# Online Sainik School Coaching|Simple Interest L-1|Anand Classes Jalandhar|@9463138669

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Sainik School Exam Solutions for Simple Interest is the best study material for those students who are finding difficulties in solving problems. Students can refer and download the PDF of Chapter Simple Interest from the given links. The solutions to all questions in Anand Classes Study Material Solutions for Sainik School Exam are given here in a detailed and step by step way to help students understand more effectively. Simple Interest is based on general transactions with interest. Anand Classes Solutions covers all the topics related to it. Some of the important topics of this lesson are listed below.

- Definition of principal
- Definition and meaning of interest
- Formula to calculate the amount
- Definition and meaning of simple interest
- Formula to calculate simple interest
- Verbally related problems on simple interest

## What is Simple Interest?

Simple Interest (S.I) is the method of calculating the interest amount for some principal amount of money. Have you ever borrowed money from your siblings when your pocket money is exhausted? Or lent him maybe? What happens when you borrow money? You use that money for the purpose you had borrowed it in the first place. After that, you return the money whenever you get the next month’s pocket money from your parents. This is how borrowing and lending work at home.

But in the real world, money is not free to borrow. You often have to borrow money from banks in the form of a loan. During payback, apart from the loan amount, you pay some more money that depends on the loan amount as well as the time for which you borrow. This is called simple interest. This term finds extensive usage in banking.

## Simple Interest Formula

The Formula for simple interest helps you find the interest amount if the principal amount, rate of interest and time periods are given.

Simple interest formula is given as:

**SI = (P × R ×T) / 100**

Where SI = simple interest

P = principal

R = interest rate (in percentage)

T = time duration (in years)

In order to calculate the total amount, the following formula is used:

**Amount (A) = Principal (P) + Interest (I)**

Where,

**Amount **(A) is the total money paid back at the end of the time period for which it was borrowed.

**Simple Interest Formula For Months**

The formula to calculate the simple interest on a yearly basis has been given above. Now, let us see the formula to calculate the interest for months. Suppose P be the principal amount, R be the rate of interest per annum and n be the time (in months), then the formula can be written as:

Simple Interest for n months = (P × n × R)/ (12 ×100)

o calculate the Principal Amount, the formula is:

P = (I × 100) / RT

To find the rate of interest, the formula will be:

R = (I × 100)/ PT

R (in decimal ) = I/PT

Thus, the rate of interest in percent is given by:

R = R * 100

To get the time, formula is:

T = (I × 100) / PR

All these formulas can be used based on the information provided in different scenarios.

### Simple Interest Problems

Let us see some simple interest examples using the simple interest formula in maths.

**Example 1:**

**Rishav takes a loan of Rs 10000 from a bank for a period of 1 year. The rate of interest is 10% per annum. Find the interest and the amount he has to the pay at the end of a year.**

**Solution**:

Here, the loan sum = P = Rs 10000

Rate of interest per year = R = 10%

Time for which it is borrowed = T = 1 year

Thus, simple interest for a year, SI = (P × R ×T) / 100 = (1000× 100 ×1) / 100 = Rs 1000

Amount that Rishav has to pay to the bank at the end of the year = Principal + Interest = 10000 + 1000 = Rs 11,000

**Example 2:**

**Namita borrowed Rs 50,000 for 3 years at the rate of 3.5% per annum. Find the interest accumulated at the end of 3 years.**

**Solution:**

P = Rs 50,000

R = 3.5%

T = 3 years

SI = (P × R ×T) / 100 = (50,000× 3.5 ×3) / 100 = Rs 5250

**Example 3:**

**Mohit pays Rs 9000 as an amount on the sum of Rs 7000 that he had borrowed for 2 years. Find the rate of interest.**

**Solution:**

A = Rs 9000

P = Rs 7000

SI = A – P = 9000 – 7000 = Rs 2000

T = 2 years

R = ?

SI = (P × R ×T) / 100

R = (SI × 100) /(P× T)

R = (2000 × 100 /7000 × 2) =14.29 %

Thus, R = 14.29%

To learn more about other topics in Maths, visit Anand Classes – The Learning App.

