## Arithmetic Chapter 2 Compound Interest Solved Example Problems

We have already noticed that interests are of two types:

**Interest: **

- Simple Interest
- Compound Interest

In this chapter, we shall study only Compound interest.

**Definition:**

If the interest of a principal for a certain period of time (e.g., 3 months, 6 months, 1 year, etc) is added to the principal to get an amount and for the next period of time the interest is calculated on that amount taking as the principal, then this type of interest is called the compound interest.

Compound interest is simply denoted by C.I.

** WBBSE Solutions for Class 10 Maths**

**Conversion period or interest period:**

**Class 10 Maths Solutions Wbbse**

The conversion period or interest period is a certain period of time at the end of which interest is calculated and added to the principal to get an amount which is treated as the new principal for the next or second period of time.

It is called the interest period. If no conversion period or interest period is noticed, then we shall have to take the interest period as 1 year.

Here, it is a must that for a certain principal for a certain period of time with a certain rate of interest, the compound interest is always greater than the simple interest.

For example, let a person takes a loan of ₹ 100 for 2 years at the rate of simple interest 10% per annum.

Now, the simple interest after 2 years = \(₹ \frac{100 \times 10 \times 2}{100}\) = ₹ 20.

So, to repay the loan, the person shall have to pay ₹ (100 + 20) = ₹ 120.

Now, if the person loans ₹100 for 1 year, then he have to pay back ₹(100 + 10) = ₹110.

But if he loans ₹100 for two years, i.e., for 1 year more, then he will have to pay back ₹110 and the simple interest of ₹110 for 1 year.

Now, the simple interest of ₹110 for 1 year at the rate of 10% per annum

= \(₹ \frac{110 \times 10 \times 1}{100}\)

Thus, after 2 years, the person will have to pay back ₹ (110 + 11) = ₹121 so as to clear his debt.

We can see that in the first case, i.e., for the simple interest, the interest is ₹ 20 and in the second case, i.e., for the compound interest, it is ₹(121- 100) = ₹21.

Clearly, in the second case the interest is ₹(21-20) =₹ 1 more.

**Class 10 Maths Solutions Wbbse**

**To determine the compound interest **

Determine the compound interest of ₹ 500 for 2 years at the rate of 4% per annum.

Solution: For the first year, the principal = ₹ 500.

∴ Interest for the first year = \(\frac{500 \times 4 \times 1}{100}\) = ₹ 20

∴ Amount of the first year = ₹ (500 + 20) = ₹ 520.

∴ For next second year the principal = ₹ 520.

∴ Interest for the second year = \(₹ \frac{520 \times 4 \times 1}{100}\) =₹ 20.8

∴ Amount at the end of 2 years = ₹ (520 + 20.8) = ₹ 540-8.

∴ The required compound interest = ₹ (540.8 – 500) = ₹ 40-8.

**Algebraic formulas regarding compound interest**

Let the principal be ₹ P and the rate of compound interest be r% per annum.

∴ Interest of the first year = \(\frac{P \times r \times 1}{100}\)

= \(₹ \frac{Pr}{100}\)

∴ Amount of the first year = \(₹\left(P+\frac{P r}{100}\right)\)

= \(₹\left(1+\frac{r}{100}\right)\)

∴ Principal of the 2nd year = \(₹\left(1+\frac{r}{100}\right)\)

∴ Interest of the 2nd year = \(₹ \frac{P\left(1+\frac{r}{100}\right) \times r \times 1}{100}\)

= \(₹ \mathrm{P}\left(1+\frac{r}{100}\right) \times \frac{r}{100}\)

∴ Amount of the 2nd year

= \(₹ \mathrm{P}\left(1+\frac{r}{100}\right)+₹ \mathrm{P}\left(1+\frac{r}{100}\right) \times \frac{r}{100}\)

= \(₹\mathrm{P}\left(1+\frac{r}{100}\right)\left(1+\frac{r}{100}\right)\)

= \(₹ \mathrm{P}\left(1+\frac{r}{100}\right)^2\)

Similarly, the amount of the 3rd year = \(₹P\left(1+\frac{r}{100}\right)^3\)

According to the above trend, we find that for the principal ₹ P at the rate of r% per annum after n years, the amount

= \(₹ \mathrm{P}\left(1+\frac{r}{100}\right)^n\)

**Maths Solutions Class 10 Wbbse**

**Formula 1**

Amount of principal ₹P on n years at the rate of r % compound interest per annum

= \(=\mathrm{P}\left(1+\frac{r}{100}\right)^n\)

**Formula 2**

Interest of principal ₹P on n years at the rate of r % compound interest per annum

= \(=₹\left\{P\left(1+\frac{r}{100}\right)^n-\mathrm{P}\right\}\)

**Formula 3**

Interest of principal ₹P on n years at the rate of r % compound interest per annum when interests are calculated m times per year

= \(=₹ P\left(1+\frac{\frac{r}{m}}{100}\right)^{m n}\)

**Formula 4 **

If m = 2 in formula 3 ie.., intersects are calculated 2 times per year which means interests are calculated after every 6 months, then total amount after n years

= \(P\left(1+\frac{\frac{r}{2}}{100}\right)^{2 n}\)

**Formula 5**

If m = 4 in formula 3 ie.., intersects are calculated 4 times per year which means interests are calculated after every 4 months, then total amount after n years

= \(=P\left(1+\frac{\frac{r}{4}}{100}\right)^{4 n}\)

**Formula 6**

If the annual rate of compound interest in 1st year be r_{1}%, in 2nd year be r_{1}%, in 3rd year be r_{1}% and ……. at last in n-th year be r_{n}%, then the total amount of principal ₹ P in total n years

= \(₹ \mathrm{P}\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right) \ldots \ldots \ldots \ldots \ldots\left(1+\frac{r_n}{100}\right)\)

**Maths Solutions Class 10 Wbbse**

**Formula 7**

If the principal be ₹ P at the rate of r% compound interest per annum and if the period of time be a fraction like 4 \(\frac{2}{3}\) years, then the amount

= \(\mathrm{P}\left(1+\frac{r}{100}\right)^4 \cdot\left(1+\frac{\frac{2}{3} r}{100}\right)\)

**Class 10 Maths Solutions Wbbse**

## Arithmetic Chapter 2 Compound Interest Multiple Choice Questions

In the following examples how different types of compound interest are calculated have been discussed thoroughly.

**Example 1. The sum of the principal and compound interest for a fixed period of time is termed as**

- Compound interest
- Amount
- Simple interest
- Total interest

Solution: 2. Amount

**Example 2. If principal is p and annual rate of compound interest is r%, then the amount for 3 years will be**

- \(₹ r\left(1+\frac{p}{100}\right)^3\)
- \(₹ 3\left(1+\frac{r}{100}\right)^p\)
- \(₹ p\left(1+\frac{r}{100}\right)^3\)
- \(₹ p\left(1+\frac{r}{100}\right)^2\)

Solution: 3. \(₹ p\left(1+\frac{r}{100}\right)^3\)

**Example 3. The amount on ₹1000 for 2 years at the rate of 5% compound interest per annum is**

- ₹1102.50
- ₹1120.50
- ₹1021.50
- ₹1202.50

Solution: 1. ₹1102.50.

