## Arithmetic Chapter 1 Simple Interest Solved Example Problems

In our daily life, we sometimes face a problem of shortage of required money to perform certain work. In such a situation, we generally borrow some money from-bank or any other monetary fund to complete the same work.

We have to pay some extra money more than the debt taken by us, after a certain time period to that very monetary fund. This very extra money paid by us to the fund is called interest, that is, we pay some interest to that monetary fund.

Moreover, we generally deposit our savings for a certain time to a bank or any other monetary fund. Here, the monetary fund pays some extra money to us after that certain period. This very extra money is also known as interest.

**Class 10 Maths Solutions Wbbse**

**Interest is of two types:**

- Simple Interest
- Compound Interest

In this chapter, we shall study only Simple Interest.

**Definition:** When interests are calculated only on the principal, then it is called simple interest.

**Some definitions regarding simple interest:**

**WBBSE Solutions for Class 10 Maths**

**Principal:** The quantity of money which is borrowed or which is invested, is known as principal.

**Interest:** After a certain time is over, the quantity of money which is given or taken, is known as interest.

**Period:** The duration of time by which debt is taken or given is known as a period.

**Amount:** The sum of principal and total interest is called the amount.

**Rate of Interest:
**

The quantity of interest paid on some money for a certain time period is known as the rate of interest, e.g., the quantity of money paid on ₹100 for 1 year is called the percentage of interest per year.

For example, Yearly rate of interest = 5 % means that the total interest paid on t 100 for 1 year is ₹5.

Sometimes, interests are calculated on the basis of half-yearly, quarterly, and even on daily basis of time period.

**Creditor: **The person or the institution who gives a debt is generally known as a creditor.

**Debtor:** The person or institution who takes a loan is generally known as a debtor.

**For example:** If any person of any institution deposits some money to a bank, then the person or the institution is known as a creditor and the bank is known as a debtor.

There is a direct ratio between the principal, period, and the rate of interest.

**Class 10 Maths Solutions Wbbse**

**Necessary formulae regarding simple interest :
**

1. Amount = Principal + Interest

2. Principal = Amount – Interest

3. Interest = Amount – Principal

Let Principal =₹P, rate of yearly simple interest = r % and period = t years, then the total interest, I = \(\frac{Prt}{100}\)

Thus, if any three of the quantities P, I, r, t are given, then the rest can be determined by this formula,

e.g., P = \(\frac{100 I}{rt}\), \(\frac{100 I}{Pt}\), \(\frac{100 I}{Pr}\)

## Arithmetic Chapter 1 Simple Interest Examples Multiple Choice Questions

Example 1. If a principal amount doubles in 10 years, then the rate of yearly simple interest is

- 10%
- 15%
- 20%
- 25%

∴ 1. 10%

Solution: Let the principal be ₹ x and the yearly rate of simple interest be r %

If the principal amounts ₹ 2x in 10 years, then the interest = ₹(2x – x) = ₹ x

∴ Total interest in 10 years =\(\frac{P rt}{100}\)= ₹ \(\frac{x \times r \times 10}{100}\) = ₹ \(\frac{x r}{10}\)

∴ \(\frac{xr}{10}\) = x

⇒ r = \(\frac{x \times 10}{x}\) = 10

∴ Rate of yearly simple interest = 10 %

**Example 2. The total interest of any principal is ₹ x in x years at the rate of yearly simple interest x%. Then the principal is**

**Class 10 Maths Solutions Wbbse**

- ₹ x
- ₹ 100 x
- ₹ \(\frac{100}{x}\)
- ₹ \(\frac{100}{x^2}\)

∴ 3. ₹ \(\frac{100}{x}\)

Solution: Let the principal = ₹ P

Rate of interest = x %

Total interest = ₹ x.

We know that I = \(\frac{P rt}{100}\)

Here, \(\frac{\mathrm{P} \times x \times x}{100}\) = x

⇒ P = \(\frac{100}{x}\)

∴ The required principal = ₹ \(\frac{100}{x}\)

**Example 3. A principal amounts double in 20 years at a certain rate of simple interest. At the same rate of interest the principal amounts triple in**

- 30 years
- 35 years
- 40 years
- None of these.

∴ 3. 40 years.

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

As per the question, the amount = ₹ 2x in 20 years.

∴ Interest = ₹ (2x – x) =₹ x

∴ \(\frac{x \times r \times 20}{100}\) =x

∴ r=\(\frac{100 \times x}{x \times 20}\) =5

∴ The required yearly rate of simple interest = 5%

If this principal amounts ₹ 3x, then the interest = ₹ (3x – x) = ₹ 2x

Let the required time = t years.

∴ \(\frac{x \times 5 \times t}{100}\) =2 x

⇒ t =\(\frac{2 x \times 100}{x \times 5}\) = 40

∴ The required time = 40 years.

**Maths Solutions Class 10 Wbbse**

**Example 4. The total interest of a principal of ₹ 2000 in 18 months at the rate of yearly simple interest 6% is**

- ₹ 160
- ₹ 180
- ₹ 200
- ₹ 220

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

Here, P = ₹ 2000

Rate of interest, r % = 6 %

Period, t = \(\frac{3}{2}\) years

∴ Total interest =₹ \(\frac{2000 \times 6 \times 3}{2 \times 100}\) = ₹ 180

∴ The required total interest = ₹ 180.

