**How to Solve Simple Interest Questions? Tips & Tricks**

How to Solve Simple Interest Questions? Tips & Tricks. In the Quantitative Aptitude section of the Bank exam consists of questions such as Simple Interest.It is a very important part of the Quantitative Aptitude section of Bank Clerk,Bank PO, SSC CGL, SSC CHSL, SSC MTS, SSC CPO and other Exams. Usually 2-3 questions are expected from this topic and they usually take less than a minute to solve. This makes it the most important topic of Quant section.

**How to Solve Simple Interest Questions Basic Tricks**

**What is simple interest?**

**Simple Interest (SI)**

If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI). (fix percentage of principal)

** ****Principal (sum)**

Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P.

** ****What is Amount?**

The Addition of Simple Interest and Principle is called the Amount.

**A = S.I + P ( Principle )**

** ****Interest**

Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed calculated on the basis of Principle.

** ****Time**

This is the duration for which money is lend/borrowed.

** ****Rate of Interest**

It is the rate at which the interest is charge on principal.

**What is Per annul means?**

“Rate of interest R% per annum” means that the interest for one year on a sum. If not stated explicitly, rate of interest is assumed to be for one year.

In simple interest, we will learn about how to calculate simple interest. We will recapitulate the formula for simple interest and know more about it. When we borrow money from any source (bank, agency, moneylender), we have to pay back the money after a certain period along with extra money for availing the facility to use the money borrowed.

*We already know what is simple interest and while calculating the time period we need to:*

Day on which money is borrowed is not counted but the day which the money is returned is counted.

When number of day is converted into year, we always divide the number of days by 365, whether it is a leap year or an ordinary year.

Here,

**Tricks & Tips of Simple Interest**

**P** = Principal

**R** = rate% per annum

**T** = time

**I** = simple interest

**A** = amount

Formula for calculating simple interest is **S.I = (P × R × T)/100**

** ****From the above formula, we can derive the followings**

P=(100×SI)/ RT

R=(100×SI)/ PT

T=(100×SI)/ PR

**Important:** Formula for calculating amount is **A = P + I**

**Some Tricks to Solve easily**

**Trick 1:**

If a sum of money becomes** “n”** times in** “T years”** at simple interest, then the rate of interest per annum can be given be

**Trick 2**:

If an amount **P1** is lent out at simple interest of **R1% **per annum and another amount **P2** at simple interest rate of **R2%** per annum, then the rate of interest for the whole sum can be given by

** **

**Trick 3: **A sum of money at simple interest **n1** itself in **t1 year**. It will become **n2 **times of itself in (If Rate is constant)

** **

**Trick 4:**

In what time will the simple interest be **“n”** of the principal at **“r %”** per annum:-

**rt =n x 100**

** ****Trick 5:**

If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are **R1, R2, … , Rn** respectively and time periods are **T1, T2, … , Tn **respectively, then the ratio in which the sum will be divided in n parts can be given by

**Examples on simple interest-**

**1. Find simple interest on $2000 at 5% per annum for 3 years. Also, find the amount.**

**Solution: **

Principal = $2000

Rate = 5% p.a.

T = 3 years

S.I = (P × R × T)/100

= (2000 × 5 × 3)/100

= $ 300

Amount = P + I

= $ ( 2000 + 300 )

= $ 2300

**2. Calculate the simple interest on $ 6400 at 10% p.a. for 9 months.**

**Solution: **

P = $ 6400

R = 10% p.a.

T = 9 months or 9/12 years

[12 months = 1 year1 months = 1/12 years

9 months = (1 × 9)/12 years]

Therefore, S.I. = (P × R × T)/100

= (6400 × 10 × 9)/(100 × 12)

= $480

**3. Mike took a loan of $20000 from a bank on 4 February 2009 at the rate of 8% p.a. and paid back the same on 6th July 2009. Find the total amount paid by Mike.**

P = $20000

R = 8 % p.a.

T = 152/365

**Solution: **

Time = February + March +April + May + Jun + July

= 24 days + 31 days + 30 days + 31 days + 30 days + 6 days

= 152 days

Therefore, S.I. = (P × R × T)/100

= (20000 × 8 × 152)/(100 × 365)

= $ (40 × 8 × 152)/73

= $ 666.30

Therefore, the amount paid = $ (20,000 + 666.30) = $ 20666.30

**4. At what per cent will $ 1500 amount to $ 2400 in 4 years?**

**Solution: **

P = $ 1500

R = ?

T= 4 years and

A = $ 2400

S.I. = A – P

= $(2400 – 1500 )

= $ 900

S.I. = (P × R × T)/100

900 = (1500 × R × 4)/100

Therefore, R = (900 × 100)/(4 × 1500) = 15%

**5. In how much time will a sum of money triple itself at 15 % p.a.?**

**Solution: **

Let P = x, then A = 3x

So, I = A – P

= 3x – x = 2x

We know that S.I = (P × R × T)/100

2x = (x × 15 × T)/100

T = (2x × 100)/(x × 15) = 40/3 = 13.3 years

**6. At what rate percent per annum simple interest will a sum of money double itself in 6 years?**

**Solution: **

Let P = x, then A = 2x

Also, S.I = A – P

= 2x – x

= x

T = 6 years

We know that S.I. = (P × R × T)/100

(x × R × 6)/100 = x

R = 100x/6x = 16.6 %

**7. A some amounted to $ 2520 at 10% p.a. for the period of 4 years. Find the sum.**

**Solution: **

Let A = $ 2520

R = 10% p.a.

T = 4 years

P = ?

Let the principal be x

S.I = (x × 10 × 4)/100 = 2x/5

A = P + I

A = x + 2x/5

A = (5x + 2x)/5 = 7x/5 [But given that A = $2520]

7x/5 = 2520

7x = 2520 × 5

x = (2520 × 5)/7 = $ 1800

**8. Ron borrowed $ 24000 from his friend at the rate of 12% per annum for 3 years. At the end of the period, he cleared the account by paying $ 10640 cash and giving the cow. Find the price of the cow.**

**Solution: **

Here, P = $24000

R = 12% p.a.

T = 3 years

S.I. = (P × R × T)/100

= (24000 × 12 × 3)/100 = $ 8640

Amount = $24000 + 8640 = $ 32640

Now, $10640 + Price of cow = $ 32460

Therefore, price of the cow = $ 32460 – 10640 = $ 22000

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