# Simple Interest Problems

Simple interest is the interest calculated on the principal amount which is borrowed. While learning how simple interest is calculated, the main terms are principal denoted by P, rate of interest denoted by R and time in years denoted by T.

The branch of commercial mathematics has one of the most important concepts, that is interest. The two types of interest are simple interest and compound interest. The idea of simple interest is based on the time value of money which has a current value, present value and future value. If invested in a deposit, earns an amount called interest. In this article, we learn about simple interest, compound interest and how to solve simple interest problems.

Following example makes these terms more clear. Malini said that she is going to buy a new refrigerator. Her father asked whether she had the money to buy it. She said her father that she is planning to take a loan from the bank. The money she borrows is the sum borrowed or the principal. To keep this money for some time, she needs to pay some extra money to the bank. That amount is the interest. So at the end of the year she has to pay back the money borrowed and the interest. This is the amount denoted by A.

Therefore, Amount = Principal + Interest.

Interest is given in percentage for a time of one year. For example 12% per annum or 12% p.a.

This means that for every 100 Rs you borrow, you have to pay Rs. 12 as interest for one year.

Example: Rajiv takes a loan of Rs. 7000 from a bank at 10% per year as rate of interest. Find the interest he has to pay at the end of one year.

Solution: Here sum borrowed, P = 7000

Rate of interest, R = 10%

This means if he borrowed Rs 100, he had to pay Rs 10 as interest. So for Rs. 7000, the interest he has to pay for one year is 7000×10/100 = Rs. 700.

So at the end of the year, the amount he has to pay back = 7000+700 = Rs. 7700

Interest for multiple years:

If money is borrowed for multiple years, the interest is calculated for the period of time it is kept. For example, if Rajiv is returning money at the end of 2 years, then he has to pay twice the interest. Rs. 700 for first year and Rs. 700 for the second year. As the number of years increase, the interest to be paid also increases. If Rs.100 is borrowed for 4 years at 5% , then the interest to be paid at the end of 4 years is 5+5+5+5 = 4×5 = 20.

So the interest paid for T years for a principal P at the rate R% is S.I = PTR/100

## How Simple Interest is different to Compound Interest

The interest calculated on the principal amount for a fixed period of time and rate of interest is called simple interest. It is used for a single period.

The interest calculated at the end of a certain fixed period and which adds to the principal so that interest can be earned in the next compounding period is called compound interest.

### SI and CI Formulas

 Simple Interest = $\frac{(P × R × T)}{100}$ Amount = SI + P A = $[\frac{PTR}{100}]+P$
 Compound Interest = CI = $P*[1+\frac{R}{100}]^{t}-P$ FA = $P*[1+\frac{R}{100}]^{t}$

where ,

CI: Compound Interest

FA: Future amount

P: Principal

R: Rate of interest per annum

T: Time in years

A: Amount

SI: Simple Interest

## Solved Examples

Example 1: Amount of Rs. 12800 was invested by Mr Rohan dividing it into two different investment schemes A and B at a simple interest rate of 11% and 14%. What was the amount in plan B if the amount of interest earned in two years was Rs. 3508.

Solution:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (12800 – x).

Then, [x . 14 . 2]/100 + [(12800 – x) . 11 . 2]/100 = 3508

28x – 22x = 350800 – (12800 x 22)

6x = 69200

x = 11533.33

So, sum invested in Scheme B = Rs. (12800 – 11533.33) = Rs. 1266.67.

Example 2: A lender claims to be lending at simple interest, but he adds the interest every 6 months in the calculation of principal. The rate of interest charged by him is 8%. What will be the effective rate of interest?

Solution:

Let the sum be Rs. 100.

