Aptitude Questions and Answers

Let,
Principal = Rs. 100.
Amount = Rs. 200.
Rate = r%
Time = 4 years.
Now,
`A = P*[1+ (r/100)]^n`;
`200 = 100*[1+(r/100)]^4`;
`2 = [1+(r/100)]^4`; ........... (i)
If sum become 8 times in the time n years,
then,
`8 = (1+(r/100))^n`;
`2^3 = (1+(r/100))^n`; ........(ii)
Using eqn (i) in (ii), we get;
`([1+(r/100)]^4)^3 = (1+(r/100))^n`;
`[1+(r/100)]^12 = (1+(r/100))^n`;
Thus, n = 12 years.

Difference between CI and SI for n^th year,
= `((Pr) /100) *[(1 +(r /100))^(n-1) -1]`
→ `7280 = ((P*20)/100) *[(1.2)^3-1]`
→ P = 50000.