The required amount

= \( ₹ 1000\left(1+\frac{5}{100}\right)^2\)

= \( ₹ 1000\left(1+\frac{1}{20}\right)^2\)

= \(₹ 1000 \times \frac{21}{20} \times \frac{21}{20}\) =₹ 1102.50

**Class 10 Maths Solutions Wbbse**

**Example 4. If the rate of compound interest for the first year is 4% and 2nd year is 5%, where the compound interest on ₹ 25000 for 2 years is**

- ₹ 3200
- ₹ 2300
- ₹ 2302
- ₹2310

Solution: 2. ₹ 2300

Amount after 2 years = ₹ 25000 \(\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\)

= \(₹ 25000 \times \frac{26}{25} \times \frac{21}{20}\) = ₹ 27300

∴ The required compound interest = ₹ 27300 – ₹ 25000 = ₹ 2300

**Example 5. At 5% compound interest per annum the compound interest on ₹ 10000 for 3 years is**

- ₹1567.25
- ₹ 1567.52
- ₹ 1657.25
- ₹ 1576.25

Solution: 4. ₹ 1576.25.

The required compound interest

\(\begin{aligned}& =₹ 10000\left(1+\frac{5}{100}\right)^3-₹ 10000 \\

& =₹ 10000\left(1+\frac{1}{20}\right)^3-₹ 10000 \\

& =₹ 10000\left\{\left(\frac{21}{20}\right)^3-1\right\} \\

& =₹ 10000\left(\frac{9261}{8000}-1\right) \\

& =₹ 10000 \times \frac{1261}{8000} \\

& =₹ \frac{10 \times 1261}{8}=₹ 1576.25

\end{aligned}\)

## Arithmetic Chapter 2 Compound Interest Very Short Answer Type Questions

**1. Write True Or False **

**Example 1. If principal = ₹ p, rate of compound interest per annum = r% and time = n years, then the amount after 2 years is \(₹ p\left(1+\frac{r}{100}\right)^2\)**

Solution: If principal = ₹p and rate of compound interest per annum for first, 2nd and 3rd year be r_{1}%, r_{2}% and r_{3}% respectively, then total amount for 3 years

Solutions:

1. True, since simple interest after 1 year = \(₹ \frac{p \times r \times 1}{100}=₹ \frac{p r}{100}\)

∴ amount after 1 year = \(₹ \mathrm{p}+₹ \frac{\mathrm{pr}}{100}=₹ \mathrm{p}\left(1+\frac{r}{100}\right)\)

Again, simple interest of new principal \(₹ \mathrm{p}\left(1+\frac{r}{100}\right)\)

for 2nd year \(₹ \frac{p\left(1+\frac{\mathrm{r}}{100}\right) \times r \times 1}{100}\)

=\(₹ \frac{\operatorname{pr}\left(1+\frac{\mathrm{r}}{100}\right)}{100}\)

∴ Amount after 2 year

= \(₹ p\left(1+\frac{\mathrm{r}}{100}\right)+₹ \frac{\operatorname{pr}\left(1+\frac{\mathrm{r}}{100}\right)}{100}\)

= \(=₹ \mathrm{p}\left(1+\frac{\mathrm{r}}{100}\right)\left(1+\frac{\mathrm{r}}{100}\right)\)

= \(₹ \mathrm{p}\left(1+\frac{\mathrm{r}}{100}\right)^2\)

Hence amount after 2 years = \(=p\left(1+\frac{r}{100}\right)^2\)

2. False, since the total amount for 3 years

= \(₹ p\left(1+\frac{r_1}{1.00}\right)\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)\)

**Maths Solutions Class 10 Wbbse**

**2. Fill in the blanks **

**Example 1. Principal and compound interest are ______ proportional.**

Solution: Directly

**Example 2. Interest = Amount.**

Solution: Principal

**Example 3. If principal = ₹ p. rate of compound interest be r% per annum and time period be n years,then _____=\(\frac{prt}{100}\) prt**

Solution: Interest

## Arithmetic Chapter 2 Compound Interest Short Answer Type Questions

**Example 1. If principal = ₹ p and rate of compound interest per annum for first, 2nd and 3rd year are r _{1}%, r_{2}% and r_{3}% respectively, then find total amount for first year.**

Solution: Interest for 1 year = \(₹ \frac{\mathrm{p} \times \mathrm{r}_1 \times 1}{100}\)

= \(₹ \frac{\mathrm{pr}_1}{100}\)

Total amount for first year = \(₹ \frac{\mathrm{pr}_1}{100}\)

**Example 2. Calculate the amount on ₹ 20000 at the rate of 5% compound interest per annum for 2 years.**

Solution: The required amount = \(₹ 20000\left(1+\frac{5}{100}\right)^2\)

= \(₹ 20000\left(1+\frac{1}{20}\right)^2\)

= \(₹ 20000 \times \frac{21}{20} \times \frac{21}{20}\) =₹ 22050

The required amount =₹ 22050

**Example 3. If the rate of compound interest of principal ₹ p is r% per annum and the interest is compounded quarterly, i.e. the number of phase of compound interest in a year is 4, then find the amount for n years.**

Solution: The required amount

= \(₹ \mathrm{p}\left(1+\frac{\frac{\mathrm{r}}{4}}{100}\right)^{4 \mathrm{n}}\)

= \(₹ \mathrm{p}\left(1+\frac{\mathrm{r}}{400}\right)^{4 \mathrm{n}}\)

The amount for n years = \(₹ \mathrm{p}\left(1+\frac{\mathrm{r}}{400}\right)^{4 \mathrm{n}}\)

**Maths Solutions Class 10 Wbbse**

**Example 4. If the compound interest of principal ₹p is a% per annum and the interest is compounded half-yearly, then find the amount in n years.**

Solution: The required amount

=\(₹ \mathrm{p}\left(1+\frac{\frac{\mathrm{a}}{2}}{100}\right)^{2 \mathrm{n}}\)

= \(₹ \mathrm{p}\left(1+\frac{\mathrm{a}}{200}\right)^{2 \mathrm{n}}\)

The amount in n years = \(₹ \mathrm{p}\left(1+\frac{\mathrm{a}}{200}\right)^{2 \mathrm{n}}\)

**Example 5. Calculate the amount on ₹5000 at the rate of 8% compound interest per annum for 3 years.**

Solution: The required amount

= \(₹ 5000\left(1+\frac{8}{100}\right)^3=₹ 5000\left(1+\frac{2}{25}\right)^3\)

= \(₹ 5000 \times\left(\frac{27}{25}\right)^3\) =₹ 6298.56

The required amount =₹ 6298.56

**Example 6. In how much time ₹64000 will amount ₹68921 at the rate of 10% compound interest per annum if interest is compounded quarterly?**

Solution: Here, p = ₹64000, A = ₹68921, r = 10.