Then,

Simple interest for 1st 6 months = Rs. [100 x 8 x 1]/[100 x 2] = Rs. 4

Simple interest for last 6 months = Rs. [104 x 8 x 1]/[100 x 2] = Rs.4.16

So, amount at the end of 1 year = Rs. (100 + 5 + 4.16) = Rs. 109.16

Effective rate = (109.16 – 100) = 9.16%

Example 3: The rate of interest offered by a bank is 10% compounded annually. A sum of Rs.10000 is deposited by a person in his account. If the sum isn’t withdrawn, then what will be the balance of his account after 4 years?

Solution:

Rs. 10000 after 4 years = 10000(1+10/100)4 = 10000(11/10)4 = Rs. 14641

Rs. 10000 after 3 years = 10000(1+10/100)3 = 10000(11/10)3 = Rs. 13310

Rs. 10000 after 2 years = 10000(1+10/100)2 = 10000(11/10)2 = Rs. 12100

Rs. 10000 after 1 year = 10000(1+10/100)1 = 10000(11/10) = Rs. 11000

Total amount after 4 years = 14641 + 13310 + 12100 + 11000 = Rs. 51051

Example 4: A town has a population of 30,000. The population decreases by 10 per thousand per year. What will be the population after 2 years?

Solution:

R = [10 × 100] / 1000 = 1%

(Because percentage is calculated for twenty per thousand)

Population after 2 years will be = P(1−R/100)T

= 30000(1−1/100)2

= 30000(99/100)2

= 29403

Example 5: The time required for a sum of money to amount to five times itself at 16% simple interest p.a. will be

Solution:

Let the sum of money be Rs. x and the time required to amount to five times itself be t years.

So, the interest in ‘t’ year should be Rs. 4x.

In case of simple interest, we know,

(P × T × r)/100 = SI

Where, P = Principal amount, T = Duration in years, i = Interest rate per year, SI = Total simple interest

Then,

x × t × 16% = 4x

⇒ t × (16/100) = 4

⇒ t = 400/16 = 25

∴ The required time = 25 years.

Example 6:  The rate of simple interest per annum at which a sum of money doubles itself in 16⅔ years is:

Solution:

Let the principal amount be P.

Now the amount A after 16⅔  years is doubled, hence amount is 2P.

I = P × R × T/100

Where,

P = principal amount

R = rate of interest

T = time in years = 162/3 = 50/3

I = simple interest

Amount A = I + P

According to question

A = P + P × R × T/100

2P = P + P × R × T/100

P = P × R × T/100

R = 100/T

R = 100 × 3/50

R = 6%

Example 7: In which year will the amount on a sum of Rs. 800 at 20% compounded half-yearly exceeds Rs.1000?

Solution:

Let the time taken for this amount to reach Rs.1000 be X.

The important thing to note is that this sum is compounded half-yearly. Hence, we use the formula:

$A = P 1 + [\frac{r}{m*100}]^{mT}$

Where, A= Amount,

P = Principal

r = interest rate

m = no. of periods within a year

T = no. of years

We need to obtain T such that the RHS should be greater than the LHS:

In this case,

A = 1000

P = 800

r = 20%

m = 2 (since it is half-yearly)

Substituting these values, we have

$1000<800(1+20 / [2*100])^{2T}=800(11/10)^{2T}\\$ $1000/800<(121/100)^T\\$

⇒ 1.25 < (1.21)T

Now, we need to use trial and error to check for the values of T.

For T = 1, 1.25 > 1.21, hence the condition is not satisfied.

For T = 2, 1.25 < 1.4641, hence the condition is met.

Therefore, it is the second year in which the amount would be greater than Rs.1000.

Example 8: In how many years will a sum of Rs. 4,000 yield a simple interest of Rs. 1,440 at 12% per annum?

Solution:

We know that, the formula for simple interest:

S.I.= [× × T] / 100

Where,

S.I. = Simple Interest = 1440

P= principal = 4000

T = Time = ?

R= Rate of Interest = 12%

Substituting the values in the formula

1440 = [4000 × 12 × T] / 100

⇒1440 = [4000 × 12 × T] / 100

⇒ T = 12/4 = 3 yrs