**Example 5. If the total interest of a principal in 6 years be 30% of the principal, then the total interest will be equal to the principal in**

- 18 years
- 20 years
- 22 years
- 24 years

∴ 2. 20 years

Solution: Let the principal be ₹ x and the rate of yearly simple interest be r %

Then, the interest in 6 years = ₹ x X \(\frac{30}{100}\) = ₹ \(\frac{3x}{10}\)

∴ The rate of interest = 5 %

Also, let the required time = t years.

\(\frac{x \times 5 \times t}{100}=x \Rightarrow t=\frac{100 \times x}{5 \times x}=20\)∴ The required time = 20 years.

**Maths Solutions Class 10 Wbbse**

## Arithmetic Chapter 1 Simple Interest Examples Very Short Answer Type Questions

**Example 1. If the rate of simple interest per annum decreases from 4% to 3 \(\frac{3}{4}\)% then the yearly income of a person decreases by ₹ 60. Find the principal of that person.
**

Solution:

**Given:**

If the rate of simple interest per annum decreases from 4% to 3 \(\frac{3}{4}\)% then the yearly income of a person decreases by ₹ 60.

(4 – 3 \(\frac{3}{4}\))% = \(\frac{1}{4}\) %

As per question, ₹ \(\frac{1}{4}\) interest decreases when principal = ₹ 100

₹ 1 interest decreases when principal = ₹ I00 x 4

₹ 60 interest decreases when principal= ₹ 100 x 4 x 60 = ₹ 24000

Hence the required principal = ₹ 24000.

**Example 2. If the rate of simple interest per annum be 6%, then find the simple interest of ₹ 3000 from 5th January 31st May of 2019.
**

Solution:

**Given:**

If the rate of simple interest per annum be 6%,

Here, the total number days are as follows:

January = 26 days

February = 28 days

March = 31 days

April = 30 days

May = 31 days

Total = 146 days

Now, 146 days = \(\frac{146}{365}\) years = \(\frac{2}{5}\) years

So, here principal, p = ₹ 3500 rate of interest, r% = 6%

Time, t = \(\frac{2}{5}\) years

We know that \(1=\frac{\text { prt }}{100}\)

= \(₹ \frac{30006 \times \frac{2}{5}}{100}\)

Hence the required simple interest = ₹ 72.

**Example 3. Determine the total amount of ₹5000 at the rate of 7 \(\frac{1}{2}\) % simple interest per annum in 5 years.**

Solution: Here, p = ₹5000, r = 7 \(\frac{1}{2}\) = \(\frac{15}{2}\) and t = 5

∴ \(I=\frac{p r t}{100}=₹ \frac{5000 \times \frac{15}{2} \times 5}{100}= ₹ 1875\)

Hence the required amount = ₹ (5000 + 1875) = ₹ 6875.

**Maths Solutions Class 10 Wbbse**

**Example 4. At what rate of simple interest in percent per annum, the ratio of principal and its interest after 20 years is 1: 1?
**

Solution: Let the rate of simple interest per annum = r %

Also, if the principal be ₹ p, then by the question, the simple interest = ₹ p.

Also, time, t = 20 years

∴ \(\begin{aligned}

& ₹ \frac{p}{8}=₹ \frac{p \times r \times 2}{100} \\

& r=\frac{100}{2 \times 8}=\frac{25}{4}=6 \frac{1}{4}

\end{aligned}\)

Hence the required rate of simple interest per annum = 5%,

**Example 5. If the interest of a principal in 2 years be \(\frac{1}{8}\) of it, then find the rate of simple interest per annum.
**

Solution: Let the principal = ₹ p and rate of interest = r %

As per question, interest = ₹ p x \(\frac{1}{8}\) = ₹ \(\frac{p}{8}\)

Time, t = 2 years.

∴ \(\begin{aligned}

& ₹ \frac{p}{8}=₹ \frac{p \times r \times 2}{100} \\

& r=\frac{100}{2 \times 8}=\frac{25}{4}=6 \frac{1}{4}

\end{aligned}\)

Hence the required rate of simple interest per annum = 6 \(\frac{1}{4}\) %

**Maths Solutions Class 10 Wbbse**

## Arithmetic Chapter 1 Simple Interest Examples Write True Or False

**Example 1. The interest given or received for a certain period of time on a certain amount of principal is called “Total interest”.**

Solution: True

**Example 2. In return to the right of using the creditor’s money for a short time, according to the condition, the debtor gives him some extra money. This money is known as Rate of interest.**

Solutions: False

## Arithmetic Chapter 1 Simple Interest Examples Fill In The Blanks

**Example 1. The person who gives a debt is generally known as a ______**

Solution: Creditor

**Example 2. The sum of the principal and total amount is called the ______**

Solution: Amount

**Example 3. The amount of ₹ 2p in t years at the rate of simple interest of \(\frac{r}{2}\) % per annum is ₹ (2p + ______ )**

Solution: \(\frac{prt}{100}\)

**Example 4. The ratio of the principal and the amount (principal along with interest) in 1 year is 8: 9, the rate of simple interest per annum is _____**

Solution: 12 \(\frac{1}{2}\) %, since let principal = ₹ 8x and amount = ₹ 9x.