Let the required time = n years

As per question,

A = \(p\left(1+\frac{\frac{\mathrm{r}}{4}}{100}\right)^{4 \mathrm{n}}\)

⇒ \(₹ 64000\left(1+\frac{\frac{10}{4}}{100}\right)^{4 n}\) =₹ 68921

or, \(\left(1+\frac{1}{40}\right)^{4 n}=\frac{68921}{64000}\)

or, \(\left(\frac{41}{40}\right)^{4 n}=\left(\frac{41}{40}\right)^3\)

⇒ 4n = 3

⇒ n = \(\frac{3}{4}\) = \(\frac{3}{4}\) x 12 months = 9 months

Hence the required time = 9 months.

**Maths Solutions Class 10 Wbbse**

**Example 7. Rina took a loan at the rate of 15% compound interest per annum. If she refunded ₹1290 after 2 years, then what sum of money did Rina take?**

Solution: Let the principal = ₹p.

As per the question,

\(₹ \mathrm{p}\left(1+\frac{15}{100}\right)^2=₹ 1290+₹ \mathrm{p}\)

or, \(\mathrm{p}\left(1+\frac{3}{20}\right)^2=1290+\mathrm{p} \text { or, } \mathrm{p}\left(\frac{23}{20}\right)^2=1290+\mathrm{p}\)

or, \(\mathrm{p}\left\{\left(\frac{23}{20}\right)^2-1\right\}=1290 \text { or, } \mathrm{p}\left(\frac{529}{400}-1\right)=1290\)

or, \(\mathrm{p} \times \frac{129}{400}=1290 \Rightarrow \mathrm{p}=\frac{1290 \times 400}{129} \Rightarrow \mathrm{p}=4000\)

Hence Rina took a loan of ₹4000.

**Example 8. At what rate of compound interest per annum ₹10000 will amount ₹****13310 in 3 years?**

Solution: Let the rate be r% per annum

As per the question,

\(₹ 10000\left(1+\frac{r}{100}\right)^3=₹ 13310\)

or, \(\left(1+\frac{r}{100}\right)^3=\frac{13310}{10000}=\frac{1331}{1000}\)

or, \(\left(1+\frac{r}{100}\right)^3=\left(\frac{11}{10}\right)^3\)

⇒ \(1+\frac{r}{100}=\frac{11}{10}\)

⇒ \(\frac{r}{100}\) = \(\frac{11}{10}-1\)

⇒ \(\frac{r}{100}=\frac{1}{10}\)

or, r=10

Hence, the required rate = 10%

**Maths Solutions Class 10 Wbbse**

**Example 9. At what rate of compound interest per annum a sum of money ₹32000 will amount ₹39753.50 in 1\(\frac{1}{2}\) years. When interest is compounded half-yearly?**

Solution: Let the required rate = r%.

As per question,

\(₹ 32000\left(1+\frac{\frac{r}{2}}{100}\right)^{2 \times \frac{3}{2}}=₹ 39753 \cdot 50\)

or, \(\left(1+\frac{\mathrm{r}}{200}\right)^3=\frac{39753 \cdot 50}{32000}\)

or, \(\left(1+\frac{\mathrm{r}}{200}\right)^3=\frac{397535}{320000}\)

or, \(\left(1+\frac{\mathrm{r}}{200}\right)^3=\left(\frac{43}{40}\right)^3\)

⇒ \(1+\frac{\mathrm{r}}{200}=\frac{43}{40}\)

or, \(\frac{\mathrm{r}}{200}=\frac{43}{40}-1\)

or, \(\frac{r}{200}=\frac{3}{40}\)

or, r=15

Hence the require rate = 15%

**Example 10. In how much time R 6400 will amount R 6561 at the rate of 5% compound interest when interest is compounded quarterly?**

Solution: Let the required time = n years.

As per question, \(₹ 6400\left(1+\frac{\frac{5}{4}}{100}\right)^{4 \mathrm{n}}=₹ 6561\)

or, \(\left(1+\frac{1}{80}\right)^{4 n}=\frac{6561}{6400}\)

or, \(\left(\frac{81}{80}\right)^{4 n}=\frac{6561}{6400}\)

or, \(\left(\frac{81}{80}\right)^{4 n}=\left(\frac{81}{80}\right)^2\)

⇒ 4n = 2

⇒ n= \(\frac{2}{4}\) = \(\frac{2}{4}\) x 12 months = 6 months

Hence the required time = 6 months.

## Arithmetic Chapter 2 Compound Interest Long Answer Type Questions

**Example 1. At the rate of compound interest 5% per annum, find the amount of ₹ 80000 in 2\(\frac{1}{2}\) years.**

Solution:

**Given:**

At the rate of compound interest 5% per annum

Here, the amount of principal (P) =₹ 80000,

at the rate of compound interest(r%) = 5% in 2 years

= \( ₹ 80000 \times\left(1+\frac{5}{100}\right)^2\)

= \(₹ 80000 \times\left(\frac{105}{100}\right)^2\)

= \(₹ 80000 \times \frac{105 \times 105}{100 \times 100}\) =₹ 88200 .

Then after 2 years the principal = ₹ 88200.

Now, at the rate of 5% per annum, the interest in next \(\frac{1}{2}\) year

= \(₹ \frac{88200 \times 5 \times 1}{2 \times 100}\) = ₹ 2205

∴ The amount in 2 \(\frac{1}{2}\) years = ₹(88200 + 2205) = ₹90405.

Hence, the required amount = ₹90405.

**Wbbse Class 10 Maths Solutions**

**Example 2. Anil has taken a loan of ₹2000 for 2 years at the rate of compound interest of 6% per annum. Find the compound interest he has to pay after 2 years. **

Solution:

**Given:**

Anil has taken a loan of ₹2000 for 2 years at the rate of compound interest of 6% per annum.

Anil has taken a loam of ₹2000.

∴ Here principal (P) = ₹2000.

Rate of compound interest per annum = 6%

Period of time = 2 years.

∴ After 2 years, the amount

= \(₹ 2000 \times\left(1+\frac{6}{100}\right)^2\)

= \(₹ 2000 \times\left(\frac{106}{100}\right)^2\)

= \(₹ \frac{2000 \times 106 \times 106}{100 \times 100}\)

= ₹ 2247

∴ The compound interest = ₹ (2247.2 – 2000) = ₹ 247.2

Hence, Anil-has to pay compound interest ₹247.2 after 2 years.

**Example 3. A poultry farmer took a loan of some money for 2 years at the rate of compound interest of 10%. If the compound interest paid by him be ₹3150, then find the quantity of loan Poultry farmer had lended.**

Solution:

**Given:**

A poultry farmer took a loan of some money for 2 years at the rate of compound interest of 10%. If the compound interest paid by him be ₹3150

Let the quantity of loan be ₹x.

∴ The amount after 2 years of ₹x at the rate of compound interest of 10% per annum.

