Question

9. A certain sum is invested at the rate of 10% per annum for 3 years. Write a
program to find and display the difference between Compound Interest (CT) and
Simple Interest (ST). Take the sum as an input.
ST
PRT
A = P(1+R/100) and CI= A-P​

Answer 1

Answer:

Explanation:

According to the problem,  

Principal ( Sum of invest)=P

Rate of interest =r

Time =n

The difference between third year and the first year intrest  

implies that,

P[(1+10/100)  

3

−1]−P[(1+10/100)  

1

−1]

=1105P[(11/10)  

3

−1]−P[(11/10)  

1

−1]

=1105P(1331/1000−1)−P(11/10−1)

=1105P(331/1000)−P(1/10)

=1105

​  

 

Now take P as common and subtract the remaining part by taking LCM

So,

$$\begin{array} { P\left( { 231/1000 } \right) =1105 } \\ { P=1105*1000/231 } \\ { P=4783.55 } \end{array}$$

Answer 2

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