∴ Interest = ₹ (9x – 8x) = ₹ x..

∴ Simple interest per annum = \(\frac{x}{8x}\) X 100% = \(\frac{25}{2}\) % = 12 \(\frac{1}{2}\) %.

**Wbbse Class 10 Maths Solutions**

## Arithmetic Chapter 1 Simple Interest Examples Short Answer Type Questions

**Example 1. The yearly income of Chandan babu becomes? 60 less when the yearly rate of simple interest becomes 3 \(\frac{3}{4}\) % decreasing from 4%. Then find the principal of Chandan babu.
**

Solution:

**Given:**

The yearly income of Chandan babu becomes? 60 less when the yearly rate of simple interest becomes 3 \(\frac{3}{4}\) % decreasing from 4%.

Let the principal of Chandan babu be ₹ x

Then the interest of 1 year at the rate of 4% = ₹ \(\frac{x \times 4 \times 1}{100}\) = ₹ \(\frac{x}{25}\)

Also, at the rate of 3 \(\frac{3}{4}\) % = \(\frac{15}{4}\) %,

the interest of 1 year = ₹ \(\frac{x \times 15 \times 1}{4 \times 100}\) = ₹ \(\frac{3 x}{80}\)

As per question, = \(\frac{x}{25}\) – \(\frac{3x}{80}\) = 60

⇒ \(\frac{16 x-15 x}{400}\) = 60

⇒ \(\frac{x}{400}\) =60

⇒ x = 400 x 60 = 24000

∴ The principal of Chandan babu = ₹ 24000.

**Example 2. Find the interest of ₹ 300 from 3rd March to 15 May 2016 at the rate of annual simple interest of 6%.
**

Solution: Time = 28 days in March + 30 days in April +15 days in May.

=73 days = \(\frac{73}{365}\) year = \(\frac{1}{5}\) year.

∴ Total interest = \(₹ \frac{300 \times 6 \times 1}{100 \times 5}=₹ \frac{18}{5}= ₹ 3.60\)

∴ The required interest = ₹ 3.60.

**Wbbse Class 10 Maths Solutions**

**Example 3. If the interest ot a principal in 10 years be \(\frac{2}{5}\) th part of itself, then find the yearly percentage of rate of simple interest.
**

Solution: Let the principal be ₹ x and the yearly rate of simple interest be r %

∴ The interest of 10 years = \( ₹ \frac{x \times r \times 10}{100}=₹ \frac{x r}{10}\)

∴ Amount = ₹ (x + \(\frac{xr}{10}\))

⇒ \(\frac{2x}{5}\) + \(\frac{2}{5}\) X \(\frac{xr}{10}\) = \(\frac{xr}{10}\)

⇒ \(\frac{2x}{5}\) = \(\frac{xr}{10}\) – \(\frac{xr}{25}\)

⇒ \(\frac{2x}{5}\) = \(\frac{5xr-2xr}{50}\)

⇒ \(\frac{2x}{5}\) = \(\frac{3xr}{50}\)

⇒ r = \(\frac{2 x \times 50}{5 \times 3 x}=\frac{20}{3}=6 \frac{2}{3}\)

Annual rate of interest =6 \(\frac{2}{3}\) %

**Example 4. Find the principal of which the monthly interest is ₹ 1 at the rate of 10% annual simple interest.
**

Solution: Let the principal be ₹ x.

Now, 1 month = \(\frac{1}{12}\) year.

∴ Simple interest = \(₹ \frac{x \times 10 \times 1}{100 \times 12}=₹ \frac{x}{120} \)

As per question, \(\frac{x}{120}\) = 1

⇒ x = 120

∴ The required principal = ₹ 120.

**Example 5. At the two different rates of annual simple interest a certain quantity of Principal becomes double in 5 years and triple in 12 years. In what case the investment is profitable?
**

Solution:

**Given:**

At the two different rates of annual simple interest a certain quantity of Principal becomes double in 5 years and triple in 12 years.

Let the principal be ₹ x

The rate of annual simple interest for the first case be a% and that for the second case be b%.

As per the question, the interest for 1st case = ₹ (2x – x) = ₹ x .

and for the 2nd case = ₹ (3x – x) = ₹ 2x.

\(\frac{x \times a \times 5}{100}=x(\text { for the } 1 \text { st case }) \Rightarrow a=\frac{100 \times x}{x \times 5} \Rightarrow a=20\)∴ For the 1st case, the rate of annual simple interest = 20%

Again, \(\frac{x \times b \times 12}{100}\) = 2x (for the 2nd case)

or, b = \(\frac{2 x \times 100}{x \times 12}\)

⇒ b = \(\frac{50}{3}\) = 16 \(\frac{2}{3}\)

For the 2nd case, the rate of annual simple interest =16 \(\frac{2}{3}\) %

Now, since the rate of annual simple, interest for the 1st case is better than that for the 2nd case.

∴ The investment is more profitable in the 1st case.

**Example 6. ₹ 10 is given as a loan to someone in such a condition that he. will pay back the loan in 11 installments at the rate of ₹ 1 per installment. Find the rate of simple interest per annum.
**

Solution:

**Given:**

₹ 10 is given as a loan to someone in such a condition that he. will pay back the loan in 11 installments at the rate of ₹ 1 per installment.

