Let Deposit be P, Let rate of interest be R (in decimal)

Compound interest earned in 1st year = P(1+R)^1 - P = PR [CI = P*(1+R)^n]

Compound interest earned in 2nd year = P(1+R)^2 - PR - P [CI for 2 years - Interest in 1st year - Principal] = P + 2PR + PR^2 - PR - P = PR + PR^2

Difference in compound interest earned in 1st and 2nd year =

PR + PR^2 - (PR)

= PR^2 = 40

If rate of interest is 3R then

(following similar procedure) Difference in Compound interest would be = P*(3R)^2 = 9PR^2

So, Difference will be 9 * 40 = 360

So, Answer will be D

Hope it helps!

desaichinmay22 wrote:

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 would have been how much?

A)40/3

B)40

C)120

D)360

E)420

_________________

Ankit

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