Subject: Compulsory Mathematics

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.

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 2^{nd }year = Rs. 1000 + interest of 1^{st} 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.

Year | Principal | Time | Rate | Interest | Amount |

1^{st} |
P | 1 year | R% | \(\frac {PR}{100}\) | \(P +\frac {PR}{100} = P((1 + \frac {R} {100})\) |

2^{nd} |
\(P (1 + \frac {R}{100})\) | 1year | R% | \(P\times(1 + \frac {R} {100})\times \frac {R}{100}\) |
\(P\times(1 + \frac {R} {100}) + P\times(1 + \frac {R} {100})\times \frac {R}{100}\) \(= P\times(1 + \frac {R} {100})\times (1 + \frac {R} {100})\) \(= P\times(1 + \frac {R} {100})^2\) |

3^{rd} |
\(P (1 + \frac {R}{100})^2\) | 1year | R% | \(P\times(1 + \frac {R} {100})^2\times \frac {R}{100}\) |
\(P\times(1 + \frac {R} {100})^2 + P\times(1 + \frac {R} {100})^2 \times\frac {R}{100}\) \(=P\times(1 + \frac {R} {100})^2 \times(1 + \frac {R} {100}) \) \(=P\times(1 + \frac {R} {100})^3\) |

4^{th} |
\(P\times(1 + \frac {R} {100})^{T-1} \) | 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

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

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

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.

- 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.

Compute compound interest on Rs 5000 for 2 years at 12% per annum.

**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*}

At what rate percent compound interest will Rs 847 in 2 years?

**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*}

What is the compound interest on Rs 50,000 for 2 years at 10% per annum payable half yearly.

**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*}

Find the compound Interest on Rs 12000 for 1 year and 6 months at rate 10% per annum?

**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*} \(\therefore Compound \: interest = Rs \: 930 \:\: _{Ans.}\)

How much will Rs 25000 amount to 2 years compounded yearly. If the rate of the successive years be 4 % and 5 % respectively.

**Solution:**

Principle (P) = Rs 25,000

Time(T) = 2 years

Rate (R_{1}) = 4%

Rate (R_{2}) = 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*}

Find the sum of money on which the compound amount is Rs 5159.68 for 1 year at 8% per annum payable half yearly?

**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 \(_{Ans.}\)

A certain sum of money amount to Rs 1323 in 2 years and to Rs 1389.15 in 3 years. Find the compound rate of interest and the sum.

**Solution:**

First compound amount (CA_{1}) = Rs 1323

Time (T_{1}) = 2 years

Second compound amount (CA_{2}) = Rs 1389.15

Time (T_{2}) = 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*} \(\therefore Principle = Rs \: 1200 \:\: and \: rate = 5\% \)

According to the system of compound interest, a sum of money in 2 years amount to Rs 7260 and is 3 years amount to Rs 7986. Find the rate of interest.

**Solution:**

First compound interest (CA_{1}) = Rs 7260

Time (T_{1}) = 2 years

Second compound interest (CA_{2}) = Rs 7986

Time (T_{2}) = 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*}

The sum of simple interest and compound interest after 2 years is Rs 202.50 and the rate of interest is 10% per annum. Find the principle.

**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 \(_{Ans.}\)

The compound interest of a sum of money in 1 year and 2 years are Rs 450 and 945 respectively. Find the rate of interest compounded yearly and sum.

**Solution:**

First compound interest (CI_{1}) = Rs 450

Time (T_{1}) = 1 year

Second compound interest (C_{2}) = Rs 945

Time (T_{2}) = 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*}

A person took a loan of Rs 46,875. If the rate of the compound interest is 4 paisa per rupee per year, in how many years will the compound interest be Rs 5,853.

**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*}

The compound amount of a sum of money is 2 years is Rs 8820 and in 3 years is Rs 9261. Find the sum and rate of interest.

**Solution:**

First compound amount (CA_{1}) = Rs 8820

Time (T_{1}) = 2 years

Second compounded amount (CA_{2}) = Rs 9261

Time (T_{2}) = 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% \(_{Ans.} \)

The compound interest on a sum of money in 3 years at 20% per annum will Rs 384 more than simple interest. Find the sum.

**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*}

Prakash lend altogether Rs 6000 to Anima and Eline for 2 years. Anima agree to pay simple interest at 10% p.a. and Eline agree to pay compound interest at the rate of 8%. If Eline paid Rs 50 more than Amina as the interest. Find how much did he lent to each.

**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

Sita borrowed Rs 170000 from Radha at the rate of 21% per annum, at the end of 1 year 6 months.

(i) How much simple interest will she have to pay?

(ii) How much compound interest will she have to pay?

**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*}

What is difference between the compound interest and simple interest on Rs 5120 for 3 years at 12.5% per annum?

**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*}

Ram borrowed Rs 4250 from Shyam at the rate of 12% per annum at the and of one year.

(i) How much simple interest will Ram have to pay?

(ii) How much interest compounded half-yearly will Ram has to pay?

**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*}

What will be the compound interest on Rs 5000 for 2 years at the rate of 20% compound per annum? At what time the same interest on the same sum at the same rate of simple interest?

**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*}

Mohan deposited Rs 5000 at 8% p.a compound interest in the bank. Find the difference between the amounts compounded yearly and half yearly in two years.

**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*}

Divide Rs 25200 into two parts such that the amount of are part in 2 years is the same as the amount of second part in 3 years the rate of compound interest being 10% per annum in both the cases.

**Solution:**

For first part

First part (P_{1}) = Rs 25200 - x

Time (T_{1}) = 2 years

Rate (R_{1}) = 10%

For second part

Second principal (P_{2}) = Rs x

Time (T_{2}) = 3 years

Rate (R_{3}) = `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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