Let the rate of annual simple interest be r %.

Here, ₹ 10 + simple interest of ₹10 for 11 months.

= ₹10 + simple interest of ₹1 for (1 + 2 + 3 + . +10) months

= ₹ 10 + simple interest of ₹1 for 55 months

= Simple interest of ₹ 1 for 55 months.

= ₹1.

∴ \(\frac{1 \times 55 \times r}{100 \times 12}=1 \Rightarrow r=\frac{240}{11}=21 \frac{9}{11}\)

∴ The rate of simple interest = 21 % per annum.

## Arithmetic Chapter 1 Simple Interest Examples Long Answer Type Questions

**Example 1. Ashim took a loan of some money from a bank for opening a dairy farm at the rate of simple interest of 12% per annum. Every month he has to repay? 450 as. interest. Determine the loan amount taken by him.
**

Solution:

**Given:**

Ashim took a loan of some money from a bank for opening a dairy farm at the rate of simple interest of 12% per annum. Every month he has to repay? 450 as. interest.

Let the loan be ₹ x.

1 month = \(\frac{1}{2}\) year.

∴ Simple interest of ₹ x = ₹ \(\frac{x \times 12 \times 1}{100 \times 12}\) = ₹ \(\frac{x}{100}\)

As per question, \(\frac{x}{100}\)= 450

⇒ x = 450 x 100 = 45000

∴ The required loan = ₹ 45000.

**Example 2. If the interest of ₹ 219 in 1 day be 5 paise, then find the rate of simple interest in percent per annum.
**

Solution: Here, principal (P) = ₹ 219

Rate of interest = x % (let)

Period (t) = 1 day =\(\frac{1}{365}\) year

Simple interest = 5 paise = ₹ \(\frac{5}{100}\) = ₹ \(\frac{1}{20}\)

**Wbbse Class 10 Maths Solutions**

**Example 3. Kamala deposited ₹ 20000 of her savings in two separate banks at the same time. The rate of simple interest per annum is of 6% in one bank and that of 7% in another bank after 2 years, if she gets ₹ 2560 in total as interest, then find the money she had deposited separately in each of two banks.**

Solution:

**Given:**

Kamala deposited ₹ 20000 of her savings in two separate banks at the same time. The rate of simple interest per annum is of 6% in one bank and that of 7% in another bank after 2 years, if she gets ₹ 2560 in total as interest,

Let Kamala deposited ₹ x in the first bank.

∴ She had deposited ₹ (20000 – x) in the second bank.

Now, at the rate of 6% per annum, the simple interest of ₹ x in 2 years = ₹ \(\frac{x \times 6 \times 2}{100}\) =₹ \(\frac{12 x}{100}\)

Again, at the rate of 7% per annum, the simple interest of ₹ (20000 – x) in 20 years

= \(₹ \frac{(20000-x) \times 7 \times 2}{100}=₹ \frac{280000-14 x}{100}\)

As per question, \(\frac{12 x}{100}+\frac{280000-14 x}{100}=2560\)

or, \(\frac{12 x+280000-14 x}{100}=2560\)

or, 28000 – 2x = 256000

or, 2x = 2800000-256000

or, 2x = 24000

or, x = \(\frac{24000}{2}\) = 12000

∴ She deposited ₹ 12000 in the first bank.

∴ In the second bank she had deposited ₹ (20000 – 12000) = ₹ 8000.

∴ Kamala had deposited ₹ 12000 and ₹ 8000 respectively in the two banks.

**Example 4. If a person get ₹ 1200 return as amount (principal along with interest) by depositing ₹ 800 in the bank at the rate of simple interest of 10% per annum, then calculate the time for which the money was deposited in the bank.**

Solution:

**Given:**

If a person get ₹ 1200 return as amount (principal along with interest) by depositing ₹ 800 in the bank at the rate of simple interest of 10% per annum

Here, principal = ₹ 800; Amount = ₹ 1200.

∴ Interest = ₹ (1200 – 800) = ₹ 400

Let the required time = t years.

∴ \(\frac{800 \times 10 \times t}{100}\) = 400

⇒ t = \(\frac{400}{80}\) = 5

∴ The required time = 5 years.

** Example 5. Sachindrababu takes a loan amount of ₹ 2,40,000 from a bank for constructing a building at the rate of simple interest of 12% per annum. After 1 year of taking the loan, he rents the house at the rate of ₹ 5200 per month, Then, determines the number of years he would take to repay his loan along with interest from the income of the house rent.**

Solution:

**Given:**

Sachindrababu takes a loan amount of ₹ 2,40,000 from a bank for constructing a building at the rate of simple interest of 12% per annum. After 1 year of taking the loan, he rents the house at the rate of ₹ 5200 per month

Let Sachindrababu repay his loan along with interest after x years of the construction of the building.

Now, the house rent per month = ₹ 5200.

∴ Houserent of 1 year =₹ 5200 x 12 x

∴ Houserent of x years = ₹ 5200 x 12 x x.

Since, the house was rented after 1 year,

∴ The loan was repaid after (x + 1) years.