### More Problems

**1. Find the simple interest, when:
(i) Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years.
(ii) Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years.
(iii) Principal = Rs 4500, Rate of Interest = 4% per annum and Time = 6 months.**

**(iv) Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months.
(v) Principal = Rs 1000, Rate of Interest = 10% per annum and Time = 73 days.**

**Solution:**

(i) Given Principal = Rs 2000, Rate of Interest = 5% per annum and Time = 5 years.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (2000 × 5 × 5)/100

= Rs 500

(ii) Given Principal = Rs 500, Rate of Interest = 12.5% per annum and Time = 4 years.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (500 × 4 × 12.5)/100

= Rs 250

(iii) Given Principal = Rs 4500, Rate of Interest = 4% per annum and Time = 6 months = ½ years

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (4500 × ½ × 4)/100

SI = (4500 × 1 × 4)/100 × 2

= Rs 90

(iv) Given Principal = Rs 12000, Rate of Interest = 18% per annum and Time = 4 months = (4/12) = (1/3) years

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (12000 × (1/3) × 18)/100

SI = (12000 × 1 × 18)/100 × 3

= Rs 720

(v) Given Principal = Rs 1000, Rate of Interest = 10% per annum and

Time = 73 days = (73/365) days

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (1000 × (73/365) × 10)/100

SI = (1000 × 73 × 10)/100 × 365

= Rs 20

**2. Find the interest on Rs 500 for a period of 4 years at the rate of 8% per annum. Also, find the amount to be paid at the end of the period.**

**Solution:**

Given Principal amount P = Rs 500

Time period T = 4 years

Rate of interest R = 8% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (500 × 4 × 8)/100

= Rs 160

Amount = Principal amount + Interest

= Rs 500 + 160

= Rs 660

**3. A sum of Rs 400 is lent at the rate of 5% per annum. Find the interest at the end of 2 years.**

**Solution:**

Given Principal amount P = Rs 400

Time period T = 2 years

Rate of interest R = 5% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (400 × 2 × 5)/100

= Rs 40 **Best Online RIMC RMS Coaching Center in Jalandhar**

**4. A sum of Rs 400 is lent for 3 years at the rate of 6% per annum. Find the interest.**

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Principal amount P = Rs 400

Time period T = 3 years

Rate of interest R = 6% p.a. **Best Online RIMC RMS Coaching Center in Jalandhar**

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (400 × 3 × 6)/100 **Best Online RIMC RMS Coaching Center in Jalandhar**

= Rs 72 **Best Online RIMC RMS Coaching Center in Jalandhar**

**5. A person deposits Rs 25000 in a firm who pays an interest at the rate of 20% per annum. Calculate the income he gets from it annually.**

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Given Principal amount P = Rs 25000 **Best Online RIMC RMS Coaching Center in Jalandhar**

Time period T = 1 year **Best Online RIMC RMS Coaching Center in Jalandhar**

Rate of interest R = 20% p.a. **Best Online RIMC RMS Coaching Center in Jalandhar**

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (25000 × 1 × 20)/100 **Best Online RIMC RMS Coaching Center in Jalandhar**

= Rs 5000 **Best Online RIMC RMS Coaching Center in Jalandhar**

**6. A man borrowed Rs 8000 from a bank at 8% per annum. Find the amount he has to pay after 4 ½ years. Best Online RIMC RMS Coaching Center in Jalandhar**

**Solution: Best Online RIMC RMS Coaching Center in Jalandhar**

Given Principal amount P = Rs 8000

Time period T = 4 ½ years = 9/2 years

Rate of interest R = 8% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (8000 × (9/2) × 8)/100

= Rs 2880

Amount = Principal amount + Interest

= Rs 8000 + 2880

= Rs 10880

**7. Rakesh lent out Rs 8000 for 5 years at 15% per annum and borrowed Rs 6000 for 3 years at 12% per annum. How much did he gain or lose?**

**Solution:**

Given Principal amount P = Rs 8000

Time period T = 5 years

Rate of interest R = 15% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (8000 × 5 × 15)/100

= Rs 6000

Principal amount P = Rs 6000

Time period T = 3 years

Rate of interest R = 12% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (6000 × 3 × 12)/100

= Rs 2160

Amount gained by Rakesh = Rs 6000 − Rs 2160

= Rs 3840

**8. Anita deposits Rs 1000 in a savings bank account. The bank pays interest at the rate of 5% per annum. What amount can Anita get after one year?**

**Solution:**

Given Principal amount P = Rs 1000

Time period T = 1 year

Rate of interest R = 5% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (1000 × 1 × 5)/100

