Question

A certain sum of money is invested at the rate of 10% p.A. Compound interest the interest compounded annually. If the difference between the interest of the third year and first year is 1105 find the sum invested

Answer 1

For Annual compound intrest,Amount A at the end of n years is given by 
  A =  where, r is rate of intrest expressed as  percentage 

Rate of intrest on a amount C =c*r where r is rate of intrest expressed as  percentage
Let , the invested amount be P
rate of intrest r=10/100=0.1 

Intrest of first year = P * 10/100 = 0.1P
Amount at the end of second year = A2=
Intrest of third year = A2*0.1

Given , difference between intrest of third year and first year is rs.1105

So, 0.1*P*  - 0.1P = 1105

0.1P() = 1105

0.1P(0.21) = 1105

P = Rs.52619

Mark Brainliest please......

Answer 2

$\large \green{ \fcolorbox{gray}{black}{ ☑ \: \textbf{Verified \: answer}}}$

• Let PRINCIPAL as “P”.
• Amount = ( 1 + r/100^n )
• Amount = ₹ 5445
• Principal = P
• Rate of interest = 10%
• Time = 10 year's

Substituting the value,

• 5445 = P × ( 1+ 10/100 )²
• 5445 = P × ( 100 + 10/100 )²
• 5445 = P × ( 110 /100 )²
• 5445 = P × ( 11 /10 )²
• 5445 = P × ( 11 × 11 / 10 × 10 )
• 5445 = P × 121 / 100
• P = 5445 × 100 / 121
• P = 45 × 100
• P = 4500

Answer = 4500

Hence, Principal = 4500