Simple interest and solved examples with shortcut method

Simple Interest
Formula • Examples • Short Tricks

Simple Interest

Principal:
When you first deposit money in a saving account or when you borrow some money from another person, bank, or any financial institution that amount is known as the principal.

Interest:
The extra amount paid by the borrower to the lender for the use of the amount lent is called interest.

Simple Interest:
When we borrow some amount from another person for a certain period of time we have to pay him some extra money is called interest. This interest is the same for the same period of time. This is called simple interest or SI.

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Formulae

S.I = (P × R × T) ÷ 100

P + SI = A

SI = A − P

Case i: If SI, R, and T are known
P = (SI × 100) ÷ (R × T)

Case ii: If SI, T, and P are known
R = (SI × 100) ÷ (P × T)

Case iii: When SI, P, and R are known
T = (SI × 100) ÷ (R × P)

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Ex.1

If SI for 6 years be equal to 30% of the Principal then in how many years it will be equal to the 2 times of principal.

Sol. Let principal = 100
30% in 6 years
1 = 6/30
200 = 200 × (6/30) = 40 years

Ex.2

By which rate any principal becomes 5 times within 20 years.

Sol. The let amount is 300 and the principal is 100
S.I = A − P = 200
200 = (10 × R × 20)/100
R = 10%

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Special Case Formula

(n1 − 1)/t1 = (n2 − 1)/t2

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Installment

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]

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Questions based on Simple Interest with short tricks

Ex.1 Find S.I on Rs. 5000 at the rate of interest of 5% per year for 5 years.

Sol. Method 1
S.I = (PRT)/100 = (5000×5×5)/100 = 1250 Rs.

Method 2
100 Rs. at 5% in 1 year = 5 Rs.
100 Rs. at 5% in 5 year = 25
In simple interest same amount is credited per year
1% of 5000 Rs. is 50.
5% interest for 5 years means 25%
Thus simple interest is 25×50 = 1250 Ans.

Ex.2 On a certain sum of money at 6% per year Simple interest is Rs. 324 for 3 years. Find a sum of money.

Sol. Method 1
P = (SI × 100)/(R × T)
= (324 × 100)/(6 × 3)
Rs. 1800 Ans.

Method 2
P    T    R
100    1 year    6
100    3 years    18
*18     *18
1800 Rs.    324 Ans

Ex.3 At what rate of interest per annum, S.I on Rs. 1000 will be Rs. 343 in 7 years?

Sol. Method 1
SI = PRT/100
343 = (1000×R×7)/100
R = 4.9%

Method 2
1000    7 year    343
1000    1 year    49
100 = 4.9% Ans

Ex.4 A certain sum of money becomes 5/2 times of itself in 5 years. Find the rate of interest?

Sol.
(n − 1) = RT/100
(5/2 − 1) = (R×5)/100
3/2 = R/20
R = 30% Ans

Ex.5 A 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?

Sol.
n = RT/100
2/5 = (R×5)/100
R = 8% Ans

Ex.6 The difference of simple interest on a certain sum of money for 4 years and 6 years is Rs. 1125 if the rate of interest is 5% p.a. Find sum of money.

Sol. Method 1
P = [(SI1 − SI2) × 100] / [R(T1 − T2)]
P = (1125 × 100)/(5 × 2)
P = Rs. 11250

Method 2
T = 2 year, R = 5%
SI = RT%
10% = Rs.1125
100% = (1125/10) × 100
Rs.11250 Ans

Ex.7 Ram lent Rs. 9000 in two parts at 5% and 8%. Combined interest rate is 7%. Find money lent at 8%.

Sol. Method 1
Let sum at 8% = x
(8×2×x)/100 + (5×2×(9000−x))/100 = (7×2×9000)/100
6x = 36000
x = Rs.6000

Method 2
P1/P2 = (8−7)/(7−5)
P2 = (2/3)×9000 = 6000 Rs. Ans

Ex.8 A sum becomes 4 times in 30 years. In how many years it becomes 6 times?

Sol.
(n1 − 1)/t1 = (n2 − 1)/t2
(4 − 1)/30 = (6 − 1)/t2
t2 = 50 years Ans

Ex.9 Ram lent money to Mohan for 5 years and to Gaurav for 3 years at 6%. Total interest Rs.1920.

Sol.
P = [(SI1 + SI2) × 100]/[(T1 + T2)R]
P = (1920 × 100)/(6 × 8)
P = Rs.4000

Method 2
T = 8 year, R = 6%
SI = RT%
48% = 1920
1% = 40
100% = 4000 Ans

Ex.10 What equal installment will discharge a debt of Rs. 848 in 4 years at 4%?

Sol.
(100×848)/(100×4 + 4×4[(4−1)/2])
= (100×848)/424
= Rs.200 Ans

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