Note on Compound Interest
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Suppose, Deepak borrows Rs. 1000 at 10% interest from Luna. The simple interest on this sum at the end of one year will be \(Rs. \frac {1000 \times 1 \times 10} {100}\) = Rs. 100. If Deepak pays this interest to Luna, Luna can get back Rs. 1100. In case, Deepak pays this interest to Luna then Luna has right to charge interest on Rs. 1100 for next year. The compound interest is infact a simple interest computed on the previous simple amount. When Deepak calculates the simple interest for 2 years then,
\(I = \frac {1000\times 2\times 10} {100}\) = Rs. 200 but when he calculates the compound interest for 2 years then compound interest for 2 years = simple interest for second year
\(=\frac{1000 \times 1 \times 10} {100} + \frac{1100\times 1 \times 10} {100}\) (Principal for 2nd year = Rs. 1000 + interest of 1st year)
=100 + 110
= Rs. 210
In this process we get Rs. 10 profit by the way of compound interest.
The following points should be remembered before calculating the compound interest.
- The compound interest for every succeeding year is always greater than the compound interest for the previous year.
- The amount of the previous year becomes the principal for the coming year.
- The final amount is equal to the sum of the original principal and the interest for all the years.
- The compound interest for the entire period is the sum of the interest for all the years that is the difference between the final amount and the original principal.
The installment is the regular interval of time in which the compound interest is calculated. The payment might be yearly, half-yearly, quarterly, monthly, daily etc. Here we use only yearly and half-yearly installments.
Derivation of yearly compound interest
| Year | Principal | Time | Rate | Interest | Amount |
| 1st | P | 1 year | R% | \(\frac {PR}{100}\) | \(P +\frac {PR}{100} = P((1 + \frac {R} {100})\) |
| 2nd | \(P (1 + \frac {R}{100})\) | 1year | R% | \(P\times(1 + \frac {R} {100})\times \frac {R}{100}\) |
\(P\times(1 + \frac {R} {100}) + P\times(1 + \frac {R} {100})\times \frac {R}{100}\) \(= P\times(1 + \frac {R} {100})\times (1 + \frac {R} {100})\) \(= P\times(1 + \frac {R} {100})^2\) |
| 3rd | \(P (1 + \frac {R}{100})^2\) | 1year | R% | \(P\times(1 + \frac {R} {100})^2\times \frac {R}{100}\) |
\(P\times(1 + \frac {R} {100})^2 + P\times(1 + \frac {R} {100})^2 \times\frac {R}{100}\) \(=P\times(1 + \frac {R} {100})^2 \times(1 + \frac {R} {100}) \) \(=P\times(1 + \frac {R} {100})^3\) |
| 4th | \(P\times(1 + \frac {R} {100})^{T-1} \) | 1year | R% | \(P\times(1 + \frac {R} {100})^{T-1} \times \frac {R} {100}\) |
\(P\times(1 + \frac {R} {100})^{T + 1-1}\) \(=P\times(1 + \frac {R} {100})^T \) |
So, the yearly compound amount for T years at R% p.a. = \(P\times \left (1 + \frac {R} {100}\right)^T \)
Compound interest for T years = Compound amount for T years - original principal
$$ = P \times \left(1 + \frac {R} {100}\right)^T - P $$
$$ =P \left \lbrace \left (1 + \frac {R} {100} \right )^T - 1 \right \rbrace$$
Half-yearly compound interest
In case of half-yearly compoundinterest, time will be double and rate will be halved.
Since yearly compound interest = \( P \left (1 + \frac {R} {100} \right ) ^ T - P \)
$$ = P \times \left(1 + \frac {R} {100}\right)^2T - P $$
$$ =P \left \lbrace \left (1 + \frac {R} {100} \right )^2T - 1 \right \rbrace$$
a. The amount of the previous year becomes the principal for the coming year.
b. The compound interest for every succeeding year is always greater than compound interest for the previous year.
c. The final amount is equal to the sum of the original principal and the interest for all the years.
d. The compound interest for the entire period is the sum of the interest for all the year that is a difference between the final amount and the original principal.
.Very Short Questions
Solution:
Principal(P) = Rs 5000,
Rate (R) = 12%
Time (T) = 2 years
Compound interest (C.I) = ?
Now,
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= 5000 \left[ \left( 1 + \frac{12}{100}\right)^2 - 1\right]\\ &= 5000[(1.12)^2 -1]\\ &= 5000[1.2544 - 1] \\ &= 5000 \times 0.2544\\ &= Rs \: 1272\: \: _{Ans.}\end{align*}
Solution:
Principal (P) = Rs 700
Compound interest (A) = Rs 847
Time (T) = 2 years
Rate of interest (R) = ?
\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ or, 847 &= 700\left(1 + \frac{R}{100}\right)^2 \\ or, \frac{847}{700} &= \left(1 + \frac{R}{100}\right)^2 \\ or, 1.21 &= \left(1 + \frac{R}{100}\right)^2\\ or,(1.1)^2 &= \left(1 + \frac{R}{100}\right)^2 \\ or, 1.1 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.1 - 1\\ or, R &= 0.1 \times 100 \\ \therefore R &= 10\% \:\:\:_{Ans.}\end{align*}
Solution:
Principal(P) = Rs 50,000
Time (T) = 2 years
Rate (R) = 10%
Compound interest (C.I) = ?