=\(₹ x \times\left(1+\frac{10}{100}\right)^2\)

= \(₹ x \times\left(\frac{110}{100}\right)^2 \cdot\)

The compound interest

= \(₹\left\{x\left(\frac{110}{100}\right)^2-x\right\}\)

= \(₹\left\{x\left[\frac{(110)^2}{(100)^2}-1\right]\right\}\)

= \(₹\left\{x\left[\frac{(110)^2-(100)^2}{(100)^2}\right]\right\}\)

= \(₹ \frac{x(110+100)(110-100)}{100 \times 100}\)

= \(₹ x \times \frac{210 \times 10}{100 \times 100}\)

As per question,

\(x \times \frac{210 \times 10}{100^2}=3150\)

⇒ x = \(₹ \frac{3150 \times 100 \times 100}{210 \times 10}\) =₹ 15000 .

Hence the required pricipal = ₹ 15000

**Example 4. Find the principal which becomes ₹2979 after getting 10% compound interest per annum for 3 years.**

Solution: Let the principal = ₹x

∴ The amount of ₹x in 3 years at the rate of compound interest of 10% per annum

= \(₹ x \times\left(1+\frac{10}{100}\right)^3\)

= \(₹ x \times\left(\frac{11}{10}\right)^3\)

= \(₹ x \times \frac{1331}{1000}\)

∴ The compound interest

= \(₹\left(x \times \frac{1331}{1000}-x\right)\)

=\(₹ \frac{1331 x-1000 x}{1000}=₹ \frac{331 x}{1000}\)

As per question, \(\frac{331 x}{1000}\) = 2979

⇒ x = \(\frac{2979 \times 1000}{331}\) = 9000

Hence, the required principal = ₹ 9000.

**Wbbse Class 10 Maths Solutions**

**Example 5. Subhasis deposited some money in State Bank at the rate of 5% compound interest and he received amount ₹46305 after 2 years. Find how much money Subhasis had deposited in State Bank.**

Solution:

**Given:**

Subhasis deposited some money in State Bank at the rate of 5% compound interest and he received amount ₹46305 after 2 years.

Let Subhasis had deposited ₹x in the State Bank.

∴ The amount of ₹x in 2 years at the rate of compound interest of 5% per annum,

= \(₹ x \times\left(1+\frac{5}{100}\right)^2=₹ x \times\left(\frac{105}{100}\right)^2\)

= \(₹ x \times \frac{105 \times 105}{100 \times 100}\)

As per question, \(x \times\left(\frac{105}{100}\right)^2=46305\)

⇒ \(x \times \frac{105 \times 105}{100 \times 100}=46305\)

⇒ \(x=\frac{46305 \times 100 \times 100}{105 \times 105}=42000\)

Hence Subhasis had deposited ₹42000 in the state bank.

**Example 6. Determine the difference between compound interest and simple interest on ₹ 8,000 for 3 years at 5% per annum**

Solution: The amount on ₹ 8,000 for 3 years at 5% compound interest per annum

= \(₹ 8,000 \times\left(1+\frac{5}{100}\right)^3\)

= \(₹ 8,000 \times\left(\frac{105}{100}\right)^3\)

= \(₹ 8,000 \times \frac{105 \times 105 \times 105}{100 \times 100 \times 100}\) =₹ 9261

∴ The compound interest = ₹(9,261 – 8,000) = ₹1,261.

Again, the simple interest for 3 years of ₹8,000 at 5% per annum

= \(=₹ \frac{8000 \times 5 \times 3}{100}\) = ₹ 1,200

Hence, the required difference = ₹ (1,261 – 1,200) = ₹ 61.

**Example 7. ****Find the sum of money if the difference between compound interest and simple interest for 2 years at the rate of 8% interest per annum is ₹ 96.**

Solution: Let the sum of money be ₹ x.

∴ At the rate of 8% interest per annum the amount of ₹ x for 2 years = \(=₹\left(1+\frac{8}{100}\right)^2\)

∴ The compound interest

= \(₹\left\{x\left(1+\frac{8}{100}\right)^2-x\right\}=₹ x\left\{\left(1+\frac{8}{100}\right)^2-1\right\}\)

= \(₹ x\left\{\left(1+\frac{8}{100}\right)^2-(1)^2\right\}=₹ x\left\{\left(1+\frac{8}{100}+1\right)\left(1+\frac{8}{100}-1\right)\right\}\)

= \(₹ x\left(2+\frac{8}{100}\right) \times \frac{8}{100}=₹ x \times \frac{208}{100} \times \frac{8}{100}\)

= \(₹ \frac{1664 x}{10000}\)

Again, at the rate of 8% interest per annum, the simple interest of ₹ x for 2 years

= \( ₹ \frac{x \times 8 \times 2}{100}\)

= \(₹ \frac{8 x}{50}\)

∴ The required sum of money = ₹15000.

**Example 8. Find the sum of money if the difference between compound interest and simple interest for 3 years becomes ₹775 at the rate of 10% per annum.**

Solution: Let the sum of money be ₹ x.

∴ At the rate of 10% compound interest per annum, the amount of ₹x for 3 years

= \(₹ x\left(1+\frac{10}{100}\right)^3\)

= \(₹ x\left(1+\frac{1}{10}\right)^3=₹ x \times\left(\frac{11}{10}\right)^3\)

= \(₹ \frac{1331 x}{1000}\)

∴ Compound interest

= \(₹\left(\frac{1331 x}{1000}-x\right)\)

= \( ₹ \frac{1331 x-1000 x}{1000}\)

= \(₹ \frac{331 x}{1000}\)

Again at the rate of 10% interest per annum, the simple interest of ₹ x for 3 years.

= \(₹ \frac{x \times 10 \times 3}{100}\)

=\(₹ \frac{30 x}{100}\)

As per question, \(\frac{331 x}{1000}-\frac{30 x}{100}=775\)

⇒ \(\frac{331 x-300 x}{1000}=775\)

⇒ 31 x=775 x 1000

⇒ x= \(\frac{775 \times 1000}{31}=25000\)

∴ The required sum of money = ₹ 25000.

**Wbbse Class 10 Maths Solutions**

**Example 9. If the rate of compound interest for the first and second year are 5% and 10% respectively, then find the compound interest on ₹5000 for 2 years.**

Solution: The principal at the beginning = ₹5000.

The rate of compound interest = 5%; period of time = 1 year.

∴ Amount after first year =\(₹ 5000\left(1+\frac{5}{100}\right)\)

= \(₹ 5000\left(1+\frac{1}{20}\right)=₹ 5000 \times \frac{21}{20}=₹ 5250\)

∴ Principal at the end of the second year = ₹5250.

The rate of compound interest = 10%

Period of time = 1 year.

∴ The amount after second year = ₹5250(1 + \(\frac{10}{100}\)

∴ Compound interest = ₹(5775 – 5000). = ₹775.

∴ The required compound interest of ₹5000 for 2 years is 775.

**Example 10. If the rate of compound interest for the first year be 3%, for the second year be 2% and for the third year be 1%, then find the sum of money, the amount of which for 3 years is ₹5305.53.**

Solution: Let the sum of money be ₹x.

