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1
Example 25: Find the compound interest on `80,000 for 1 years at 10% p.a. when compounded
half yearly. 2
1
Solution: Here, P = `80,000, R = 10% p.a. or 5% half-yearly, T = 1 years or 3 half-years.
2
R

We know, A = P 1 + 100   T


  5  3   21 21 21

So, A = ` 80 000 1,  +   = ` 80 000 × × ×  = `92,610
,

   100    20 20 20 
So, Compound Interest = `(92,610 – 80,000) = `12,610.
Example 26: Find the compound interest on `10,000 for 1 year at the rate of 8% per annum when
the interest is compounded quarterly.
Solution: Here, P = `10,000, Rate = 8% per annum or 1 × 8% = 2% quarterly and n = 1 year or
1 × 4 = 4 quarters. 4 [As, there are 4 quarters in a year]

 R  n  2  4  51 4
A = P 1 + 100  = `10,000 1 + 100  = `10,000   50





51 51 51 51
= `10,000 × × × × = `10,824.32
50 50 50 50
So, Compound Interest = `10,824.32 – `10,000 = `824.32
Thus, the required compound interest is `824.32.
Example 27: Find the amount on `2,00,000 after 2 years, if the interest is compounded annually,
the rate of interest being 4% p.a. during the first year and 5% p.a. during the second year. Also,
find the compound interest.
Solution: We know that if R % and R % are the rates of interest for the first and second year,
2
1
R 

R 
respectively, and P is the Principal, then the amount for 2 years is given by A = P 1 + 100  ×    1 + 100 
2
1


[This formula may be extended (or shortened) for any number of years.]
Here, P = `2,00,000; R = 4% p.a. and R = 5% p.a.
1
2
 4   5  26 21
So, Amount after 2 years = 2,00,000 1 + 100  ×  1 + 100 = 2,00,000 × 25 × 20 = `2,18,400




Thus, amount after 2 years = `2,18,400, and compound interest = `(2,18,400 – 2,00,000) = `18,400
Inverse cases based on Compound Interest
Example 28: What sum will become `6000 after 2 years at 5% per annum when the interest is
compounded annually?
Solution: A = `6000, R = 5% p.a., Time (T) = 2 years, P = ?
5 

R 

Amount (A) = P 1 + 100  n ⇒ `6000 = P1 + 100 2





 105  2 P × 105 × 105
⇒ `6000 = P ×   100   ⇒ `6000 = 100 × 100
`6000 × 100 × 100
⇒ P = ⇒ P = `5442.18
105 × 105
So, the sum is `5442.18.
183 Comparing Quantities