Now, the simple interest of ₹ 240000 in (x + 1) years

= ₹ \(\frac{240000 \times 12 \times(x+1)}{100}\)

= ₹ 2400 x 12 (x +1)

∴ Amount = ₹ {240000 + 2400 x 2 (x + 1)}

= ₹ 2400 [100 + 12x +12]

= ₹ 2400 (112 + 12x)

∴2400 (112 + 12x) = 5200 x 12x

⇒112 + 12x = \(\frac{5200 \times 12 x}{2400}\)

⇒ 14x = 112

⇒ x = \(\frac{112}{14}\) = 8

∴ The required time = (8 + 1) years = 9 years.

**Wbbse Class 10 Maths Solutions**

**Example 6. Chandan got ₹ 100000 when he retired from his service. He deposited some of that money in the bank and rest of his money in the post office and got ^ 5400 in total per year as interest. If the rates of simple interest per annum in the bank and in the post office are 5% and 6% respectively, then find the money he had deposited in the bank and post office.
**

Solution:

**Given:**

Chandan got ₹ 100000 when he retired from his service. He deposited some of that money in the bank and rest of his money in the post office and got ^ 5400 in total per year as interest. If the rates of simple interest per annum in the bank and in the post office are 5% and 6% respectively

Let Chandan deposited ₹ x in the bank.

∴ He deposited ₹ (100000 -x)

Now, the simple interest of ₹ x in 1 year = \( ₹ \frac{x \times 5 \times 1}{100}=₹ \frac{5 x}{100}\)

Again, the simple interest of ₹ (100000 – x) in 1 year at the rate of 6% per annum

= \( ₹ \frac{(100000-x) \times 6 \times 1}{100}=₹ \frac{600000-6 x}{100}\)

As per question,

\(\begin{aligned}& \frac{600000-6 x}{100}+\frac{5 x}{100}=5400 \\

& \frac{600000-6 x+5 x}{100}=5400

\end{aligned}\)

⇒ 600000 – X = 540000 ⇒ x = 600000 – 540000 = 60000.

∴ Chandan deposited ₹ 60000 in the bank.

∴ He deposited ₹ (100000 – 60000) = ₹ 40000 in the Post Office.

∴Chandan deposited ₹ 60000 in the bank and ₹ 40000 in the post-office.

**Example 7. A bank gives 5% simple interest per annum. In that bank, Bhubanbabu deposits ₹ 15,000 at the beginning of the year, but withdraws ₹ 3000 after 3 months, and then again, after 3 months he deposits ₹ 8000. Determine the amount (principal along with interest) Bhubanbabu will get at the end of the year.**

Solution:

**Given:**

A bank gives 5% simple interest per annum. In that bank, Bhubanbabu deposits ₹ 15,000 at the beginning of the year, but withdraws ₹ 3000 after 3 months, and then again, after 3 months he deposits ₹ 8000

3 months = \(\frac{3}{12}\) year = \(\frac{1}{4}\) year

Now, simple interest of 7 15000 in 3 months at the rate of 5% per annum

= \(₹ \frac{15000 \times 1 \times 5}{10.0 \times 4}\)

=₹ \(\frac{375}{2}\)

= ₹ 187.50

Withdrawing ₹ 3000, the principal becomes = ₹ (15000 – 3000) = ₹ 12000.

∴ Then the simple interest of ₹ 12000 in 3 months at the rate of 5% per annum

= \( ₹ \frac{12000 \times 5 \times 1}{100 \times 4}=₹ 150\)

The rest time = 1 year – ( 3 months + 3 months)

= 12 months – 6 months = 6 months

= \(\frac{6}{12}\) year = \(\frac{1}{2}\)

∴ The principal of the last months = ₹ (120000 + 8000) = ₹ 20000

∴ The simple interest of ₹ 20000 in 6 months at the rate of 5% per annum.

= \(₹ \frac{12000 \times 5 \times 1}{100 \times 4}= ₹ 150\)

∴ The total interest = ₹ (500 + 150 + 187.50) = ₹ 837.50.

∴ Total amount = ₹(20000 + 837.50) = ₹ 20837.50.

∴ Bhubanbabu will get an amount ₹ 20837.50.

**Class 10 Maths Wbbse Solutions**

**Example 8. Ratanbabu deposits the money for each of his two sons in such a way that when the ages of each of his sons will be 18 years each one will get ₹ 120000. The rate of simple interest per annum in the bank is 10% and the present ages of his sons are 13 years and 8 years respectively. Determine the money, he had deposited separately in the bank for each of his sons.
**

Solution:

**Given:**

Ratanbabu deposits the money for each of his two sons in such a way that when the ages of each of his sons will be 18 years each one will get ₹ 120000. The rate of simple interest per annum in the bank is 10% and the present ages of his sons are 13 years and 8 years respectively.

Let Ratanbabu deposit ₹ x for his 13 yrs aged son and ₹ y for his 8 years aged son in the bank.

∴ Simple interest of ₹ x in (18 – 13) years = 5 years at the rate of 10% per annum

=₹ \(=\frac{x \times 10 \times 5}{100}=₹ \frac{x}{2}\)

Again, simple interest of ₹ y in (18 – 8) years = 10 years at the rate of 10% per annum

= \(₹ \frac{y \times 10 \times 10}{100}=₹ y\)

∴ The son of aged 13 will get at end of 18 years = \(=\left(x+\frac{x}{2}\right)=₹ \frac{3 x}{2}\)

Again, the sons of aged 8 will get at the end of 1.8 years = ₹ (y + y) = ₹ 2y.