∴ At the rate of 3% interest per annum, the amount of ₹x after first year

= \(₹ x \times\left(1+\frac{3}{100}\right)\) = \(₹ \frac{103 x}{100}\)

for the second year, the prinicipal = ₹ \(\frac{103 x}{100}\)

Again, at the rate of 2% compound interest per annum,

the mount of, ₹ \(\frac{103 x}{100}\) for 1 year

= \(₹ \frac{103 x}{100}\left(1+\frac{2}{100}\right)\)

= \(₹ \frac{103 x}{100} \times \frac{102}{100}\)

∴ For the third year the pricipal = \(₹ \frac{103 x}{100} \times \frac{102}{100}\)

At last, at the rate of 1% compound interest per annum,

the amoun of, \(₹ \frac{103 x}{100} \times \frac{102}{100}\) for year

= \(₹ \frac{103 x}{100} \times \frac{102}{100} \times\left(1+\frac{1}{100}\right)\)

= \(₹ \frac{103 x}{100} \times \frac{102}{100} \times \frac{101}{100}\)

= \(₹ \frac{101 \times 102 \times 103}{100 \times 100 \times 100} \cdot x\)

As per question, \(\frac{101 \times 102 \times 103}{100 \times 100 \times 100} . x=5305.53\)

⇒ x= \(\frac{5305.53 \times 100 \times 100 \times 100}{101 \times 102 \times 103}\)

⇒ x = 5000

∴ The required sum of money = ₹5000.

**Wbbse Class 10 Maths Solutions**

**Example 11. If the simple interest of a certain sum of money for 1 year is X50 and compound interest for 2 years is X 102, then find the sum of money and the rate of interest.**

Solution: Let the principal be X xand the rate of interest be r% per annum,

∴ at the rate of r %interest per annum, the simple interest of 1 year

= \(₹ \frac{x \times r \times 1}{100}\) = \(₹ \frac{r x}{100}\)

As per question, \(\frac{rx}{100}\) = 50

⇒ rx = 5000…..(1)

Again, at the rate of r% compound interest per annum of ₹x in 2 years, the amount

= ₹\(x\times\left(1+\frac{r}{100}\right)^2\)

∴ The compound interest

= \(₹\left\{x \times\left(1+\frac{r}{100}\right)^2-x\right\}\)

= \(₹ x\left[\left(1+\frac{r}{100}\right)^2-1\right]\)

= \(₹ x\left[\left(1+\frac{r}{100}+1\right)\left(1+\frac{r}{100}-1\right)\right]\)

= \(₹ x\left[\left(2+\frac{r}{100}\right) \times \frac{r}{100}\right]\)

= \(₹ \frac{x r}{100}\left(2+\frac{r}{100}\right)\)

= \(₹ \frac{5000}{100}\left(2+\frac{r}{100}\right) \quad[\text { by (1)] }\)

= \(₹ 50\left(2+\frac{r}{100}\right)=₹\left(100+\frac{r}{2}\right)\)

As per question, 100 + \(\frac{r}{2}\)=102

⇒ \(\frac{r}{2}\) =102-100

⇒ \(\frac{r}{2}\) = 2

⇒ r = 4

from (1) we get, 4x = 5000 [r = 4]

⇒ x = \(\frac{5000}{4}\) ⇒ x = 1250

∴The required sum of money = ₹1250 and rate of interest = 4% per annum.

**Example 12. If simple interest and compound interest of a certain sum of money for two years are ₹8400 and ₹8652 respectively, then find the sum of money and the rate of interest.**

Solution: Let the principal be ₹x and the rate of interest be r% per annum.

∴ The simple interest of ₹x in 2 years at the rate of r% per annum

= \(₹ \frac{x \times r \times 2}{100}=₹ \frac{2 x r}{100}\)

Again, the amount of ₹x for 2 years at the rate of r % per annum

= \(₹\left\{x \times\left(1+\frac{r}{100}\right)^2-x\right\}\)

= \(₹ x\left\{\left(1+\frac{r}{100}\right)^2-(1)^2\right\}\)

= \(₹ x\left(1+\frac{r}{100}+1\right)\left(1+\frac{r}{100}-1\right)\)

= \(₹ x\left(2+\frac{r}{100}\right) \times \frac{r}{100}\)

As per question, \(\frac{2xr}{100}\) = 8400

⇒ \(\frac{xr}{100}\) = 42000 ….(1)

and \(\frac{xr}{100}\) \(\left(2+\frac{r}{100}\right)\) = 8652 [by (1)]

⇒ \(4200\left(2+\frac{r}{100}\right)=8652\)

⇒ \(2+\frac{r}{100}=\frac{8652}{4200}\)

⇒ \(\frac{r}{100}=\frac{8652}{4200}-2\)

⇒ \(\frac{r}{100}=\frac{8652-8400}{4200}\)

⇒ \(r=\frac{252 \times 100}{4200}\)

⇒ r=6

Then, from(1) we get, \(\frac{x \times 6}{100}\)=4200[r=6]

⇒ \(x=\frac{4200 \times 100}{6}\)

⇒ x=70000

Hence, the required sum of money =₹70000 and the rate of interest = 6% per annum.

**Example 13. Divide T 6305 into three parts in such a way that at the rate of compound interest of 5% per annum, the amounts of 1st part in 2 years, the amounts of 2nd part in 3 years and the amounts of 3rd part in 4 years are all equal.**

Solution: Let the three parts be ₹x, ₹y and ₹ z respectively.

∴ x+y+z= 6305.. (1)

At the rate of 5% compound interest per annum,

The amount ₹x in 2 years

= \(₹ x\left(1+\frac{5}{100}\right)^2\)

= \(₹ x \times\left(\frac{21}{20}\right)^2\)

The amount ₹y in 3 years

= \(₹ y\left(1+\frac{5}{100}\right)^3\)

= \(₹ y\times\left(\frac{21}{20}\right)^3\)

The amount ₹z in 4 years

= \(₹ z\left(1+\frac{5}{100}\right)^4\)

= \(₹ z\times\left(\frac{21}{20}\right)^4\)

As per the question,

\(₹ x \times\left(\frac{21}{20}\right)^2\) = \(₹ y\times\left(\frac{21}{20}\right)^3\)

= \(₹ z\times\left(\frac{21}{20}\right)^4\) = k(let}

∴ \(₹ x \times\left(\frac{21}{20}\right)^2\) = k

⇒ \(x=k \times\left(\frac{20}{21}\right)^2=\frac{400 k}{441}\)

∴ \(₹ y\times\left(\frac{21}{20}\right)^3\) = k

⇒ \(y=k \times\left(\frac{20}{21}\right)^3=\frac{8000 k}{9261}\)

∴ \(₹ z\times\left(\frac{21}{20}\right)^4\)

⇒ \(z=k \times\left(\frac{20}{21}\right)^4=\frac{160000 k}{194481}\)

Then from(1) we get,

\(\frac{400 k}{441}+\frac{8000 k}{9261}+\frac{160000 k}{194481}\)=6305

⇒ \(k \times \frac{400}{441}\left(1+\frac{20}{21}+\frac{400}{441}\right)\)=6305

⇒ \(\frac{k \times 400}{441} \times\left(\frac{441+420+400}{441}\right)\)=6305

⇒ \(\frac{k \times 400}{441} \times \frac{1261}{441}\)=6305

⇒ \(k=\frac{6305 \times 441 \times 441}{400 \times 1261}\)

=\(\frac{441 \times 441}{80}\)

∴ \(x=\frac{400}{441} \times \frac{441 \times 441}{80}\)=2205

y= \(\frac{8000}{9261} \times \frac{441 \times 441}{80}\) =2100

z = \(\frac{160000}{194481} \times \frac{441 \times 441}{80}\) = 2000

∴ The required three parts of ₹6305 are ₹2205, ₹2100, ₹2000.

