# Tips And Tricks And Shortcuts To Solve Compound Interest

### Tips and Tricks and Shortcuts on Compound Interest

Here, are quick and easy tips and tricks on Compound Interest. Learn the tricks and concept of compound interest.

- There are mainly 5 types of questions asked in exams. However, these questions can be twisted to check the students understanding level of the topic but it can be solved using tips and tricks.

### Compound Interest Tricks and Tips and Shortcuts

- A sum of money placed at compound interest becomes x time in ‘a’ years and y times in ‘b’ years. These two sums can be related by the following formula:

(x)^{1/a}= (y)^{1/b} - If an amount of money grows up to Rs x in t years and up to Rs y in (t+1) years on compound interest, then

r% = (y-x)/x * 100 - A sum at a rate of interest compounded yearly becomes Rs. A
_{1}in t years and Rs. A_{2}in (t + 1) years, then

P = A_{1}(A_{1}/A_{2})t

## Type 1: Tips and Tricks to find the compound interest (Yearly, Quarterly, and Half-yearly)

### Question 1.

**Find the amount if Rs 50000 is invested at 10% p.a. for 4 years**

Options:

A. 73205

B. 7320.5

C. 73250

D. 73502

#### Solution:

We know that, Amount = P (1+r/100)^{t}

A = 50000 (1 + 10/100)^{4}

A = 50000 (1.1)^{4}

A = 73205

#### Correct option: A

## Type 2: Tips and Tricks and Shortcuts for Compound Interest (To find the Rate of Interest, Time period, Principal)

### Question 1.

**The Compound Interest on a sum of Rs 576 is Rs 100 in two years. Find the rate of interest.**

Options:

A. 7.33%

B. 4.33%

C. 8.33%

D. 5.33%

#### Solution:

We know that,

Amount = Compound Interest + Principal

A = 576 + 100 = 676

Amount = P (1+r/100)^{t}

676 = 576 (1+ r/100)^{2}

676/576 = (1+ r/100)^{2}

(26/24)^{2} = (1+ r/100)^{2}

(26/24) = (1+ r/100)

(26/24 – 1) =r/100)

1.0833 – 1 = r/100

0.0833 = r/100

r = 8.33%

#### Correct option: C

## Type 3: Tips and Tricks to Find sum or rate of interest when compound interest and S.I are given

### Question 1.

**The difference between the CI and SI on a certain amount at 14% p.a. for 2 years is Rs. 100. What will be the value of the amount at the end of three years if compounded annually? **

Options:

A. 7558.89

B. 7500.45

C. 7558

D. 7554.56

#### Solution:

We know that, Difference between CI and SI = P(R^{2})/(100^{2})

100 = P(14^{2})/(100^{2})

P = (100*100^{2})/14^{2}

P = 1000000/196

P = 5102.04

Now calculate the CI on Rs. 5102.04

A = 5102.04 (1+14/100)^{3}

A = 5102.04 (114/100)^{3}

A = 5102.04 * 1.48

A= 7558.89

#### Correct option: A

## Type 4: Shortcuts of Compound Interest (When sum of money becomes x time in ‘a’ years and y times in ‘b’ years.)

### Question 1.

**A sum of money placed at compound interest doubles itself in 8 years. In how many years will it amount to 32 times itself?**

Options:

A. 40 years

B. 50 years

C. 32 years

D. 35 years

#### Solution:

We know that,

(x)^{1/a} = (y)^{1/b}

2^{(1/8)} = 32^{(1/x)}

2^{(1/8)} = 2^{(5/x)}

1/8 = 5/x

x = 40 years

#### Correct option: A

## Type 5: Tips and Shortcuts for Compound Interest (When Rates are different for different years)

**Question 1.**

** Find the compound interest on a sum of money of Rs.10000 after 2 years, if the rate of interest is 2% for the first year and 4% for the next year?**

Options:

A. Rs. 304

B. Rs. 608

C. Rs. 1000

D. Rs. 710

#### Solution:

We know that,

Amount = P (1+r_{1}/100) (1+r_{2}/100) (1+r_{3}/100)

Here, r_{1} = 2% r_{2} = 4% and p = Rs.10000,

CI = A – P

CI = 10000 (1 + 2/100) (1 + 4/100) – 10000

CI = 10000 * (102/100)(104/100) – 10000

CI = 10000 * (51 * 52/2500) – 10000

CI = 10000 * (2652 / 2500) – 10000

CI = 10000 * (1.06) – 10000

CI = 10608 – 10000

CI = 608

Hence, the required compound interest is Rs. 608.

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