 = 0.01
COST = DOWNPAYMENT + PRESENT VALUES OF PAYMENTS
We need to find present values of loan payments.
Use loan payment formula:
i
PMT = PV ------------
1- (1+i)-n
Plug in all the values known:
0.01
30 = PV -----------------
1 -(1+0.01)-36
Solve the equation for PV:
(1 1.01-36) 0.301075
PV = 30 --------------- = 30 ------------ = $903.23 present values of $30 payments within 3 yrs
0.01 0.01
COST = 600 + 903.23 = $1,503.23
9. (10 points) A new car is priced at $11,000. The buyer must pay $3,000 down and pay the balance within next 4 years in monthly payments at 12% compounded monthly.
Find the amount of each payment.
10000 3000 = 8000 amount of loan, or the present value of future payments.
i = r/m = 0.12/12 = 0.01
n = mt = 12 (4) = 48 periods
PV = 8000
Use formula for loan payments:
i
PMT = PV ------------
1- (1+i)-n
Plug in all the values known:
0.01 0.01
PMT = 8000 --------------- = 8000 ---------- = $210.67 monthly payment on the car loan
1 -(1+0.01)-48 0.3797
10. The Smiths buy a house for $285,000. They pay $35,000 down and take out a 30-year mortgage for the balance at 8% compounded monthly.
a) Find their monthly mortgage payment.
285000 35000 = $250,000 amount of a mortgage loan, or present value of future payments
i = r/m = 0.08/12 = 0.06667
n = mt = 12 (30) = 360 periods
PV = 250,000
Use formula for loan payments:
i
PMT = PV ------------
1- (1+i)-n
Plug in all the values known:
0.06667 0.06667
PMT = 250.000 ----------------- = 250,000 -------------- = $1,834.41 monthly mortgage 1 -(1+0.06667)-360 0.908557 payment
b) Find the total of interest they will pay.
TOTAL INTEREST = LOAN PAYMENTS LOAN AMOUNT
$1,834.41 (360) = $660,387.60 LOAN payments
660,387.60 - 250,000 = $410,387.60 total interest paid
c) Find the part of the first payment that is interest and the part that is applied to balance reduction.
Interest is earned on the unpaid balance (amount owed).
During the first month the amount owed is total 250,000, interest portion of the payment will be:
I = Prt = 250000 (0.08) (1/12) = $1,666.67
1,834.41 1,666.67 = $167.43 - balance reduction portion (or principal portion)
BONUS (5 points). As the prize in a contest, you are offered $1000 now or $1300 in 5 years. If money can be invested at 6% compounded annually, which is larger?
We need to compare two sums of money, which are given at different times.
So to make comparison we should adjust them to the same time.
Present value of the first sum is P1 = 1000,
i = 0.06 (annual conpounding),
n = mt = 1(5) = 5 periods
S = P(1 + i)n = 1000(1 + 0.06)5 = $1,338.23 the future value (in 5 years) of $1000 given today.
Now we can compare these 2 numbers: $1,338.23 and $1300.
Answer: 1000 today is better than 1300 in 5 years!
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