Compound Interest Questions and Answers

Given: A = 58694 P = 50000 n = 2 years r=?
A = P (1 + r/100)ⁿ
=>58694 = 50000 (1 + r/100)²
=>58694/50000 = (1 + r/100)²
=>(1+ r/100) = sqrt (1.17)=1.08
=>r/100=1.08-1=0.08
=>r = 0.08 x 100 = 8%

=9000(1+15/100)2⅓

=9000(1+3/20)² x (1+3/20x3)

=9000* (23/20)  x (23/20)  x  (21/20)

=12497.625

∴ Compound Interest = Rs. (12497.625– 9000) = Rs. 3497.625

Given: P = Rs.8000, rate of interest = 8%, T = 4years, n = 2

calculation: half-yearly, therefore, r=   8 %=4%

nT=2 x 4=8

A=8000( 1 + 8/100 x 2)⁸ =8000× (26/25)⁸

=8000 x 1.3685=10948

Interest = Amount - Principal = `10948 - 8000 = 2948

Let the sum be `100.

Therefore, SI=100×10×2/100 = `20

and CI=100 (1 + 10/100)² - 100 = 121 − 100 = 21

Difference of CI and SI = 21 − 20= 1

Now if the difference is 1 , then the total= 100

So if the difference is 20 then the total will be= 100× 20 = INR 2000

13560(1 + 7/100)³

Required population: 13560(1.07)³ = 16612

Measurement shows that 100 at 9% compounded semi-annually will grow to

A=100 (1 + .09/2)² = 100(1.045)²

=100(1.09)=109.20

So the actual rate=109.20 - 100=9.20. Thus if we get 9.20 % 100 in 1 year with yearly compounding, then the rate will be 9.20/100=0.092=9.20 so the actual rate = 9.20%

Given r=.049 and m= 12.

Value of `p after n years = p (1+ r/100)ⁿ

Effective rate is re = (1 + .049/12)¹² − 1

=1.050115575-1=.0501

or 5.01%

Down payment = INR 2000

So at the completion of 1^st-year x will become INR 1500.

So, 1500 = x (1 + 10/100) or

x = (1500×100/110) = 1363.63

Similarly, 1050 = y (1 + 10/100)² or y= (1050×20×20/22×22) = 420,000/484=867.76

and z = (930×20×20x20/22×22×22) = 698.72

Hence, CP = 2000+1363.63+867.76+698.72 = 4930.11 or 4930

As given, P+ Compound Interest of 3 years = 5500.......(i)
P+ Compound Interest of 2 yrs = 4645.......(ii)
By solving the equations we get Compound Interest of 3rd year = 5500-4645 = 855
Alternate method:
Difference of sum after n years and n + 1 years × 100 /Amount after n years
Here n=2
Rate = [(5500−4645) × 100] /4645= (855×100)/4645 = 18.40% or 18%.

After first year the amount =

28750 (1 + (4/100)) = 28750 (104/100)

After 2 nd year the amount = 28750 (104/100) (108/100)

=28750 (26/25) (27/25) = (1.12 x 28750)=32292

CI = 32292-28750 = 3542

According to the question

Taking P = x becomes 9x in the span of 2 years when compounded annually

A = P\left ( 1 + \frac{r}{100}\right )^{t}

9x = x\left ( 1 + \frac{r}{100}\right )^{2}

\frac{9x}{x} = \left (\frac{100+r}{100} \right )^{2}

3^{2} = \left (\frac{100+r}{100} \right )^{2}

\frac{100+r}{100} =3

100 + r = 300

r = 200

Required difference = (pr²)(300+r)/(100)³
(4525×8×8)(300+8)/(100×100x100) = 89.19