Compound Interest

Subject: Compulsory Mathematics

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Overview

Compound interest can be defined as interest calculated on the initial principal and also on the accumulated interest of previous periods of a deposit or loan. Depreciation definition, decrease in value due to wear and tear, decay, the decline in price, etc.

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)

Scholarships after +2 Abroad Studies Opportunities

=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.

  1. The compound interest for every succeeding year is always greater than the compound interest for the previous year.
  2. The amount of the previous year becomes the principal for the coming year.
  3. The final amount is equal to the sum of the original principal and the interest for all the years.
  4. 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 = \( P \left (1 + \frac {R} {100} \right ) ^ T - P \)

$$ = P \times \left(1 + \frac {R} {100}\right)^2T - P $$

$$ =P \left \lbrace \left (1 + \frac {R} {100} \right )^2T - 1 \right \rbrace$$

Things to remember

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.

  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Compound Interest
Compound Interest Type 1 and Explanation
Compound Interest Type 2
Compound Interest Type 3
Compound Interest Type 4
Introduction to compound interest
Questions and Answers

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 \(_{Ans.}\)

Solution:

First compound amount (CA1) = Rs 1323
Time (T1) = 2 years
Second compound amount (CA2) = Rs 1389.15
Time (T2) = 3 years
We know that,

\begin{align*} CA &= P \left(1 + \frac{R}{100}\right)^T \\ 1323 &= P \left(1 + \frac{R}{100}\right)^2 \: \: \: \: \: ........ (1)\\ or, 1389.15 &= P \left(1 + \frac{R}{100}\right)^3 \:\:\:\:\: ........(2) \\ \end{align*}

\(\text{The equation (2 )is divided by equation (1) } \)

\begin{align*} \frac{1389.15}{1323} &= \frac{P \left(1 + \frac{R}{100}\right)^3 }{P \left(1 + \frac{R}{100}\right)^2 }\\ or, 1.05 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.05 - 1\\ or, R &= 0.05 \times 100\%\\ R &= 5\% \end{align*}

\(\text{Putting Value of R in}\:\: eq^n (1)\)

\begin{align*}P \left(1 + \frac{R}{100}\right)^2 &= 1323 \\ or, P \left(1 + \frac{5}{100}\right)^2 &=1323 \\ or, P \times 1.1025 &= 1323\\ P &= \frac{1323}{1.1025}\\ \therefore P &= 1200\\ \end{align*} \(\therefore Principle = Rs \: 1200 \:\: and \: rate = 5\% \)

Solution:

First compound interest (CA1) = Rs 7260
Time (T1) = 2 years
Second compound interest (CA2) = Rs 7986
Time (T2) = 3 years
We know that,

\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ 7260 &= P\left( 1 + \frac{R}{100}\right)^2 \: \: \: \: ....... (1) \\ 7986 &= P\left( 1 + \frac{R}{100}\right)^3 \:\:\:\: ....... (2)\\ \end{align*}

\( \text{The equation (2) is divide d by equation (1)}\)

\begin{align*} \frac{7986}{7260} &= \frac{P\left( 1 + \frac{R}{100}\right)^3 }{P\left( 1 + \frac{R}{100}\right)^2}\\ or, 1.1 &= 1 + \frac{R}{100}\\ or, \frac{R}{100} &= 1.1 - 1 \\ or, \frac{R}{100} &= 1.1 -1 \\ or, R &= 0.1 \times 100 \\ R &= 10\% \end{align*}

Solution:

Let, Principle (P) = Rs x
Time (T) = 2 years
Rate (R) = 5%

\begin{align*} Simple \: Interest \: (S.I) &= \frac{P\:T\:R}{100} \\ &= \frac{x \times 2 \times 5}{100} \\ &= x \times 0.1 \end{align*}

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= x \left[ \left( 1 + \frac{5}{100}\right)^2 - 1\right] \\ &= x [(1.05)^2 - 1] \\ &= x[1.1025 - 1]\\ &= x \times 0.1025 \end{align*}