= Rs 50

Total amount paid after 1 year = Principal amount + Interest

= Rs 1000 + Rs 50

= Rs 1050

**9. Nalini borrowed Rs 550 from her friend at 8% per annum. She returned the amount after 6 months. How much did she pay?**

**Solution:**

Given Principal amount P = Rs 550

Time period T = ½ year

Rate of interest R = 8% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (550 × ½ × 8)/100

= Rs 22

Total amount paid after ½ year = Principal amount + Interest

= Rs 550 + Rs 22

= Rs 572

**10. Rohit borrowed Rs 60000 from a bank at 9% per annum for 2 years. He lent this sum of money to Rohan at 10% per annum for 2 years. How much did Rohit earn from this transaction?**

**Solution:**

Given Principal amount P = Rs 60000

Time period T = 2 years

Rate of interest R = 10% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (60000 × 2 × 10)/100

= Rs 12000

Principal amount P = Rs 60000

Time period T = 2 years

Rate of interest R = 9% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (60000 × 2 × 9)/100

= Rs 10800

Amount gained by Rohit = Rs 12000 − Rs 10800

= Rs 1200

**11. Romesh borrowed Rs 2000 at 2% per annum and Rs 1000 at 5% per annum. He cleared his debt after 2 years by giving Rs 2800 and a watch. What is the cost of the watch?**

**Solution:**

Given Principal amount P = Rs 2000

Time period T = 2 years

Rate of interest R = 2% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (2000 × 2 × 2)/100

= Rs 80

Principal amount P = Rs 1000

Time period T = 2 years

Rate of interest R = 5% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (1000 × 2 × 5)/100

= Rs 100

Total amount that he will have to return = Rs. 2000 + 1000 + 80 + 100 = Rs. 3180

Amount repaid = Rs. 2800

Value of the watch = Rs. 3180 – 2800 = Rs. 380

**12. Mr Garg lent Rs 15000 to his friend. He charged 15% per annum on Rs 12500 and 18% on the rest. How much interest does he earn in 3 years?**

**Solution:**

Given Principal amount P = Rs 15000

Time period T = 3 years

Rate of interest R = 15% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (15000 × 3 × 15)/100

= Rs 6750

Rest of the amount lent = Rs 15000 − Rs 12500 = Rs 2500

Rate of interest = 18 % p.a.

Time period = 3 years

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (2500 × 3 × 18)/100

= Rs 1350

Total interest earned = Rs 6750 + Rs 1350 = Rs 8100

**13. Shikha deposited Rs 2000 in a bank which pays 6% simple interest. She withdrew Rs 700 at the end of first year. What will be her balance after 3 years?**

**Solution:**

Given Principal amount P = Rs 2000

Time period T = 1 year

Rate of interest R = 6% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (2000 × 1 × 6)/100

= Rs 120

So amount after 1 year = Principal amount + Interest = 2000 + 120 = Rs 2120

after 1 year, amount withdrawn = Rs 700

Principal amount left = Rs 2120 − Rs 700 = Rs 1420

Time period = 2 years

Rate of interest = 6% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (1420 × 2 × 6)/100

Interest after two years = Rs 170.40

Total amount after 3 years = Rs 1420 + Rs 170.40 = Rs 1590.40

**14. Reema took a loan of Rs 8000 from a money lender, who charged interest at the rate of 18% per annum. After 2 years, Reema paid him Rs 10400 and wrist watch to clear the debt. What is the price of the watch?**

**Solution: Best Online Sainik School Coaching center in jalandhar**

Given Principal amount P = Rs 8000

Time period T = 2 years Best Online Sainik School Coaching center in jalandhar

Rate of interest R = 18% p.a. Best Online Sainik School Coaching center in jalandhar

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (8000 × 2 × 18)/100 Best Online Sainik School Coaching center in jalandhar

= Rs 2880 Best Online Sainik School Coaching center in jalandhar

Total amount payable by Reema after 2 years = Rs 8,000 + Rs 2,880

= Rs 10,880 Best Online Sainik School Coaching center in jalandhar

Amount paid = Rs 10,400 Best Online Sainik School Coaching center in jalandhar

Value of the watch = Rs 10,880 − Rs 10,400 = Rs 480

**15. Mr Sharma deposited Rs 20000 as a fixed deposit in a bank at 10% per annual. If 30% is deducted as income tax on the interest earned, find his annual income.**

**Solution: Best Online Sainik School Coaching center in jalandhar**

Given Principal amount P = Rs 20000

Time period T = 1 year Best Online Sainik School Coaching center in jalandhar

Rate of interest R = 10% p.a.

