Shortcut tricks on compound interest are one of the most important topics in exams. Time takes a huge part in competitive exams. If you know time management then everything will be easier for you. Most of us miss that part. We provide examples on Compound Interest shortcut tricks here in this page below. These shortcut tricks cover all sorts of tricks on Compound Interest. We request all visitors to read all examples carefully. These examples here will help you to better understand shortcut tricks on compound interest.

Before doing anything we recommend you to do a math practice set. Then find out twenty math problems related to this topic and write those on a paper. Solve first ten math problems according to basic math formula. You also need to keep track of Timing. After finish write down total time taken by you to solve those ten maths. Now read our examples on compound interest shortcut tricks and practice few questions. After doing this go back to the remaining ten questions and solve those using shortcut methods. Again keep track of timing. The timing will be surely improved this time. But this is not all you need. You need to practice more to improve your timing more.

You all know that math portion is very much important in competitive exams. That doesn’t mean that other topics are less important. But only math portion can leads you to a good score. You can get good score only by practicing more and more. All you need to do is to do math problems correctly within time, and you can do this only by using shortcut tricks. But it doesn’t mean that without using shortcut tricks you can’t do any math problems. You may have that potential to do maths within time without using any shortcut tricks. But other peoples may not do the same. Here we prepared compound interest shortcut tricks for those people. Here in this page we try to put all types of shortcut tricks on Compound Interest. But it possible we miss any. We appreciate if you share that with us. Your little help will help so many needy.

**Find Compound Interest Tricks**

In case of Compound Interest the interest is vary according to time but the first year it is equal that is 1st year Interest is

**Compound Interest = Simple Interest.**

But after that year it is increases. So we can find Compound interest using tricks. This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

**Here is principal is given rate percent and Time is given and find the Compound interest.**

**Example 1:** What would be compound interest obtained on an amount of Rs 8000 at the rate of 10% p.a after 2 year ?

**Answer :** **Short cut tricks :**

For 1st year interest is = 8000 x 10 / 100 = Rs. 800

Second year = 800+ 800+80 = Rs. 1680

After 2 years compound interest is 1680.

**Example 2 :** What would be the compound interest to be obtained on an amount of Rs. 8000 at the rate of 10% p.a after 2 years ?

**Answer :** **Formula :** C.I = P [(1 + r / 100)^{n} – 1]

8000 x [(1 + 10 / 100)^{2} – 1]

= Rs.1680

**Example 3:** Raju invest and amount of Rs. 8460 /- 6% for 2 years, What approx amount he would obtain at the end of year two years?

**Answer:** 8460 x 106 x 106 / 100 x 100 = 9505/-(approx)

**Example 4:** What will be the compound interest on a sum of Rs .6500/- at the rate of 6 p.c.p.a for 2 years ?

**Answer :** (6500 x 53 x 53 / 50 x 50 -1)

= Rs. 803.4/-

**Example 5:** Find the amount of Rs. 18,000 at 5% per annum compound interest in 3 yrs.

**Answer :** 18000 x 21 x 21 x 21 / 20 x 20 x 20 = 20837.25

**Example 6:** Find the compound interest obtained on an amount Rs. 8000 at the rate of 12% p.a after 2 years.

**Answer :** CI = 8000 [ (112 / 100)^{2} – 1 ]

= Rs.2035.2

**Example 7:**

The simple interest deposited on sum of certain principle is Rs. 8400/- in seven years at the rate of 12 p.c.p.a. What should be the C.I deposited on that principle at the rate of 6 p.c.p.a in 2 years ?

**Answer :** X x 7 x 12 / 100 = 8400

x = Rs. 10,000/-

So, Compound Interest is = ( 10,000 x 105 x 105 / 100 x 100 ) – 10,000

= 11236 – 10,000 = Rs. 1236

**Example 8:**

Principal is 15000 at rate percent 4% p.c.p.a for 2 years, and compound annually,

Find the C.I.

**Answer:**

We apply the formula to obtain C.I that is

C.I = p [(1 + r / 100)n – 1]

P = 15000.

R = 4%.

Time = 2 years.

= 15000 x [( 1 + 4 / 100)2 – 1].

= 15000 x [ 26 x 26 / 25 x 25 – 1]. [ as we put down (26 / 25)2 ]

= 15000 x 51 / 625.

= 1224.

**Example 9:**

The **Simple Interest** accrued on an **amount of Rs. 22,500 **at the end of 3 years is** Rs. 10800. **what would be the **Compound Interest accrued** on the **same amount** at the **same rate** at the **end of two years ?**

**Answer :**

Here is given** amount = 22,500**, **Time = Years 3** and** S.I = 10800.**

So, we need to find Rate percent.

**Step 1:** we know S.I = P x R x T / 100

10,800 = 22,500 x R x 3 / 100

R = 1080000 / 67500 = 16 % So R = 16 %.

**Step 2:** Compound Interest accrued on the **same amount** at the **same rate **at the **end of two years**

is we apply the formula

C.I = 22500 x (116 / 100 x 116 / 100 – 1 )

= 22500 x( 116 x 116/ 10000 – 1) = 22500 x( 13456 / 1000) – 1 = 22500 x (1.3456 – 1) = 22500 x 0.3456 =7776

So the C.I end of two years is 7776.