**Example 14. Divide ₹3903 between A and B in such a way that at the rate of compound interest of 4% per annum, the amount obtained by A after 7 years is equal to the amount obtained by B after 9 years.**

Solution: Let the sum of money given to A be ₹x and that given to B be ₹y.

Then, the amount obtained by A after 7 years at the rate of compound interest of 4% per annum

= \(₹ x \times\left(1+\frac{4}{100}\right)^7\)

= \(₹ x \times\left(\frac{26}{25}\right)^7\)

Similarly, the amount obtained by B after 9 years at the same rate of interest

= \(₹ y \times\left(1+\frac{4}{100}\right)\)

= \(₹ y \times\left(\frac{26}{25}\right)^9\)

As per the question,

= \( x \times\left(\frac{26}{25}\right)^7\)

= \( y \times\left(\frac{26}{25}\right)^9\)

⇒\( x=y \times\left(\frac{26}{25}\right)^{9-7}\)

⇒ \( x=y \times\left(\frac{26}{25}\right)^2\)

⇒ \(\frac{x}{y}=\frac{676}{625}\)

x: y = 676: 625

∴ Sum of money of A = ₹ 3903 x \(\frac{676}{676+625}\)

= ₹ 3903 x \(\frac{676}{1301}\)

= ₹ 3 x 676 = ₹2028

and sum of money of B = ₹ 3903 x \(\frac{625}{676+625}\)

= ₹ 3903 x \(\frac{625}{676+625}\)

= ₹ 3 x 625 = ₹ 1875

Hence, the required sum of money is ₹ 2028 obtained by A and the sum of money is ₹ 1875 obtained by B.

**Example 15. A sum of money becomes 2 times at the rate of compound interest in 15 years. In how many years, it will become 8 times at the same rate of compound interest?**

Solution: Let the principal be ₹x and the rate of compound interest per annum be r%

∴ The amount of ₹x in 15 years at the rate of compound interest of r% per annum

= \(₹ x\left(1+\frac{r}{100}\right)^{15}\)

As per question, \(x\left(1+\frac{r}{100}\right)^{15}\) = 2x

⇒ \(\left(1+\frac{r}{100}\right)^{15}\)=2 …(1)

Now, let the principal becomes 8 times in t years at the sane rate of compound interest

∴ \(x\left(1+\frac{r}{100}\right)^t\) =8 x

⇒ \(\left(1+\frac{r}{100}\right)^t\) =8

⇒ \(\left(1+\frac{r}{100}\right)^t=2^3\)

⇒ \(\left(1+\frac{r}{100}\right)^t\)

= \(\left\{\left(1+\frac{r}{100}\right)^{15}\right\}^3\)

\(\left[because 2=\left(1+\frac{r}{100}\right)^{15} \text { by (1) }\right]\)**Wbbse Class 10 Maths Solutions**

⇒ \(\left(1+\frac{r}{100}\right)^t\)

= \(\left(1+\frac{r}{100}\right)^{45}\)

⇒ t=45 .

Hence, the required time = 45 years.

**Example 16. The simple interest of a sum of money at the rate of 4% per annum in 3 years is ₹303.60. Then find the compound interest of the same principal at the same rate of compound interest in the same period of time.**

Solution:

**Given:**

The simple interest of a sum of money at the rate of 4% per annum in 3 years is ₹303.60.

Let the principal be ₹x.

∴ The simple interest of ₹x at the rate of 4% simple interest per annum in 3 years.

= \(₹ \frac{x \times 4 \times 3}{100}\)= \(₹ \frac{3 x}{25}\)

As per question, \(\frac{3x}{25}\) = 303.60

⇒ x=\(\frac{303 \cdot 60 \times 25}{3}\)

⇒ x = 2530.

Hence the principal = ₹2530.

Now, the compound interest of ₹x in 3 years at the rate of 4% compound interest per annum.

= \( ₹ 2530 \times\left(1+\frac{4}{100}\right)^3-₹ 2530\)

= \(₹ 2530 \times\left\{\left(\frac{26}{25}\right)^3-1\right\}\)

= \( ₹ 2530 \times\left(\frac{17576}{15625}-1\right)\)

= \(₹ 2530 \times\left(\frac{17576-15625}{15625}\right)\)

= \(₹ 2530 \times \frac{1951}{15625}\)

= ₹315.905 = ₹315.91 (approx)

Hence, the required compound interest = ₹315.91 (approx).

**Example 17. Atanu borrowed ₹64000 from a Cooperative bank. If the rate of interest per annum be 2.5 paisa per ₹1, then how much compound interest should Atanu have to pay to the Cooperative bank after 3 years?**

Solution:

**Given:**

Atanu borrowed ₹64000 from a Cooperative bank. If the rate of interest per annum be 2.5 paisa per ₹1

The rate of interest = 2.5 paisa per ₹1

= (2.5 x 100)paisa per ₹100

= ₹2.5 per ₹100

= 2.5% per annum.

∴ At the rate of 2.5% compound interest per annum, the amount of ₹64000 in 3 years.

= \(₹ 64000\left(1+\frac{2 \cdot 5}{100}\right)^3\)

= \(₹ 64000\left(1+\frac{25}{1000}\right)^3\)

= \(₹ 64000\left(\frac{41}{40}\right)^3\)

= ₹ 68921

∴ The compound interest = ₹(68921 – 64000) = ₹4921.

Hence, the required compound interest = ₹ 4921.

**Example 18. Anil took a loan of ₹20000 from a bank under the condition that he would repay the loan by three equal instalments. If the rate of compound interest of the bank be 5% per annum, then find the amount of money in each instalment.**

Solution:

**Given:**

Anil took a loan of ₹20000 from a bank under the condition that he would repay the loan by three equal instalments. If the rate of compound interest of the bank be 5% per annum

Let each of the instalments be ₹100 and the principal of the first instalment be ₹x.

