Question

Question 7

If money is worth 12% per annum, compounded monthly, how long will it take the principal P to double?

[1] 69,66 years

[2] 8,33 years

[3] 7,27 years

[4] 6,12 years

[5] None of the above

Question 8

A savings account pays interest at the rate of 5% per year, compounded semi-annually. The amount that should be deposited now so that R250 can be withdrawn at the end of every six months for the next ten years is

[1] R3 144,47.

[2] R6 386,16.

[3] R1 930,43.

[4] R3 897,29.

[5] none of the above.

Question 14

If 15% per year, interest is compounded every two months, then the equivalent weekly compounded rate is

[1] 14,464%.

[2] 14,837%.

[3] 14,484%.

[4] 14,816%.

[5] none of the above.

Answer 1

7)expected net cash flows are as follows:

Here $10,000 is assumed to be the principal amount.

Hence, it would be take 69.66 months for the principal to double which makes 5.805 years. Since this is not an option available so (I) is the correct choice.

(8) Given Amount to be withdrawn every six months is R250 = x

Rate of interest, i = 5%

For six months , i = (5/2)% = 2.5% = 0.025

Given Time, n = 10

For six months, n = 10*2 = 20

We know that present value of series of payments is P = x[1-(1+i)^-n]/i

Now P = 250[1-(1+0.025)^-20]/0.025

P = 250[1-0.61]/0.025

P = 250(0.39)/0.025

P = 250*15.589

P = R3897.291

Thus, R3897.29 should be deposited so that R250 can be withdrawn at the end of every six months for the next ten years.

So, option 4 is correct.

(14) $A = P(1 + \frac{r}{n})^{nt}$

A=final amount

P=initial principal balance

r=interest rate

n=number of times interest applied per time period

t=number of time periods elapsed

Let's take an example where p=$2000 t=10years

$A=2000(1+\frac{0.15}{6})^{(6\times10)}$

$=2000(1+0.025)^{60}$

$=\$ 8799.58$

The rate will be

$8799.58=2000(1+\frac{r}{52})^{(52×10)}$

$\frac{8799.58}{2000}=(1+\frac{r}{52})^{(520)}$

$4.39979=(1+\frac{r}{52})^{520}$

R=14.837%

Hence option [2] 14,837%. Is correct.