We know that simple interest = (P × T × R)/100

On substituting these values in above equation we get

SI = (20000 × 1 × 10)/100

= Rs 2000 Best Online Sainik School Coaching center in jalandhar

Amount deducted as income tax = 30% of 2000 = (30 × 2000)/100

= Rs 600 Best Online Sainik School Coaching center in jalandhar

Annual interest after tax deduction = Rs 2,000 − Rs 600 = Rs 1,400

## Frequently Asked Questions on Simple Interest – FAQs

### What is simple interest and example?

### What is simple interest and compound interest?

### What are the types of simple interest?

### What are the 2 types of interest?

### How do I calculate S.I.?

SI = (PTR)/100

Here,

SI = Simple interest

P = Principal (sum of money borrowed)

R = Rate of interest p.a

T = Time (in years)

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**AISSEE – **All India Sainik School Entrance Exam acronym as AISSEE** **is conducted to admit boys to various Sainik School (All the Sainik Schools are affiliated to CBSE, New Delhi.) throughout India. The entrance exam is held only for students aspiring to join VI and IX standards.

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**ALL INDIA SAINIK SCHOOL ENTRANCE EXAM ELIGIBILITY**

**Eligibility for VI Class**

**Educational Qualification**

The aspiring students must have completed their previous years’ education from recognized school to get admission to Sainik School. For SC candidates 15% and for ST candidates 7.5% of total seats are reserved. 25% of seats are reserved for the children of service personnel and ex-servicemen.

**Age Limits: **

- Boys aged between 10 and 11 years are eligible to appear for the entrance exam to VI standard
- Boys aged between 13 and 14 years are eligible for IX standard.

**ALL INDIA SAINIK SCHOOL ENTRANCE EXAM SYLLABUS**

**Syllabus for Class IX**

Subjects |
Marks |

Mathematics and Science | 200+75 |

English and Social Studies | 100+75 |

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Syllabus for Class VI**

Subject |
Marks |

Mathematics and Language Ability Test | 200 |

Intelligence Test | 100 |

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**All India Sainik School Entrance Exam Pattern**

The medium of examination for Class IX is English. However students are allowed to write exam in any of their preferred languages.

For the Class VI students, question papers will be prepared in English, Hindi and all other recognized regional official languages.

Question paper pattern keep on changing every year. The syllabus for the VI and IX entrance exams is usually based on the NCERT books of V and VIII respectively.

**HOW TO APPLY FOR ALL INDIA SAINIK SCHOOL ENTRANCE EXAM**

Aspiring Candidates can obtain prospectus, application form and question papers from the Principal of the respective Sainik schools at the residing place of the candidate.

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**ALL INDIA SAINIK SCHOOL ENTRANCE EXAM DATE**

All India Sainik School Entrance usually held in the month of April.

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**ALL INDIA SAINIK SCHOOL EXAM CONTACT DETAILS**

CBSE – Regional Office, PS 1-2, Institutional Area

I.P. EXTN., Patparganj

New Delhi – 110 090

Ph No: 91-11-22509252 – 59

Website: http://cbse.nic.in/

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**LIST OF SAINIK SCHOOL IN INDIA**

There are 18 Sainik Schools in various state of India these are **:**

- Sainik School – Kanjpura (Haryana)
- Sainik School – Kapurthala (Punjab)
- Sainik School – Chittorgarh (Rajasthan)
- Sainik School – Satara (Maharashtra)
- Sainik School – Balachadi (Gujarat)
- Sainik School – Korukonda (Andhra Pradesh)
- Sainik School – Bhubaneshwar (Orissa)
- Sainik School – Purulia (West Bengal)
- Sainik School – Kazhakootam (Kerala)
- Sainik School – Amravatinagar (Tamil Nadu)
- Sainik School – Rewa (Madhya Pradesh)
- Sainik School – Bijapur (Karnataka)
- Sainik School – Tilaiya (Bihar)
- Sainik School – Gopalpara (Assam)
- Sainik School – Gorakhpur (Uttar Pradesh)
- Sainik School – Nagrota (Jammu and Kashmir)
- Sainik School – Imphal (Manipur) and
- Sainik School – Sujanpur Tira (Himachal Pradesh)

* All Sainik Schools are affiliated to CBSE, New Delhi*

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