∴ \( x \times\left(1+\frac{5}{100}\right)\)=100

⇒ \( x \times\left(\frac{100+5}{100}\right)\)=100

⇒ \( x \times \frac{105}{100}=100 \)

⇒ \( x=\frac{100 \times 100}{105}=\frac{2000}{21}\)

Similarly, the principal of the second instalment

= \(₹ \frac{100 \times 100}{105} \times \frac{100}{105}\)= \(₹ \frac{40000}{441}\)

And the principal of the third instalment

= \(₹ \frac{100 \times 100}{105} \times \frac{100}{105} \times \frac{100}{105}\)

= \(₹\frac{800000}{9261}\)

∴ The total principle

= \(₹\left(\frac{2000}{21}+\frac{40000}{441}+\frac{800000}{9261}\right)\)

= \(₹\frac{2000}{21}\left(1+\frac{20}{21}+\frac{400}{441}\right)\)

= \(₹\frac{2000}{21}\left(\frac{441+420+400}{441}\right)\)

= \(₹\frac{2000}{21} \times \frac{1261}{441}\)

= \(₹\frac{2000000}{9261}\)

∴ If the principal be ₹ \(₹\frac{2000000}{9260}\) 2, then the instalment is ₹100

∴ If the principal be ₹1, then the instalment is \(₹\frac{100 \times 9261}{2000000}\)

∴ If the principal be ₹20000 then the instalment is \(₹\frac{100 \times 9261 \times 20000}{2000000}\)

= ₹9261

Hence, the amount of each instalment is ₹9261.

**Wbbse Class 10 Maths Solutions**

**Example 19. The simple interest and compound interest of a certain sum of money for 2 years are ₹420 and ₹434.70 respectively. Find the sum of money and the rate of interest.**

Solution:

**Given:**

The simple interest and compound interest of a certain sum of money for 2 years are ₹420 and ₹434.70 respectively.

Let the principal be ₹x and the rate of interest be r%

Then the simple interest of ₹x in 2 years

= \(₹ \frac{x \times r \times 2}{100}\)

As per question, \(\frac{2xr}{100}\) = 420

⇒ \(\frac{xr}{100}\) 210 ….(1)

Again, the compound interest of ₹x in 2 years

= \(₹\left\{x \times\left(1+\frac{r}{100}\right)^2-x\right\}\)

= \(₹ x \times\left\{\left(1+\frac{r}{100}\right)^2-(1)^2\right\}\)

= \(₹ x \times\left(1+\frac{r}{100}+1\right)\left(1+\frac{r}{100}-1\right)\)

= \(₹ x \times\left(2+\frac{r}{100}\right) \times \frac{r}{100}\)

=\( ₹ \frac{x r}{100}\left(2+\frac{r}{100}\right)\)

= \(₹ 420\left(2+\frac{r}{100}\right)\) [because by(1), \(\frac{x r}{100}\)=420]

As per question, \(210\left(2+\frac{r}{100}\right)\) =434.70

or, \(2+\frac{r}{100}\) =\(\frac{434 \cdot 70}{210}\)

or, \(\frac{200+r}{100}\) = \(\frac{43470}{210 \times 100}\)

or, 200+r = \(\frac{43470}{210}\)

or, 200 + r = 207

or, r = 207 – 200

or, r =7

∴ From(1) we get, \(\frac{x \times 7}{100}\) = 210

or, \(x=\frac{210 \times 100}{7}\) = 3000

Hence, the sum of money = ₹3000 and the rate of interest per annum is 7%

**Class 10 Maths Wbbse Solutions**

**Example 20. Calculate at what rate of compound interest ₹5000 amount to ₹5408 in 2 years.**

Solution: Let the compound interest be r%.

∴ The amount of ₹5000 in 2 years = \(₹ 5000 \times\left(1+\frac{r}{100}\right)^2\)

As per the question, \(5000 \times\left(1+\frac{r}{100}\right)^2\) = 5408

or, \(\left(1+\frac{r}{100}\right)^2=\frac{5408}{5000}\)

or, \(\left(1+\frac{r}{100}\right)^2=\frac{676}{625}\)

or, \(\left(1+\frac{r}{100}\right)^2=\left(\frac{26}{25}\right)\)

or, \(1+\frac{r}{100}=\frac{26}{25}\) (by taking square rooty of both sides)

or, \(\frac{r}{100}=\frac{26}{25}-1\)

or, \(\frac{r}{100}=\frac{1}{25}\)

or, \(r=\frac{1 \times 100}{25}=4\)

Hence, the rate of compound interest is 4% per annum.

**Example 21. A person invested a sum of money in a bank under the condition that the bank will offer him compound interest. If the compound interest of two consecutive years be ₹225 and ₹238.50 respectively, then find the rate of compound interest per annum.**

Solution:

**Given:**

A person invested a sum of money in a bank under the condition that the bank will offer him compound interest. If the compound interest of two consecutive years be ₹225 and ₹238.50 respectively

The compound interest in the second year = ₹238.50 and in the first year = ₹225.

∴ the interest of ₹225 in 1 year = ₹(238.50 – 225) = ₹13.50

= \(₹\frac{135}{10}\) = ₹ \(\frac{27}{2}\)

Let the rate of interest be r%

∴ \(225 \times\left(1+\frac{r}{100}\right)-225\)=\(\frac{27}{2}\)

⇒ \(225 \times\left(1+\frac{r}{100} \cdot 1\right)\)=\(\frac{27}{2}\)

⇒ \(225 \times \frac{r}{100}\)=\(\frac{27}{2}\)

⇒ \( r=\frac{27 \times 100}{2 \times 225}\)

⇒ r=6

Hence, the required rate of compound interest is 6% per annum.

**Class 10 Maths Wbbse Solutions**

**Example 22. Calculate the compound interest on ₹5000 for 1 year at the rate of 8% compound interest per annum compounded at the interval of 6 months.**

Solution: Here, the principal is ₹5000 and the rate of compound interest is 8% per annum.

Since the interest is compounded at the interval of 6 months, the period of time is 2 and the time is 1 year.

∴ The required amount

= \(₹ 5000 \times\left(1+\frac{\frac{8}{2}}{100}\right)^{2 \times 1}\)

= \(₹ 5000 \times\left(1+\frac{1}{25}\right)^2\)

=\(₹ 5000 \times\left(\frac{26}{25}\right)^2\)

= \(₹ 5000 \times \frac{26}{25} \times \frac{26}{25}\)=₹ 5408

∴ The compound interest = ₹(5408 – 5000) = ₹408.

Hence, the compound interest =₹408.

**Example 23. Calculate the compound interest on ₹5250 for 9 months at the rate of 10% compound interest per annum compounded at the interval of 3 months.**

Solution: Here the principal is ₹5250.

The rate of compound interest is 10%. Time = 9 months = \(\frac{9}{12}\) year = \(\frac{3}{4}\) year.

Since the interest is compounded at the interval of 3 months, the period is 4.

∴ The amount of ₹5250 in 9 months at the rate of 10% per annum

= \(₹ 5250 \times\left(1+\frac{\frac{10}{4}}{100}\right)^{\frac{3}{4} \times 4}\)

= \(₹ 5250 \times\left(1+\frac{10}{4 \times 100}\right)^3\)

= \(₹ 5250 \times\left(\frac{40+1}{40}\right)^3=₹ 5250\left(\frac{41}{40}\right)^3\)

= \(₹ 5250 \times \frac{41 \times 41 \times 41}{40 \times 40 \times 40}\)

=₹ 5653.67 (approx)

∴ The compound interest = ₹(5653.67 – 5250) = ₹403.67.