As per question, \(\frac{3x}{2}\) = 120000

⇒ x = \(\frac{120000 \times 2}{3}\) = 80000

Again, 2y = 120000

⇒ y= \(\frac{120000}{2}\) = 60000

∴ Ratanbabu deposited ₹ 80000 and ₹ 60000 for the sons respectively in the bank.

**Example 9. At the same rate of simple interest in percent per annum, if a principal becomes the amount of ₹7100 in 7 years and of ₹6200 in 4 years. Determine the principal and rate of simple interest in percent per annum.
**

Solution:

∴ The interest of 4 years = ₹ \(\frac{900 \times 4}{3}\) = ₹ 1200

∴ Principal = ₹ (6200 – 1200) = ₹ 5000.

Let the rate of simple interest in percent per annum be r %.

∴ \(\frac{5000 \times r \times 4}{100}=1200\)

⇒ r =\(\frac{1200 \times 100}{5000 \times 4}=6\)

Principal = ₹ 5000 and rate of interest in percent per annum = 6%

**Class 10 Maths Wbbse Solutions**

**Example 10. Soma auntie deposits ₹ 6,20,000 in such a way in three banks at the rate of simple interest of 5% per annum for 2 years, 3 years, and 5 years respectively so that the total interests in the 3 banks are equal. Calculate the money deposited by Soma’s auntie in each of the three banks.
**

Solution:

**Given:**

Soma auntie deposits ₹ 6,20,000 in such a way in three banks at the rate of simple interest of 5% per annum for 2 years, 3 years, and 5 years respectively so that the total interests in the 3 banks are equal.

Let Soma auntie deposit ₹ x, ₹ y, and ₹ z respectively in the 1st, 2nd, and 3rd banks.

As per the question, x + y + z = 620000 …….. (1)

The simple interest of ₹ x in 2 years at the rate of 5% per annum

= \(₹ \frac{x \times 5 \times 2}{100}=₹ \frac{x}{10}\)

At the same rate, simple interest of ₹ y in 3 years

= \(₹ \frac{y \times 5 \times 3}{100}=₹ \frac{3y}{20}\)

Also, at the same rate, simple interest of ₹ z in 5 years

= \(₹ \frac{z \times 5 \times 5}{100}=₹ \frac{z}{4}\)

As per question, \(\frac{x}{10}\) = \(\frac{3y}{20}\) =k (let)

⇒ 2x = 3y = 5z = k [Multiplying by 20]

⇒ x= \(\frac{k}{2}\),y = \(\frac{k}{3}\) and z = \(\frac{k}{5}\)

Now, putting these values of x, y, and z in (1) we get,

\(\frac{k}{2}\) + \(\frac{k}{3}\) + \(\frac{k}{5}\)= 620000

or, \(\frac{15 k+10 k+6 k}{30}\) = 620000

or, 31 k = 620000 x 30

or, k = \(\frac{620000 \times 30}{31}\)

or, k = 20000 x 30 = 600000

∴ x = \(\frac{600000}{2}\) = 300000

y = \(\frac{600000}{3}\) = 200000

z = \(\frac{600000}{5}\) = 120000

∴ Soma aunties deposited ₹ 300000, ₹ 200000, and ₹ 120000 respectively in three banks.

**Example 11. Amitava deposits ₹ 1000 on the first day of every month in a monthly savings scheme. In the bank, if the rate of simple interest is 5% per annum, then determine the amount, Amitava will get at the end of 6 months.**

Solution:

**Given:**

Amitava deposits ₹ 1000 on the first day of every month in a monthly savings scheme. In the bank, if the rate of simple interest is 5% per annum

As per the question, the money of Amitava that was deposited in 1st, 2nd, 3rd, ……. 6th month have to earn interest for 6 months, 5 months, 4 months, 3 months, ………, 1 month respectively.

∴ The total interest in 6 month

\(=₹\left(\frac{1000 \times 5 \times 6}{100 \times 12}+\frac{1000 \times 5 \times 5}{100 \times 12}+\frac{1000 \times 5 \times 4}{100 \times 12}+\ldots \ldots+\frac{1000 \times 5 \times 1}{100 \times 12}\right)\)

= \( ₹ \frac{1000 \times 5}{100 \times 12}(6+5+4+3+2+1)\)

= \( ₹ \frac{1000 \times 5}{100 \times 12} \times 21=₹ \frac{175}{2}=₹ 87 \cdot 50\)

∴Amitava will get an amount at ₹(6 x 1000 + 87.5)

= ₹ 6087.5 after 6 months

∴ Required amount = ₹ 6087.5

**Class 10 Maths Wbbse Solutions**

**Example 12. Ashisbabu deposited ₹ 50000 for his son of age 10 years, in a monetary fund. The fund investing that money at the rate of 4% simple interest per annum, gave the son a quantity of ₹ 1200 at the end of every year. The annual expenses of the fund is ₹ 300. After fulfilling 18 years, what amount the boy will get from the fund?
**

Solution:

**Given:**

Ashisbabu deposited ₹ 50000 for his son of age 10 years, in a monetary fund. The fund investing that money at the rate of 4% simple interest per annum, gave the son a quantity of ₹ 1200 at the end of every year. The annual expenses of the fund is ₹ 300.