From question,

\begin{align*} S.I + C.I &= Rs \: 202.50 \\ x \times 0.1 + x \times 0.1025 &= Rs \: 202.50\\ or, x(0.1 + 0.1025) &= Rs \: 202.50\\ x &= \frac{202.50}{0.2025} \\ \therefore x &= Rs \: 10000 \end{align*}

\(\therefore\) Principle (P) = Rs 10000 \(_{Ans.}\)

Solution:

First compound interest (CI1) = Rs 450
Time (T1) = 1 year
Second compound interest (C2) = Rs 945
Time (T2) = 2 years
Let Principal = P and rate = R
We know that,

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ C.I_1&= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right] \\ 450 &= P \left[ \left( 1 + \frac{R}{100}\right)^1 - 1\right] \: \: ....... (1) \\ C.I_2&= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right] \\ 945 &= P \left[ \left( 1 + \frac{R}{100}\right)^2 - 1\right] \:\:\: ....... (2)\end{align*}

Equation (2) is divided by equation (1)

\begin{align*} \frac{ P \left[ \left( 1 + \frac{R}{100}\right)^2 - 1\right]}{ P \left[ \left( 1 + \frac{R}{100}\right)^1 - 1\right]} &= \frac{945}{450}\\ or, \frac{\left[ \left( 1 + \frac{R}{100}\right) + 1\right] \left[ \left( 1 + \frac{R}{100}\right) - 1\right]}{\left[ \left( 1 + \frac{R}{100}\right)^1 - 1\right]} &= 2.1 \\ or, 2 + \frac{10}{100} &= 2.1\\ or, \frac{R}{100} &= 2.1 - 2 \\ or, R &= 0.1 \times 100\% \\ \therefore R &= 10\%\end{align*}

Putting the value of R in equation (1)

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right) - 1\right] \\ or, P \times \frac{10}{100} &= 450\\ or, P &= 450 \times 10\\ P &= 4500 \\ \\ \therefore The \: sum = 4500, rate (R) = 10\% \end{align*}

Solution:

Principal (P) = Rs. 46875
Compound interest (C.I.) = Rs. 5853
In 1-year compound interest of Rs 1 is 4 paisa.
In 1-year compound interest of Rs 100 is 4 × 100 paisa.
In 1-year compound interest of Rs 100 is Rs. \( \frac{4 \times 100}{100} \) = Rs 4

Interest rate (R) = 4%
Time (T) =?
We know that,

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ 5853 &= 46875 \left[ \left( 1 + \frac{4}{100}\right)^T - 1\right] \\ \frac{5853}{46875} &=\left( 1 + \frac{4}{100}\right)^T - 1\\ or, 0.124864 + 1 &=(1.04)^T\\ or, (1.04)^3 &= (1.04)^T \\ \therefore Time &= 3 \: years \end{align*}

Solution:

First compound amount (CA1) = Rs 8820
Time (T1) = 2 years
Second compounded amount (CA2) = Rs 9261
Time (T2) = 3 years
Let principal = P , Rate = R

\begin{align*} Compound \: amount \: (CA) &= P \left(1 + \frac{R}{100}\right)^T \\ 8820 &= P \left(1 + \frac{R}{100}\right)^2\:\:\: ...... (1)\\ 9261 &= P \left(1 + \frac{R}{100}\right)^3 \:\:\: ..... (2) \end{align*}

Equation (2) is divided by equation (1)

\begin{align*} \frac{P \left(1 + \frac{R}{100}\right)^3}{ P \left(1 + \frac{R}{100}\right)^2} &= \frac{9261}{8820} \\ or, 1 + \frac{R}{100} &= 1.05\\ or, \frac{R}{100} &= 1.05 - 1 \\ or, R &= 0.05 \times 100\% \\ R &= 5\% \end{align*}