Hence, the compound interest of ₹5250 in 9 months is ₹403.67 (approx).

**Class 10 Maths Wbbse Solutions**

**Example 24. Calculate at what rate of interest per annum will ₹60000 amount to ₹69984 in 2 years. **

Solution: Let the compound interest be r%** **per annum.

Here the principal is ₹60000 and the time is 2 years.

∴ The amount of ₹60000 in 2 years = ₹60000 x \(\left(1+\frac{r}{100}\right)^2\)

As per question,

\(60000 \times\left(1+\frac{r}{100}\right)^2=69984\)⇒ \(\left(1+\frac{r}{100}\right)^2=\frac{69984}{60000}\)

⇒ \(\left(1+\frac{r}{100}\right)^2=\frac{729}{625}\)

⇒ \(\left(1+\frac{r}{100}\right)^2\)

=\(\left(\frac{27}{25}\right)^2=\)

⇒ \(1+\frac{r}{100}=\frac{27}{25}\)

⇒ \(\frac{r}{100}=\frac{27}{25}-1 \Rightarrow \frac{r}{100}\)

=\(\frac{27-25}{25}\)

⇒ \(\frac{r}{100}=\frac{2}{25}\)

⇒ \(r=\frac{2 \times 100}{25}\)

⇒r=8

Hence, the rate of compound interest is 8% per annum.

**Example 25. Calculate the principal which amounts ₹9826 after 18 months at the rate of compound interest 2.5% per annum when interest is compounded at the interval of 6 months.**

Solution: Let the principal be ₹x.

The rate of compound interest per annum is 2 • 5% = \(\frac{5}{2}\) %

Time = 18 months = \(\frac{18}{12}\) years = \(\frac{3}{2}\)years

Since interest is compounded at the interval of 6 months, the period of time is 2.

∴ The amount of ₹x in \(\frac{3}{2}\)

= \(₹ x \times\left(1+\frac{\frac{5}{2 \times 2}}{100}\right)^{\frac{3}{2} \times 2}\)

= \(₹ x \times\left(1+\frac{1}{80}\right)^3\)

= \(₹ x \times\left(\frac{80+1}{80}\right)^3\)

= \(₹ x \times\left(\frac{81}{80}\right)^3\)

As per question

\(x \times\left(\frac{81}{80}\right)^3\) =9826

⇒ \(x \times \frac{81 \times 81 \times 81}{80 \times 80 \times 80}\) =9826

⇒ \(x=\frac{9826 \times 80 \times 80 \times 80}{81 \times 81 \times 81}\)

⇒ x = 9466.54(approx)

Hence, the required principal is ₹9466-54 (approx).

**Class 10 Maths Wbbse Solutions**

**Example 26. Calculate in how many years will ₹300000 amount to ₹399300 at the rate of 10% compound interest per annum.**

Solution: Here principal = ₹300000.

Rate of compound interest = 10% per annum, let the required time be t years.

∴ The amont of ₹300000 after t years =

= \(₹ 300000 \times\left(1+\frac{10}{100}\right)^t\)

= \(₹ 300000 \times\left(1+\frac{1}{10}\right)^t \)

= \(₹ 300000 \times\left(\frac{11}{10}\right)^t\)

As per question, \(₹ 300000 \times\left(\frac{11}{10}\right)^t\) = 399300

or, \(\left(\frac{11}{10}\right)^t=\frac{399300}{300000}\)

= \(\frac{1331}{1000}=\left(\frac{11}{10}\right)^3\)

= \(\left(\frac{11}{10}\right)^t=\left(\frac{11}{10}\right)^3\)

⇒ t =3

Hence the required time = 3 years.

Hence, the required time = 3 years.

**Example 27. A sum of money was invested in a monetary fund for 2 years at the rate of compound interest of 20% per annum. If interest had been compounded at the interval of 6 months, then the compound interest would be ₹482 more of the same principal at the same rate of interest. Find the sum of money invested.**

Solution:

**Given:**

A sum of money was invested in a monetary fund for 2 years at the rate of compound interest of 20% per annum. If interest had been compounded at the interval of 6 months, then the compound interest would be ₹482 more of the same principal at the same rate of interest.

Let the sum of money invested be ₹x

Then for the first case, the amount of ₹x after 2 years

= \(₹ x \times\left(1+\frac{20}{100}\right)^2\)

= \(₹ x \times\left(1+\frac{1}{5}\right)^2\)

= \(₹ x \times\left(\frac{5+1}{5}\right)^2\)

= \(₹ x \times\left(\frac{6}{5}\right)^2\)

= \(₹\frac{36x}{25}\)

Also, for the second case, the amount of ₹x after 2 yeras, the amount

= \(₹ x \times\left(1+\frac{\frac{20}{2}}{100}\right)^{2 \times 2}\)

= \(₹ x \times\left(1+\frac{1}{10}\right)^4=₹ x \times\left(\frac{11}{10}\right)^4\)

= \(₹ \frac{14641 x}{10000}\)

The sum of money invested = \(₹ \frac{14641 x}{10000}\)

**Class 10 Maths Wbbse Solutions**

**Example 28. At the rate of compound interest of 8% per annum in how many years will the principal ₹40000 amount ₹46656?**

Solution: Here, the principal is ₹40000.

The rate of compound interest = 8%

Let the required time be t years.

∴ The amount of ₹40000 in t years

= \(₹ 40000 \times\left(1+\frac{8}{100}\right)^t\)

= \(₹ 40000 \times\left(1+\frac{2}{25}\right)^t\)

= \(₹ 40000 \times\left(\frac{27}{25}\right)^t\)

As per question,

\(40000 \times\left(\frac{27}{25}\right)^t\) =46656

or,\(\left(\frac{27}{25}\right)^t\)=\(\frac{46656}{40000}\)

or, \(\left(\frac{27}{25}\right)^t\) =\(\frac{729}{625}\)

= \(\left(\frac{27}{25}\right)^2\)

⇒ t = 2

Hence, the required time = 2 years.

**Example 29. Calculate the compound interest and amount on ₹1600 for 1 \(\frac{1}{2}\) years at the rate of 10% compound interest per annum compounded at the interval of 6 months. **

Solution: Here principal = ₹1600, rate of interest = 10%, time = 1 \(\frac{1}{2}\) years

= \(\frac{3}{2}\) years, period = \(\frac{12}{6}\) = 2

Hence the required amount

= \(₹ 1600\left(1+\frac{\frac{10}{2}}{100}\right)^{\frac{3}{2} \times 2}\)

= \(₹ 1600\left(1+\frac{1}{20}\right)^3\)

= \(₹ 1600\left(\frac{21}{20}\right)^3\)

= \( ₹ \frac{1600 \times 21 \times 21 \times 21}{20 \times 20 \times 20}\)=₹1852.20

Required compound interest = ₹(1852.20 – 1600) = ₹252.20

Hence the required compound interest = ₹252.20 and the required amount = ₹1852.20.