At the rate of 4% per annum, the interest of ₹ 50000 in 1 year

= \(₹ \frac{50000 \times 4 \times 1}{100}=₹ 2000\)

The annual expenses of the fund =₹ (1200 + 300) = ₹ 1500

∴ Rest of the money = ₹ (2000 – 1500) = ₹ 500 .

∴ After completion of 18 years, the total quantity of simple interest = ₹ (18 – 10) x 500 = ₹ 4000.

The total amount the boy will get from the fund at the end of 18 years = ₹ (50000 + 4000) = ₹ 54000.

∴ The required amount = ₹ 54000.

**Example 13. In a bank if ₹ 4750 is deposited in simple interest, it becomes an amount of ₹ 6650 after a period of time of 4 years. According to the same rate of interest, in how many days ₹ 85000 will amount ₹ 106250?
**

Solution:

**Given:**

In a bank if ₹ 4750 is deposited in simple interest, it becomes an amount of ₹ 6650 after a period of time of 4 years.

Let the rate of simple interest be r % per annum.

In the first case, the amount = ₹ 6650, and the principal =₹ 4750.

∴ Interest = ₹ (6650 – 4750) = ₹ 1900.

Accordingly, we get, \(\frac{4750 \times r \times 4}{100}=1900\)

\( r=\frac{1900 \times 100}{4750 \times 4}=10 \)∴ The rate of interest in percent per annum = 10%.

Also, let ₹ 85000 amounts ₹ 106250 in t years,

i.e., principal = ₹ 85000; Amount = ₹ 106250

∴ Interest = ₹ (106250 – 85000) = ₹ 21250.

∴ Rate of interest per annum = 10% and period of time = t years.

∴ \(\frac{85000 \times 10 \times t}{100}=21250\)

t=\(\frac{21250 \times 100}{85000 \times 10}=\frac{5}{2}=2 \frac{1}{2}\)

∴ ₹ 85000 amounts ₹ 106250 in 2 \(\frac{1}{2}\) years.

**Example 14. A certain money amounts ₹ 9440 on simple interest in 3 years. If the rate of simple interest be 25% increased per annum, then the money will become ₹ 9800 after same period of time. Find the principal and the rate of interest in percent per annum.
**

Solution:

**Given:**

A certain money amounts ₹ 9440 on simple interest in 3 years. If the rate of simple interest be 25% increased per annum, then the money will become ₹ 9800 after same period of time.

The difference between two given amounts = ₹ (9800 – 9440) =₹ 360.

Since the principal and period of time are the same, we can say that due to 25% increase of interest, the increase in amount is ₹ 360.

∴ 25% of interest =₹ 360. .

1% of interest = ₹ \(\frac{100}{25}\)

100% of interest = ₹ \(\frac{360 \times 100}{25}\)

∴ Simple interest in 3 years = ₹ 1440.

So, the principal = ₹ (9440- 1440) = ₹8000.

Now, let the rate of interest be r% in percent per annum

∴ \(\frac{8000 \times r \times 3}{100}=1440 \)

r= \(\frac{1440 \times 100}{8000 \times 3}=6\)

∴ The required principal = ₹ 8000 and rate of interest in percent per annum = 6%.

**Class 10 Maths Wbbse Solutions**

**Example 15. If a debt of ₹ 4600 at a rate of simple interest 10% per annum, is to be repaid in 4 years, then what will be the installment per annum?
**

Solution:

**Given:**

If a debt of ₹ 4600 at a rate of simple interest 10% per annum, is to be repaid in 4 years,

Let the installment per annum be ₹ x.

As per question,

\(\left(x+\frac{x \times 10 \times 3}{100}\right)+\left(x+\frac{x \times 10 \times 2}{100}\right)+\left(x+\frac{x \times 10 \times 1}{100}\right)+x=4600\)

⇒ \(\left(x+\frac{3 x}{10}\right)+\left(x+\frac{2 x}{10}\right)+\left(x+\frac{x}{10}\right)+x=4600\)

⇒ \(\frac{13 x}{10}+\frac{12 x}{10}+\frac{11 x}{10}+x=4600\)

⇒ \(\frac{13 x+12 x+11 x+10 x}{10}=4600\)

⇒ \(\frac{46 x}{10}=4600 \Rightarrow x=\frac{4600 \times 10}{46}\)

⇒ x = 1000

∴ The installment per annum = ₹ 1000.

**EXample 16. In a certain bank, the rate of simple interest in percent per annum during first 2 years is 3%, 6% during the next 3 years, and 9% during the rest of the years. If a person deposits some money to that bank, he gets an interest of ₹2760 after 10 years. Find the principal of the person.
**

Solution:

**Given:**

In a certain bank, the rate of simple interest in percent per annum during first 2 years is 3%, 6% during the next 3 years, and 9% during the rest of the years. If a person deposits some money to that bank, he gets an interest of ₹2760 after 10 years.

Let the person deposited ₹ x in the bank.