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^2T - 1\right]\\ &= 50,000\left[ \left( 1 + \frac{10}{100}\right)^{2 \times 2} - 1\right] \\ &= 50,000 [(1 + 0.05)^4 - 1]\\ &= 50,000 [1.2155 -1] \\ &= 50,000 \times 0.2155 \\ &= Rs \: 10775 \:\: \: _{Ans.} \end{align*}
Solution:
Principle (P) = Rs 12,000
Time (T) = 1 year
Month(M)=6 month
Rate (R) = 10%
Compound interest (C.I) = ?
\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T \left( 1 + \frac{MR}{1200}\right)- 1\right]\\ &= 12,000 \left[ \left( 1 + \frac{10}{100}\right)^1 \left( 1 + \frac{6 \times 10}{1200}\right) - 1 \right] \\ &= 12,000 [(1.1) (1.05) - 1]\\ &= 12,000\: [1.155 - 1] \\ &= 12,000\times 0.155 \\&= Rs 930\\ \end{align*} \(\therefore Compound \: interest = Rs \: 930 \:\: _{Ans.}\)
Solution:
Principle (P) = Rs 25,000
Time(T) = 2 years
Rate (R1) = 4%
Rate (R2) = 5%
Compound Amount (CA) = ?
\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right)
\\ &= 25000\left(1 + \frac{4}{100}\right) \left(1 + \frac{5}{100}\right) \\ &= 2500 \times 1.04 \times 1.05 \\ &= Rs \: 27300 \:\:\: _{Ans.} \end{align*}
Solution:
Compound Amount (CA) = Rs 5191.68
Time (T) = 1 year
Rate (R) = 8%
Principal (P) = ?
\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{200}\right)^{2T} \\ or, 5191..68 &= P \left(1 + \frac{8}{200}\right)^2\\ or, 5191.68 &= P(1.04)^2\\ or, P &= \frac{5191.68}{1.0816}\\ &= \: Rs \: 4800 \end{align*}
\(\therefore\) The principal (P) = Rs \: 4800 \(_{Ans.}\)
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Santosh borrowed Rs130000 from Suresh at the rate of 21% per annum. Find the simple interest at the end of 2 years and interest compounded yearly.
Rs 40000, Rs45000
Rs 60500, Rs 70000
Rs 46000, Rs 48222
Rs 54600,Rs 60333
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Ram borrowed Rs. 150000 from Sita at the rate of 21% per year. At the end of nine months how much compound interest should he pay in nine months compounded half yearly?
Rs 22678
Rs 20000
Rs 24451.87
Rs 21745
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A man deposited Rs 2000 in the fixed deposit account of the bank for two years at the rate of 12% per annum. The interest is compounded is semi-annually.How much will be the amount and the compound interest at the end of 2 years?
Rs 2289,489.3
Rs 2222.22, Rs 421.67
Rs 2524.95, Rs 524.95
Rs 2124.44,Rs 505.65
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The compound interest and simple interest on Rs. 7500 in 2 years at 10% pa.
Rs 65
RS 70
Rs 75
Rs 60
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Simple interest and annual compound interest on Rs 18000 for 2 yrs. at the rate of 15% pa.
Rs 405
Rs 305
Rs 205
Rs 505
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The difference between the compound interest and the simple interest on a certain sum at 10% per annum for 3 years is Rs 155.
Rs 48000
Rs 4660
Rs 45000
Rs 5000
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The compound interest on the sum of money in two years at the rate of10% per annum will be Rs 420 more than simple interest.Find the sum.
Rs 38000
Rs 4000
Rs35000
Rs 42000
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Bishal deposited Rs.10,000 in 'A' bank and Rs 10,000 in 'B' bank at the rate of 6% per annum.'A' bank pays interest compounded half-yearly whereas 'B' bank pays interest compounded yearly. Calculate the interest which bank had paid more and much at the end of 2years.
Rs 18
Rs 20
Rs 19
Rs 15
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If the difference between the compound interest compounded half yearly and yearly on a sum of money for 2 years at the rate of 20% per year is Rs 289.20, find the sum.
Rs 11500
Rs 13000
Rs 13500
Rs 12000
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The difference between the annual and semi-annual compound interest on a sum of money is Rs 482 at the rate of 20% per annum for 2 years.Find the sum.
Rs 16000
Rs 20000
Rs 14000
Rs 12000
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The compound interest on a certain sum for 2 years at 8% per annum compounded annually is Rs 312 Find the simple interest for the same time and the same rate .
Rs 275
Rs 150
Rs 300
Rs 200
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Accordingly to the yearly compound interest in what will the compound interest Rs 400000 at the rate of interest 6.5% per annum be Rs 53690?
6 years
2 years
5 years
1 years
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Accordingly to the yearly compound interest in what time will the compound interest on Rs 800000 at the rate of interest 5% annum be Rs 126100?
6 years
3years
2 years
10 years
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At what percent rate of compound interest per annum will the compound interest on Rs 125000 be Rs 91000 in 3 years?
20%
10%
17%
15%
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The simple interest on a sum of money in 2 yrs is Rs 36 less than the compound interest compounded annually. if the rate of interest is 12% p.a find the sum.
Rs 1200
Rs 2500
Rs 1500
Rs 2000
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PRATIK ACHARYA
if SI=1000 and CI=2050 and T=2years find principal and rate
Jan 24, 2017
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suraj gautam
The simple interest and the compound interest of a sum of money in 2 years are Rs 1000 and Rs 1050 respectively. Find annual rate of interest and the principal
Jan 21, 2017
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