Compound Interest - Shortcuts & Tricks for Placement Tests, Job Interviews & Exams

Share

Summary

This video provides a comprehensive guide to understanding and solving compound interest problems, a frequently tested topic in various competitive exams and job interviews. It introduces the main formula for compound interest and then categorizes different types of problems, offering specific formulas and methods for each, including population-related questions.

Highlights

Introduction to Compound Interest Formula
00:00:00

The video begins by emphasizing the importance of compound interest for competitive exams and introduces the core formula: Amount = Principal * (1 + Rate/100)^Number of Years. It clarifies that this formula calculates the total amount (principal + compound interest).

Types of Compound Interest Problems and Formulas
00:02:49

The presenter outlines four main types of compound interest problems. The first is direct application of the main formula. The second involves the difference between compound interest (CI) and simple interest (SI) for 2 years (Difference = P * (R/100)^2). The third type deals with the difference between CI and SI for 3 years (Difference = P * (R/100)^2 * (R/100 + 3)). The fourth type, installments, is covered in a separate video. Population problems are also discussed as a variation of the first type.

Population Change Formulas
00:08:03

The video explains how the compound interest formula can be adapted to calculate population changes. For future population, it's Present Population * (1 +/- Rate/100)^Number of Years, with '+' for increase and '-' for decrease. For past population, it's Present Population / (1 +/- Rate/100)^Number of Years.

Example 1: Compound Interest Compounded Quarterly
00:11:46

This example demonstrates calculating compound interest when it's compounded quarterly. The key is to adjust the rate (divide by 4) and the number of years (multiply by 4) when the compounding period is not annual. The problem involves finding the interest earned on ₹6,400 for 6 months at 25% compounded quarterly, resulting in ₹825 CI.

Example 2: Compound Interest for Fractional Years
00:18:49

This section addresses how to handle fractional years in compound interest calculations. The formula is modified to P * (1 + R/100)^N * (1 + (A/B)*R/100), where N is the whole number of years and A/B is the fraction. The example calculates the CI for ₹5,000 in 3 years 10 months at 30% per annum.

Example 3: Variable Interest Rates Per Year
00:22:52

The video explains how to calculate compound interest when the interest rate changes each year. The formula adapts by multiplying terms for each year with its specific rate: Amount = P * (1 + R1/100) * (1 + R2/100) * (1 + R3/100). The example finds the principal amount for an accumulated sum with varying annual rates.

Example 4: Time to Quadruple and Multiply an Amount
00:25:52

This problem is solved using a logical approach rather than a formula. If an amount quadruples in 6 years, to become 64 times (4^3), it will take 3 cycles of 6 years, totaling 18 years.

Example 5: Simple Interest to Compound Interest Conversion
00:26:49

This example shows how to find compound interest given simple interest for the same sum, rate, and time. It first calculates the principal from the simple interest information and then applies the compound interest formula. It also reiterates that for the first year, simple interest equals compound interest.

Example 6: Finding Principal and Rate from Consecutive Amounts
00:28:39

This problem uses the unique property that the difference between the amounts at consecutive years under compound interest for one year effectively represents the simple interest on the previous year's amount. This allows for calculating the rate and then the initial principal.

Example 7: Future Population Calculation
00:35:15

A direct application of the population formula for future growth is shown. If a city's population is 50,000 and increases at 10% per annum, its population in 3 years will be 66,550.

Example 8: Past Population Calculation
00:36:32

This example demonstrates finding the population in the past. The formula is Present Population / (1 + R/100)^N. The population 4 years ago for a city with a current population of 50,000, increasing at 10%, is approximately 34,151.

Example 9: Difference in Past Population
00:38:34

Building on the past population concept, this problem asks for the difference in population between 3 years ago and 2 years ago for a village. Both past populations are calculated using the division method, and then the difference is found.

Example 10: Difference Between CI and SI for 2 Years
00:41:44

This problem directly applies the formula for the difference between CI and SI for 2 years: Difference = P * (R/100)^2. Given the difference, rate, and time, the principal amount is quickly calculated as ₹625.

Example 11: Converting CI to SI for 2 Years
00:43:10

This example uses an extended formula for the difference between CI and SI for 2 years: CI - SI = SI * (R/200). Given the compound interest, rate, and time, the corresponding simple interest is found to be ₹100.

Example 12: Difference Between CI and SI for 3 Years
00:45:08

The final example applies the formula for the difference between CI and SI for 3 years: Difference = P * (R/100)^2 * (R/100 + 3). Given the difference, rate, and time, the principal amount is calculated as approximately ₹15,755.38. The video concludes by reiterating the importance of memorizing these key formulas for efficient problem-solving.

Recently Summarized Articles

Loading...