Compound Interest

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.

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

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

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

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.

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*}

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

0%
• 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 54600,Rs 60333

Rs 40000, Rs45000

Rs 60500, Rs 70000

Rs  46000, Rs 48222

Rs 21745

Rs 22678

Rs 20000

Rs 24451.87

• 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 2222.22, Rs 421.67

Rs 2289,489.3

Rs 2124.44,Rs 505.65

Rs 2524.95, Rs 524.95

Rs 75

Rs 65

Rs 60

RS 70

Rs 405

Rs 505

Rs 305

Rs 205

Rs 45000

Rs 48000

Rs 4660

Rs 5000

Rs 42000

Rs 4000

Rs35000

Rs 38000

Rs 18

Rs 15

Rs 20

Rs 19

Rs 12000

Rs 11500

Rs 13000

Rs 13500

Rs 16000

Rs 14000

Rs 12000

Rs 20000

Rs 150

Rs 300

Rs 200

Rs 275

1 years

5 years

6 years

2 years

3years

2 years

6 years

10 years

17%

10%

20%

15%

Rs 2500

Rs 2000

Rs 1500

Rs 1200