3-3.
The simple interest is charged only on the borrowed principal.
Thus the amount of interest in 3 years = 2000*0.10*3 = $600
Thus her payment after 3 years would be 2000+600 = $2600.
 

3-5.
Here F = 200, i = 10%, n = 4, Q is the present sum.
So, Q = F*(1+i)^-n
So, Q = 200*(1+0.1)^-4 = 200*1.1^-4 = 200*0.683 = $136.60
Alternatively,
Q = 200*(P/F,10%,4) = 200*(0.6830) = $136.60.

3-7.
P, i, and n are given, F = ?.
So, F = P*(1+i)ⁿ = 750*1.08³=750*1.2597 = $944.78
Alternatively,
F = 750*(F/P,8%,3) = 750*1.260 = $945.

3-8.
In this problem F = 8250, i = 4% and n = 2*2 = 4, find P.

P = 8250*(P/F,4%,4) = 8250*0.8548 = $7052.

3-10.
 2% interest will be paid per six months period. The question is, if P amount is deposited now after how many periods it will become 2P.

Here P, F and i is given find n

Formula: F = P*(1+i)ⁿ

In this case plugging the values in the formula: 2P = P*(1+.02)ⁿ

Taking log in both sides (log 2) = n (log 1.02)

Dividing both side by (log 1.02) n = (log 2)/log(1.02)

            = 0.30103/ 0.0086

So it would take 35 six months to double the money, which is equal to 17.5 years.

 3-15
In this problem P = 600, F = 29,152,000, n=1995-1903=92 years, i=?

So, 29152000 = 600*(1+i)⁹²
that is, 29152000/600 = 48586.67 = (1+i)⁹²
Then 1+i = 48586.67^(1/92) = 1.12445
Thus, i = 1.12445-1 = 0.12445 = 12.45%