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# Tips, Tricks And Shortcuts Of Compound Interest

## Tips, Tricks And Shortcuts Of Compound Interest

Compound interest is the interest calculated on the initial principal and also includes all the interest of previous periods on a deposit or loan

Compound Interest can be calculated , CI = P (1+ \mathbf{\frac{r}{100}})^{n} –

Total Amount can be calculated, A = P (1+ \mathbf{\frac{r}{100}})^{n}

### 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:

\mathbf{x^{\frac{1}{a}}=y^{\frac{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% =**\mathbf{\frac{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**\mathbf{\frac{A_{1}}{A_{2}}}_{1}(**)**

### 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+ \frac{r}{100})^{n}

A =50,000(1+ \frac{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+ \frac{r}{100})^{n}

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

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

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

\frac{26}{24}=(1+ \frac{r}{100})

(\frac{26}{24} – 1) =\frac{r}{100}

1.0833 – 1 =\frac{r}{100}

0.0833 = \frac{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 for 2 years , Difference = \frac{P(R)^{2}}{(100)^{2}}

100 = \frac{P(14)^{2}}{(100)^{2}}

P =\frac{100(100)^{2}}{(14)^{2}}

P =\frac{1000000}{196}

P = 5102.04

Now calculate the CI on Rs. 5102.04

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

A =5102.04 ( \frac{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^{\frac{1}{a}} = y^{\frac{1}{b}}

2^{\frac{1}{8}} = 32^{\frac{1}{x}}

2^{\frac{1}{8}} = 2^{\frac{5}{x}}

\frac{1}{8} = \frac{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+\frac{r_{1}}{100}) (1+\frac{r_{2}}{100}) (1+\frac{r_{3}}{100})

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

CI = A – P

CI = 10000 (1 + \frac{2}{100}) (1 + \frac{4}{100}) – 10000

CI = 10000 * (\frac{102}{100})(\frac{104}{100}) – 10000

CI = 10000 * (51 * \frac{52}{2500}) – 10000

CI = 10000 * (\frac{2652}{2500}) – 10000

CI = 10000 * (1.06) – 10000

CI = 10608 – 10000

CI = 608

Hence, the required compound interest is Rs. 608.

**Correct option: B**

**Read Also – Formulas for Simple interest **

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