During first 2 years, he gets simple interest = \(₹ \frac{x \times 3 \times 2}{100}\)

= \(₹ \frac{3 x}{50}\)

During the next 3 years, he gets simple interest = \(₹ \frac{x \times 6 \times 3}{100}=₹ \frac{9 x}{50}\)

During the last (10-2-3) years = 5 years he gets simple interest = \(₹ \frac{x \times 9 \times 5}{100}\)

=\(₹ \frac{9 x}{20}\)

As per question,

\(\frac{3 x}{50}+\frac{9 x}{50}+\frac{9 x}{20}\)=2760

\(\frac{6 x+18 x+45 x}{100}=2760\)

∴ The required principal = ₹ 4000.

**Class 10 Maths Wbbse Solutions**

**Example 17. The simple interest of a principal in 1 year and 9 months at the rate of simple interest of 5% per annum is ₹ 63 more than that of the same principal in 2 years and 4 months at the rate of simple interest of 4 \(\frac{1}{2}\) % per annum. Find the principal.
**

Solution:

**Given:**

The simple interest of a principal in 1 year and 9 months at the rate of simple interest of 5% per annum is ₹ 63 more than that of the same principal in 2 years and 4 months at the rate of simple interest of 4 \(\frac{1}{2}\) % per annum.

1 year and 9 months = \(\left(1+\frac{9}{12}\right) \text { years }=\frac{7}{4} \text { years }\)

Similarly, 2 years and 4 months = \(\left(2+\frac{4}{12}\right) \text { years }=\frac{7}{3} \text { years }\)

Now, let the principal be ₹ x.

So, the simple interest of ₹ x in \(\frac{7}{4}\) years at the rate of 5% simple interest per annum

= ₹ \(\frac{x \times 5 \times 7}{100 \times 4}=₹ \frac{7 x}{80}\)

Again, the simple interest of ₹ x in \(\frac{7}{3}\) years at the rate of 4 \(\frac{1}{2}\) % simple interest per annum

= \(₹ \frac{x \times 9 \times 7}{100 \times 2 \times 3}\)

= \(₹ \frac{21 x}{200}\)

As per the question,

= \(\frac{21 x}{200}-\frac{7 x}{80}=63 \Rightarrow \frac{42 x-35 x}{400}=63\)

or, \(\frac{7x}{400}\)

or, 7x = 63 x 400

or, x = \(\frac{63 \times 400}{7}\)

or, x = 3600

∴ The required principal = ₹ 3600.

**Example 18. A gives some money to B at a rate of 5% and to C ₹ 800 more at a rate of 7% per annum. They repaid both loans after 5 years and so that C have to pay ₹ 1240 more than B. Find the quantities of money which both of them had debted.
**

Solution:

**Given:**

A gives some money to B at a rate of 5% and to C ₹ 800 more at a rate of 7% per annum. They repaid both loans after 5 years and so that C have to pay ₹ 1240 more than B

Let A have given a loan of ₹ x to B and ₹ (x + 800) to C.

∴ The simple interest of ₹ x in 5 years at the rate of 5% per annum = \(₹ \frac{x \times 5 \times 5}{100}=₹ \frac{x}{4}\)

∴ B have to pay \(₹\left(x+\frac{x}{4}\right)=₹ \frac{5 x}{4}\) after 5 years.

Again, the simple interest of ₹ (x + 800) in 5 years at the rate of 7% per annum

= \( ₹ \frac{(x+800) \times 7 \times 5}{100}\)

= \( ₹ \frac{35(x+800)}{100}=₹ \frac{35 x+28000}{100}\)

∴ C have to pay \(₹\left\{x+800+\frac{35(x+800)}{100}\right\}\)

= \( ₹ \frac{100 x+80000+35 x+28000}{100} \)

= \( ₹ \frac{135 x+108000}{100} \text { after } 5 \text { years }\)

As per question, \(\frac{135 x+108000}{100}-\frac{5 x}{4}=1240\)

⇒ \( \frac{135 x+108000-125 x}{100}=1240\)

⇒ 10 x+108000=124000

⇒ 10 x = 124000-108000

⇒ 10x = 16000

⇒ x = 1600.

∴ B had debted Rs. 1600 and C had debted x (1600 + 800) = ₹ 2400.

**Example 19. A person took a loan of X 40000 at the rate of 10% simple interest per annum to construct a house. After 2 years he gave return ₹20000 to the bank. After more 2 years what amount should he return to the bank so as to repay his loan completely?
**

Solution:

**Given:**

A person took a loan of X 40000 at the rate of 10% simple interest per annum to construct a house. After 2 years he gave return ₹20000 to the bank.

At the rate of 10% per annum,

The simple interest of ₹ 40000 in 2 yeras = ₹ \(\frac{40000 \times 10 \times 2}{100}=₹ 8000\)

When the person gave ₹ 20000 to the bank, it will cut ₹ 8000 from ₹ 20000 as interest, and the rest of the money,

i.e. ₹(20000 – 8000) =₹ 12000 will be subtracted from the principal ₹ 40000.

Thus, for the next 2 years, the person will have to pay interest on the principal ₹(40000 – 12000) = ₹ 28000.

Now, the simple interest of ₹ 28000 in 2 years at the rate of 10% per annum = \(₹ \frac{28000 \times 10 \times 2}{100}\) = ₹ 5600.

∴ The total amount = ₹ (28000 + 5600) =₹ 33600.

Hence, the person have to pay ₹ 33600 to the bank after more 2 years so as to repay his loan completely.