Simple interest and solved examples with shortcut method
Simple Interest
When you first deposit money in a savings account, or when you borrow money from another person, a bank or any financial institution, that amount is known as the principal.
The extra amount paid by the borrower to the lender for the use of the amount lent is called interest.
When we borrow some amount from another person for a certain period of time, we have to pay him some extra money, which is called interest. When this interest is the same for every equal period of time, it is called simple interest or SI.
Formulae
S.I = (P × R × T) ÷ 100A = P + SISI = A − P
- If SI, R and T are knownP = (SI × 100) ÷ (R × T)
- If SI, T and P are knownR = (SI × 100) ÷ (P × T)
- If SI, P and R are knownT = (SI × 100) ÷ (R × P)
Solved examples
Ex.1If the SI for 6 years is equal to 30% of the principal, in how many years will the SI be equal to 2 times the principal?
Let the principal = 100.
SI of 30% is earned in 6 years, so 1% takes 6/30 years.
SI equal to 2 times the principal = 200%, so time = 200 × (6/30) = 40 years.
Ans: 40 years
Ex.2At what rate of interest will any principal become 5 times itself in 20 years?
Let the principal = 100, so the amount = 5 × 100 = 500.
SI = A − P = 500 − 100 = 400
400 = (100 × R × 20)/100, so R = 20%.
Ans: 20% per annum
Special case formula
If a sum becomes n1 times itself in t1 years and n2 times itself in t2 years, then:
(n1 − 1)/t1 = (n2 − 1)/t2
Installment
An installment is a sum of money due as one of several equal payments for something, spread over an agreed period of time.
Equal installment = (100A) / [100T + RT × (T − 1)/2]
Questions based on Simple Interest with short tricks
Ex.3Find the S.I on Rs. 5000 at the rate of 5% per year for 5 years.
Method 1
SI = (P × R × T)/100 = (5000 × 5 × 5)/100 = Rs. 1250
Method 2
In simple interest the same amount is earned every year.
1% of Rs. 5000 = Rs. 50
5% for 5 years means 25%, so SI = 25 × 50 = Rs. 1250
Ans: Rs. 1250
Ex.4On a certain sum of money at 6% per year, the simple interest for 3 years is Rs. 324. Find the sum.
Method 1
P = (SI × 100)/(R × T) = (324 × 100)/(6 × 3) = Rs. 1800
Method 2
For P = 100: in 1 year the SI is 6, so in 3 years it is 18.
So SI of 18 corresponds to P of 100.
Multiply both by 18: SI = 18 × 18 = 324 and P = 100 × 18 = Rs. 1800
Ans: Rs. 1800
Ex.5At what rate of interest per annum will the S.I on Rs. 1000 be Rs. 343 in 7 years?
Method 1
SI = PRT/100, so 343 = (1000 × R × 7)/100 and R = 4.9%
Method 2
Rs. 1000 earns Rs. 343 in 7 years, so Rs. 49 in 1 year.
Rs. 1000 earns Rs. 49, so Rs. 100 earns Rs. 4.9, i.e. R = 4.9%
Ans: 4.9%
Ex.6A certain sum of money becomes 5/2 times itself in 5 years. Find the rate of interest.
If a sum becomes n times itself, then (n − 1) = RT/100.
(5/2 − 1) = (R × 5)/100
3/2 = R/20, so R = 30%.
Ans: 30%
Ex.7A sum of money is lent at simple interest. After 5 years its S.I becomes 2/5 of the principal. Find the rate of interest.
SI/P = RT/100
2/5 = (R × 5)/100, so R = 8%.
Ans: 8%
Ex.8The difference between the simple interest on a certain sum for 4 years and for 6 years is Rs. 1125. If the rate is 5% p.a., find the sum.
Method 1
P = [(SI1 − SI2) × 100]/[R(T1 − T2)] = (1125 × 100)/(5 × 2) = Rs. 11250
Method 2
The difference is for T = 2 years at R = 5%, so SI = RT% = 10% of P.
10% = Rs. 1125, so 100% = (1125/10) × 100 = Rs. 11250
Ans: Rs. 11250
Ex.9Ram lent Rs. 9000 in two parts, one at 5% and the other at 8%. The combined rate of interest is 7%. Find the money lent at 8%.
Method 1
Let the sum lent at 8% = x.
(8 × 2 × x)/100 + (5 × 2 × (9000 − x))/100 = (7 × 2 × 9000)/100
16x + 90000 − 10x = 126000, so 6x = 36000 and x = Rs. 6000
Method 2 (alligation)
P(5%) : P(8%) = (8 − 7) : (7 − 5) = 1 : 2
Money lent at 8% = (2/3) × 9000 = Rs. 6000
Ans: Rs. 6000
Ex.10A sum becomes 4 times itself in 30 years. In how many years will it become 6 times itself?
(n1 − 1)/t1 = (n2 − 1)/t2
(4 − 1)/30 = (6 − 1)/t2
t2 = 50 years
Ans: 50 years
Ex.11Ram lent the same sum of money to Mohan for 5 years and to Gaurav for 3 years at 6% p.a. The total interest received is Rs. 1920. Find the sum.
Method 1
P = [(SI1 + SI2) × 100]/[(T1 + T2) × R] = (1920 × 100)/(6 × 8) = Rs. 4000
Method 2
Total T = 5 + 3 = 8 years at R = 6%, so SI = RT% = 48% of P.
48% = 1920, so 1% = 40 and 100% = Rs. 4000
Ans: Rs. 4000
Ex.12What equal installment will discharge a debt of Rs. 848 in 4 years at 4%?
Installment = (100 × A)/[100T + RT(T − 1)/2]
= (100 × 848)/[100 × 4 + 4 × 4 × (4 − 1)/2]
= 84800/(400 + 24) = 84800/424
Ans: Rs. 200
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