Compound Interest:
(1) Principal (P)- It is the money borrowed or lent out for a certain time.
(2) Interest (I)- It is the additional money that the lender receives from the borrower in consideration of the borrower using the money.
(3) Amount (A)- The sum of principal and interest is called amount.
(4) Rate of Interest (R)- Money paid per $100 per year is called rate of interest.
(5) Time (T)- The period for which the money is borrowed or lent out is called time.
(6) Simple Interest (SI)- When the principal remains the same for the entire duration for which the money is lent (or borrowed) interest is calculated on the original principal for the given time and rate, this interest is known as simple interest. In other words, if the interest on a sum is calculated uniformly on the same sum, then it is called simple interest.
| Formulae for Simple Interest: Principal = P, Time = T years, Rate = R% p.a., Amount = A Then, SI = [(P x R x T) / 100] P = (SI X 100) / (R x T) R = (SI X 100) / (P x T) T = (SI X 100) / (P x R) A = P + SI |
(7) Compound Interest (CI)- When the interest is earned over a fixed time (a year or any specified period such as six months, three months, etc.) is added to the principal for the next time period and this process is repeated for several time periods, the difference between the amount at the end of the last time period and the original principal is called the compound interest for the whole period.
The time period after which the interest is added to the original principal to form a new principal is called the conversion period. This period may be one year, six months, three months, or one month. Thus correspondingly, the interest is said to be compounded annually, semi-annually, quarterly, or monthly.
Note: In the case of simple interest, the principal remains the same for the whole duration whereas in the case of compound interest, the principal keeps changing periodically.
If the interest is payable yearly, then the compound interest for 1 year will be the same as the simple interest for 1 year.
The compound interest on a principal for two years is more than the simple interest on the same principal for two years.
Formulae for Compound Interest:
Compound Interest Examples:
| Example 1: Andrew invested $2500 at the rate of compound interest for 2 years. He got $3600 after the specified period. Find the rate of interest. Solution: Here, P = $2500, n = 2 yr, A = $3600 Now, according to the formula, A = P (1 + r/100)n ⇒ 3600 = 2500 (1 + r/100)2 ⇒ 3600/2500 = [(100 + r)/100]2 ⇒ (6/5)2 = [(100 + r)/100]2 ⇒ 6/5 = (100 + r)/100 ⇒ 100 + r = (6/5) x 100 ⇒ 100 + r = 120 ⇒ r = 120 – 100 ∴ r = 20% |
| Example 2: Find the compound interest on $4000 in 2 yr at 5% pa, the interest being compounded half-yearly. Solution: Here, P = $4000, r = 5% pa, n = 2yr Now, according to the formula, A = P [1 + (r/2×100)]2n ⇒ A = 4000 (1 + 5/200)4 ⇒ A = 4000 (205/200)4 ⇒ A = 4000 x 41/40 x 41/40 x 41/40 x 41/40 ⇒ A = 4415.26 ∴ Compound Interest = A – P = 4415.26 – 4000 = $415.26 |
| Example 3: Find the compound interest on $6000 at 10% pa for 9 months, compounded quarterly. Solution: Here, P = $6000 n = 9 months = 3/4 yr r = 10% pa Now, according to the formula, A = P [1 + (r/4×100)]4n ⇒ A = 6000 (1 + 10/400)4×3/4 ⇒ A = 6000 (1 + 1/40)3 ⇒ A = 6000 (41/40)3 ⇒ A = 6000 x 41/40 x 41/ 40 x 41/40 ⇒ A = $6461.35 ∴ Compound Interest = A – P = $6461.35 – 6000 = $461.25 |
| Example 4: Find the compound interest for 1 year 3 months on $4500 lent at 10% pa, reckoned annually. Solution: Interest for 1st year: P = $4500, n = 1 yr, r = 10% pa ∴ I = (4500 x 1 x 10)/100 = $450 A = P + I = 4500 + 450 = $4950 Interest for the next 3 months i.e. 1/4 year: P = $4950, n = 1/4 yr, r = 10% pa ∴ I = (4950 x 1/4 x 10)/100 = $123.75 A = P + I = 4950 + 123.75 = $5073.75 ∴ Compound Interest = A – P = 5073.75 – 4500 = $573.75 |
| Example 5: Find the compound interest on $25,000 for 3 years if the rates for the 3 years are 2%, 6%, and 9% respectively. Solution: Here, P = $25,000 r for 3 years = 2%, 6%, 9% Now, according to the formula, A = 25,000 (1 + 2/100) (1 + 6/100) (1 + 9/100) ⇒ A = 25,000 (102/100) (106/100) (109/100) ⇒ A = 29,462.7 ∴ Compound Interest = A – P = 29,462.7 – 25,000 = $4462.7 |
| Example 6: A sum of money becomes $5832 in 2 years at a compound interest rate of 8% pa. Find the original sum and the principal at the beginning of the second year. Solution: Let the sum be $P. R = 8% pa T = 1 yr Interest = (P x R x T)/100 = (P x 8 x 1)/100 = $8P/100 Amount at the end of the first year = P + 8P/100 = $108P/100 ∴ Principal at the beginning of the second year = $108P/100 ∴ Interest for the second year = [(108P/100) x 8 x 1] / 100 = $864P/10000 Amount at the end of the second year = 108P/100 + 864P/10000 = $11664P/10000 = $5832 (given) ⇒ P = 5832 x 10000/11664 ∴ Original sum = $5000 Principal at the beginning of the second year = $108P/100 = $(108 x 5000)/100 = $5400 |
| Example 7: Catherine purchased six years of NSC for $1000. After six years she got $2015. Find the rate of interest, if the interest is compounded half-yearly. [Given that (2.015)1/12 = 1.06012] Solution: P = $1000, A = $2015, n = 6yr. When the interest is compounded half-yearly, then the time is 12 half-years. Let the half-yearly rate of interest be r%. Then A = P (1 + r/100)n ⇒ 2015 = 1000 (1 + r/100)12 ⇒ (1 + r/100)12 = 2015/1000 = 2.015 ⇒ (1 + r/100) = (2.015)1/12 = 1.06012 ⇒ r/100 = 1.06012 – 1 = 0.06012 ⇒ r = 0.06012 x 100 = 6.012 ∴ Rate of interest per annum = 2r% = 2 x 6.012% = 12.024% ≅ 12% Hence, the required rate of interest is 12% pa |
| Example 8: The compound interest on $30,000 at 7% pa for a certain time is $4347. Find the time. Solution: Amount (A) = 30,000 + 4347 = $34,347 Principal (P) = 30,000 Rate (r) = 7% Let the time be t years. Now, according to the formula, A = P (1 + r/100)t ⇒ 34,347 = 30,000 (1 + 7/100)t = 30,000 (107/100)t ⇒ (107/100)t = 34,347/30,000 = 11449/10000 ⇒ (107/100)t = (107/100)2 ∴ t = 2 Hence, the required time is 2 years. |
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