Tips & Tricks to Solved CI, SI Question, Simple Interest Compound Interest Examples

Tips & Tricks to Solved CI, SI Question, Simple Interest & Compound Interest Examples, Simple Interest and Compound Interest Tricks PDF | Bank Exams, Concept and Short Trick on C.I and S.I

Simple Interest (SI)

Principal/: – The money borrowed or lent out for certain period is called the principal or the Sum.

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Interest: – Extra money paid for using other money is called interest.

If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.

Tips & Tricks to Solved CI, SI Question, Simple Interest Compound Interest Examples

यदि कुछ अवधि के लिए उधार राशि पर ब्याज समान रूप से माना जाता है, तो इसे साधारण ब्याज कहा जाता है।

Let Principal = P, Rate = r % per annum (p.a.), and Time = t years then

Simple Interest (SI)= ((principal ×rate × time))/100

Using this formula we can also find out

P=(100×Simple Interest)/( rate × time);

Rate=(100×SI)/( Principal ×time);

Time =(100×SI)/( Principal ×rate).

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Simple Interest Tips and Tricks for Bank, SSC, SBI Exam

Compound interest concept

  • When compound interest is applied, interest is paid on both the original principal and on earned interest.
  • So for one year Simple interest and Compound interest both are equal.
  • Suppose if you make a deposit into a bank account that pays compounded interest, you will receive interest payments on the original amount that you deposited, as well as additional interest payments.

This allows your investment to grow even more than if you were paid only simple interest.

So Amount at the end of 1st year (or Period) will become the principal for the 2nd year (or Period) and

Amount at the end of 2nd year (or Period) becomes the Principal of 3rd year.

Amount = Principal + Interest

A= P (1+r/100) ^n

A= Amount,

P= Principal,

r= Rate %,

n= number of years.

So Compound Interest (CI) = [Principal (1+r/100) ^ n – Principal]

= Principal [(1+r/100) ^ n – 1]


1.When  interest is compounded annually,

Amount = Principal (1+r/100)^n

2.When interest  is compounded half yearly,

Amount = Principal (1+(r/2)/100)^2n

3.When interest is compounded Quarterly,

Amount = Principal (1+(r/4)/100)^4n

4.When interest is compounded annually but time is in fraction, say 3 whole 2/5 year

Amount = Principal (1+r/100)^3×(1+(2r/5)/100)

5.When Rates are different for different years, say r1%, r2%, and r3% for 1st, 2nd and 3rd year respectively.


Amount = Principal (1+r1/100)×(1+r2/100)×(1+r3/100).

Present worth of Rs. x due n years hence is given by:

Present Worth = x/(1+r/100)

Difference between Compound Interest & Simple interest Concept For Two years

Compound Interest – SI =P(r/100)^2

For Three Year

Compound Interest – Simple Interest =P(r^2/(100^2 ))×(300+r)/100)

For Two year

Compound Interest / Simple Interest =(200+r)/200

Difference between Simple Interest and Compound Interest

In case the same principle P is invested in two schemes, at the same rate of interest r and for the same time period t, then in that case:

Simple Interest {SI}= (P x R x T)/100
Compound Interest {CI}= P [(1+R/100)T – 1]
 So, the difference between them is
= P*R*T/100 – P*[(1+R/100)T -1]
= P [(1+r/100)T -1-RT/100]

Two shortcuts which we can use:
Difference between CI and SI when time given is 2 years = P(R/100)2
Difference between CI and SI when time given is 3 years = P[(R/100)3 + 3(R/100)2]

Application of Compound Interest for concepts of population growth

Case 1: When population growing in a constant rate
If the rate growth of population increased with a constant, rate then the population after T years will be = P (1+R/100)t

In fact, this is nothing else but an application of the fundamentals of compound interest.
It is actually similar to finding the compound amount after time T years
Net population after T years = = P (1+R/100)t
Net population increase = P [(1+R/100)t– 1]

When the population growing with different rates and for different intervals of time 
If the rate growth of population increased with different, rate and for different intervals of time then the population after T years will be =
P (1+R1/100)t1 x  (1+R2/100)t2……………………………..  (1+RN/100)tn

When the population is decreasing with rate R
Population after a time period of T years=P (1-R/100)t
Where T is the total population
““““““`R is rate at which the population is decreasing

C.I. and S.I. Quiz based on and Tricks

Example 1 Raman invested a certain sum in a simple interest bond whose value grew to Rs. 300 at the end of 3 Year and to Rs. 400 at the end of another 5 Year. The rate of interest was

(a) 12%

(b) 12.5%

(c) 8.33%

(d) 6.67%

Solution: –

(Principal *3)/100 +P = 300…….(1)

(Principal *8)/100 +P= 400…….(2)

subtract (1) from (2)

Principal p X Rate = 2000 ………..(3)

put (3) in (1)

P= 240, then Rate = 8.33%

Example 2: – If Rs. 450 amount to Rs. 540 in 4 years, what will it amount to in 6 years at the same rate %?

(a) Rs. 600

(b) Rs. 585

(c) Rs. 700

(d) Rs. 640

Solutions: – SI per year = (540-450)/4 = 22.5
so after 6 year = 22.5 x 6 = 135
So Amount 450+135 = Rs.585

Example: –  The population of a city two years ago was 1,25,000. Due to migration from cities, it decreases every year at the rate of 4% per annul. Find its present population. How many person have migrated in last 2 years?

(a) Rs. 8,400

(b) Rs. 9,800

(c) Rs. 6,600

(d) Rs. 5,600

Solution: –

Population 2 year ago = 125000 = p

for two years, rate of decrease = 4% p.a.

Population after 2 year = present population = p(1-R/100)^2

Present population = 125000(1-4/100)^2

= 115200

So number of persons migrated during last 2 years= 125000-115200 = 9800

Example: – The difference between simple and compound interest on a certain sum of money for 3 years at 5 per cent per annum is Rs. 325. The sum of money is :

(1) Rs. 42623

(2) Rs. 46255

(3) Rs. 43260

(4) Rs. 35990

Solution: –

Difference of CI and SI for 3 years = pr^2(300+r)/(100)^3

325 = (p*25*305)/1000000

so p = 42623

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