Interest (Simple & Compound) Test - 3

Amount after 3 years = P(1 + R/100)^T
=> 40000(1 + 11/100)³
=> 40000(111/100)³
=> 40000[(111*111*111)/(100*100*100)]
=> (4*111*111*111)/100 
=> 54705.24
Compound Interest = 54705.24 - 40000 
= Rs. 14705.24

[15000×(1+R/100)² − 15000] - (15000×R×2)/100 = 96
⇒15000[(1+R/100)² − 1 − 2R/100]=96
⇒ 15000[(100+R)² − 10000−(200×R)]/10000 = 96
⇒ R2 = (96×23) = 64
⇒ R = 8
∴ Rate = 8%

Amount = Rs.(30000+4347) = Rs.34347
let the time be n years
Then,30000(1+7/100)ⁿ = 34347
(107/100)ⁿ = 34347/30000 
= 11449/10000 
= (107/100)²
n = 2years

Let the rate be R% per annum
P(1+R/100)T = 1573.04
= 1400(1+R/100)2 = 1573.04
= (1+R/100)2 = 1573.04/1400 = 157304/140000 = 11236/10000
(1+R/100) = (√11236/10000)
= (√11236)/(√10000)
= 106/100
R/100 = (106/100 − 1)
=> 6/100
R = 6%

Simple interest = P * r * t /100, where, P is the Principal, r is the rate of interest and t is the time period such that the rate of interest and the time period have mutually compatible units like r % per annum and t years.

Here, in the first case, let the Principal be Rs.P and rate of interest be r % per annum. Time period is given as 8 years . Interest is given as 80 P / 100
So, 80 P / 100 = P * r * 8 / 100
Or, 80 = 8 r
So, r = 10 % per annum.

Now, in the second case, Principal is given as Rs. 14,000, rate of compound interest is 10 % per annum as determined above; and t is 3 years.
Compound interest = P * ( 1 + r/100)t - P 
= 14,000 * ( 1+ 10 /100)3 - 14,000
= 14,000 * ( 1 + 0.1)3 - 14,000
= 14,000 *(1.1)3 - 14,000
(14,000 * 1.331) - 14,000
= 18,634 - 14,000 = 4,634
So, the compound interest on Rs.14,000 at the same rate of interest as in case 1 in 3 years will be Rs. 4,634.