The difference in compound interest earned on a deposit (compounded annually) in year 1 and year 2 is $ 40. Had the interest rate been three times its present value, the difference z would have been how much?
Select
A) 40/3
B) 40
C) 120
D) 360
E) 420
CASE 1:   Deposit = $x;   Rate of increase = r.   Interest yearned in 1 year = xr. Deposit in 1 year = x + xr.   Interest yearned in 2 year = (x + xr)r.   The difference z= (x + xr)r - xr = xr^2 = 40.   CASE 2:   Deposit = $x;   Rate of increase = 3r.   Interest yearned in 1 year = x(3r). Deposit in 1 year = x + 3xr.   Interest yearned in 2 year = (x + 3xr)3r.   The difference = (x + 3xr)3r - 3xr = 9xr^2. Since from CASE 1 we know that xr^2 = 40, then 9xr^2 = 9*40= 
<gadget id="calculator">9*40</gadget>
<output>360</output>
360.   Answer: 360.
<result>D</result>
The difference in compound interest earned on a deposit (compounded annually) in year 1 and year 2 is $ 40. Had the interest rate been three times its present value, the difference z would have been how much?