Note for questions:(If you solve this problem with algebra round intermediate ca

Question

Note for questions:(If you solve this problem with algebra round intermediate calculations to 6 decimal places, in all cases round your final answer to the nearest penny.)

1. Mr. A is saving to buy a house in five years’ time. She plans to put 20 percent down at that time and she believes that she will need $9,050 at that time. If Tracy can invest in a fund that pays 6.50 percent annually, how much should she invest today?

Amount to be invested today=

2. A aunt is planning to invest in a bank deposit that will pay 9 percent interest semi-annually. If she has $13,000.00 to invest, how much will she have at the end of four years?

Value of investment after 4 years=

3. The future value of $1000 invested at a 7% annual rate, compounded quarterly for 3 years is:

4. If you invested $1000 for 5 years and your return was 12%, how much interest on interest did you earn?

5. If you can invest $10,000 at a 7% annual rate, compounded monthly, how long will it take to have $13,050?

Explanation / Answer

Question 1. Present Value of 9050 = 9050 x (1/(1+6.50/100)^5) Therefore PV = 9050 x 0.729881 Or, PV= 6605.42 Answer: Amount to be invested today = $6,605.42 Question 2. After 4 year she will have 13000 x (1+9/200)^(4x 2) = 13000 x 1.422101 = 18487.31 Value of Investment after 4 years = $18,487.31 Question 3. Future Value of 1000 = 1000 x (1+ 7/400)^(3x4) = 1000 x 1.231439 =$1231.44 Answer: $1231.44 Question 4. If simple interest earned, then interest = 1000 x 12% x 5 =$600 For Compound interest, Amount after 5 years = 1000 x (1.12)^5 =$1762.34 So Interest on interest =1762.34-1000-600 = $162.34 Answer: $162.34 Question 5. Let it take month to become $13,050 So 13050 = 10000 x (1+7/1200)^n Or, 1.005833^n=13050/10000=1.305 Apply on both side we get, n log 1.005833 = log 13.05 or n x0.0025 = 0.1155 or n = 46 month Answer: 46 month

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