**Example 10:** What would be the compound interest on an amount 6000 at the rate 12 p.c.p.a for 2 years ?

**Answer :**

= 6000[ (112 / 100)^{2} – 1]

= 6000 x 2544 / 10000

= 1526.4

**Example 11:**

What will be the compound interest on a sum of Rs. 4800/- at the rate of 6 p.c.p.a for 2 years ?

**Answer:**

Compound Interest = P[1+R/100]^{n}-1

4800[( 1 + 6 / 100 )^{2} – 1]

4800 [53 x 53 / 50 x 50 – 1 ]

=593.28

So the compound interest is 593.28

**Example 12:**

What would be the compound interest obtained on an amount of Rs. 1,600 at the rate of 8 p.c.p.a after two years ?

**Answer:**

Amount = SI +Principal

compound Interest = p ( 1 + r / 100)^{n} – 1600

1600 ( 1 + 8 / 100 )^{2} – 1600

( 1600 x 27 x 27 / 25 x 25 ) – 1600

= ( 1866.24 – 1600 )

= 266.24

**Example 13:**

What would be the compound interest obtained on an amount of Rs. 6000 at the rate of 10% p.a. after 2 years ?

**Answer :**

6000[ (1 + 10 / 100) ]^{n}-1

= 6000 [ (11 / 10 )^{2} – 1]

= 6000 x 21 / 100

= 1260

So the compound interest is 1260.

**Example 14: **

What would be the compound interest obtained on an amount of Rs.8850 at the rate of 12 p.c.p.a after two years?

**Answer :**

Amount = 8850

rate = 12

Time = 2 years

compound interest = ?

A = 8850 ( 1 + 12 / 100 )^{2}

= 8850 x 28 x 28 / 25 x 25

= 11101.44

So Amount = Rs.11101.44

C.I = ( 8850 – 11101.44 ) = 2251.44

**Example 15:**

What would be the compound interest obtained on an amount of Rs.7500 at the rate of 6 p.c.p.a after two years ?

**Answer :**

Amount = 7500

rate = 6

Time = 2 years

compound interest = ?

= 7500 [ 106 x 106 / 100 x 100 ] – 7500

= 7500 x 11236 / 10000 -7500

= 8427 – 7500

= 927

So, compound interest after 2 years is 927.

**Example 16:** In after the period of 2 years What would be the compound interest get on a Principal amount of 6400/- at the rate of 8 p.c.p.a.

**Answer :**

**Shortcut tricks :**

In first year always same so,6400 x 8 / 100 = 512

In Second year compound interest is 512 + 40.96. [ So, 512 x 8 / 100 = 40.96 ]

So total compound interest is 512 + 512 + 40.96 = **1064.96**

**Example 17:** What compound interest accrued on an amount of Rs. 18000 at the rate of 10 p.c.p.a for the 2 years ?

**Answer:** Formula: A = P(1 + R / 100)^{t}

= 18000 ( 1 + 10 / 100)^{2}

= 18000 x 22 x 22 / 20 x 20

= 21780

C.I. = (21780 – 18000)

= Rs. 3780.

### More Shortcut tricks on Simple and Compound Interest

- Find Simple Interest based question
- Find the rate % based question
- Find Principle or Sum based question
- Compound Interest shortcut tricks
- Difference between CI and SI of Three Years question
- << Go back to SI and CI main page

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sir can you explain the compound interest short cut Technic how you solved

p.a.=6400

r=8

for 1st year t=1

s.i.=6400*1*8/100=512

for second year u need 6400’s s.i. + 512’s s.i.

for one year now t=1

6400’s s.i. for one year=512

512,s =512*1*8/100=40.96

then add all

Its really good and quite helpful

Thanks 1

Why do we use a minus 1 in the above formulae?

to determine compountd interest directly use minus 1 else the answer we get is amount.so to calculate directly CI they used

amount=p(1+r/100)^n

compound interest =p((1+r/100)-1)^n

i am not sufficient for this method .Don’t use the easy method .Time is very important …

compound interest simple method fpr example

P.a=10000

R=10%

T=3yr

solve:

1000*3=3000

100*3=300

10*1=10

total =3310

compound interest=3310Ans

to determine difference on CI AND SI for 2years what short trick is used

P*(R/100)^2 = Difference

(CI)2 yrs -( SI)2 yrs =12,R%=20%,P=?

(SI)2 yrs=20+20=40% ;

(CI)2 yrs=(20+20+(20*20/100))% (shortcut)

=(40+(400/100))%

=(40+4)%

=44%

(CI)2 yrs -(SI)2 yrs =44-40 =4

we know that, P=(CI)2 yrs -(SI)2 yrs

given that, (CI)2 yrs -(SI)2 yrs =12

P = 44/100 -40/100

P=100

we need to multiply by 3 becoz in questn given 12

P=300

If % increases or a% and b% overall % increases is given by (a+b+ab/100)%

ex:20% increase for 2 yrs is

=( 20+20+20*20/100)%=44%