Putting value of R in equation (1)

\begin{align*} P \left(1 + \frac{R}{100}\right)^T &= 8820 \\ or, P \left(1 + \frac{5}{100}\right) &= 8820\\ or, P \times (1.05)^2 &= 8820 \\ or, P \times 1.1025 &= 8820\\or,P &= \frac{8820}{1.1025}\\ P &= Rs \: 8000 \end{align*}

\(\therefore \) The principal = Rs 8000 and Rate = 5% \(_{Ans.} \)

Solution:

Time (T) = 3 years
Rate (R) = 20%
Principal (P) = ?
Compound Interest (C.I) - Simple Interest (S.I) = Rs. 384
We know that,

\begin{align*} S.I &= \frac{P \: T \: R}{100}\\ &= \frac{P \times 3 \times 20}{100}\\ &= P \times 0.6 \end{align*}

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= P \left[ \left( 1 + \frac{20}{100}\right)^3 - 1\right]\\ &=P\: [(1.2)^3 - 1]\\ &= P[1.728 - 1]\\ &= P \times 0.728 \end{align*}

From question,

\begin{align*} C.I - S.I &= Rs\: 384\\ P \times 0.728 - P \times 0.6 &= Rs\: 384 \\P\:(0.728 - 0.6)&= 384\\ or, P \times 0.128 &= 384\\ or, P &= \frac{384}{0.128}\\ \therefore P &= Rs\:3000 \:\:\:\: _{Ans.} \end{align*}

Solution:

For Anima,
Let, Principal (P) = Rs x,
Time (T) = 2 years
Rate (R) = 10%

\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{x \times2 \times10 }{100}\\ &= x \times 0.2 \end{align*}

For Eline,
Principal (P) = Rs 6000 - x
Time (T) = 2 years
Rate (R) = 5%

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= (6000 - x)\left[ \left( 1 + \frac{5}{100}\right)^2 - 1\right]\\ &= (6000 - x ) [(1.08)^2 - 1] \\ &=Rs \: 998.4 \times Rs \: 0.1664x \end{align*}

From question,

\begin{align*} C.I. - S.I &= Rs \: 50\\ or, 998.4 - x \times 0.1664 - x \times 0.2 &=50\\ or, 998.4 - x \times 0.3664&=50\\ or, -x \times 0.3664&=50 - 998.4\\ or, x &= \frac{-948.4}{-0.3664}\\ &= Rs\: 2588.43\end{align*}

For Anima sum = Rs 2588.43
For Eline sum = (6000 - 2588.43) = Rs 3411.57

Solution:

Principal (P) = Rs 170000
Rate (R) = 21%
Time (T) = 1 year 6 month = \( 1 + \frac{6}{12} = \frac{3}{2} \: years \)

\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{170000\times3 \times21 }{100\times 2}\\ &= Rs\: 53550 \:\:\:\:_{Ans.} \end{align*}

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ C.I &= 170000 \left[ \left( 1 + \frac{21}{100}\right)^{\frac{3}{2}} - 1\right] \\&= 170000 \left[ \left( \frac{121}{100}\right)^{\frac{3}{2}} - 1\right]\\ &= 170000\left[ \left( \frac{11}{10}\right)^{\frac{2 \times3}{2}} - 1\right] \\&= 170000[(1.1)^3 - 1]\\ &=170000[1.331 - 1]\\ &= Rs\: 170000 \times 0.331\\ &= Rs\: 56270 \:\:\:_{Ans.} \end{align*}

Solution:

Principle (P) = Rs 5120
Time (T) = 3 years
Rate (R) = 12.5%


\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{5120 \times3 \times12.5 }{100} \\ &= Rs\:1920\end{align*}

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= 5120 \left[ \left( 1 + \frac{12.5}{100}\right)^3 - 1\right]\\ &=5120 [1.4238281 - 1] \\ &= 5120 \times 0.4238281 \\&= Rs\: 2170 \end{align*}

\begin{align*}Difference &= C.I - S.I\\ &= 2170-1920\\ &= Rs\: 250 \end{align*}

Solution:

Principal (P) = Rs 4250
Rate (R) = 12%
Time (T) = 1 year

\begin{align*} Simple \: Interest \: (S.I) &= \frac{P \: T \: R}{100} \\ &= \frac{4250 \times 1 \times12 }{100}\\ &=Rs\: 510 \end{align*}

For half yearly,

\begin{align*} C.I &= P \left[ \left( 1 + \frac{R}{200}\right)^{2T} - 1\right]\\ &= 4250\left[ \left( 1 + \frac{12}{200}\right)^{2\times 1} - 1\right] \\ &= 4250[(1.06)^2 -1]\\ &= 4250 [1.1236-1]\\ &=4250 \times 0.1236\\ &= Rs \:525.30 \:\:_{Ans}\end{align*}

Solution:

Principal (P) = Rs 5000
Time (T) = 2 years
Rate (R) = 20%

\begin{align*} Compound\:Interest\:(C.I) &= P \left[ \left( 1 + \frac{R}{100}\right)^T - 1\right]\\ &= 5000\left[ \left( 1 + \frac{20}{100}\right)^2 - 1\right]\\ &= 5000[(1.2)^2 - 1]\\ &= 5000 \times 0.44 \\ &= Rs\:2200 \end{align*}

Let, Simple interest (S.I) = Rs 2200
Principle (P) = Rs 5000
Rate (R) = 20%
Time (T) = ?

\begin{align*} T &= \frac{I \times 100}{PR}\\ &= \frac{2200 \times 100}{5000 \times 20}\\ &= 2.2 \: years \end{align*}

Solution:

Principal (P) = Rs 5000
Rate (R) = 8%
Time (T) = 2 years
Compound amount (CA) = ?


\begin{align*} For\: yearly,\\ CA &= P \left(1 + \frac{R}{100}\right)^T \\ &= 5000 \left(1 + \frac{8}{100}\right)^2\\ &=5000 \times \frac{108}{100} \times \frac{108}{100} \\ &= Rs \: 5832 \:\:\: _{Ans.} \end{align*}

\begin{align*} For\: half \:yearly \: \\CA &= P \left(1 + \frac{R}{200}\right)^{2T} \\ &= 5000 \left(1 + \frac{8}{200}\right)^{2\times 2}\\ &= 5000 (1.04)^4\\ &= 5849.29 \end{align*}

\begin{align*} Difference \: between &= 5849.29 - 5832\\ &= Rs \:17.29 \:\:\: _{Ans.} \end{align*}

Solution:

For first part
First part (P1) = Rs 25200 - x
Time (T1) = 2 years
Rate (R1) = 10%

For second part
Second principal (P2) = Rs x
Time (T2) = 3 years
Rate (R3) = `10%

\begin{align*} Compound \: amount \: (CA_1) &= P \left(1 + \frac{R}{100}\right)^T \\ &=(25200 - x) \left(1 + \frac{10}{100}\right)^2\\ &= (25200 - x) (1.1)^2\\ &= (25200-x) \times 1.21 \end{align*}

\begin{align*} Compound \: amount \: (CA_2) &= P \left(1 + \frac{R}{100}\right)^T \\ &= x \left(1 + \frac{10}{100}\right)^3\\ &=x (1.1)^3 \\ &= x \times 1.331 \end{align*}

From Question,

\begin{align*} CA_1 &=CA_2 \\ (25200-x) \times 1.21 &= x\times 1.331 \\ or, 30492 - 1.21x &= x \times 1.331\\ or, 1.331x+1.21x&=30492\\ or, 2.541x &= 30492\\ x&= \frac{30492}{2.541}\\ \therefore x&=12000 \end{align*}

\begin{align*} 1^{st} \: principle &= 25200 -x\\&= 25200-12000\\&= Rs\:13200 \:\:\:_{Ans.} \end{align*}

\(2^{nd }\: principal = x= Rs 12000 \:\:\:_{Ans.}\)

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