Questions on Extra Amount Received on Extra Interest Rate
Duration: 8 min
This video lesson is available to enrolled students.
AI Summary
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This educational video, presented by Yash Jain, is a tutorial on solving problems related to simple and compound interest. The video begins with an introductory slide featuring a graph and a house made of money, setting the theme of financial growth. The main content consists of two worked examples. The first problem asks to find the principal given that a 3% increase in the interest rate over 2 years results in an extra Rs. 300. The instructor demonstrates a step-by-step algebraic solution using the formula for simple interest (SI = PRT/100), showing that the difference in interest (SI2 - SI1) is equal to the extra amount. The second problem is similar, where a 3% increase in the rate over 2 years leads to an extra interest of Rs. 72, and the instructor uses a direct formula: Principal = (Extra Interest * 100) / (Extra Rate * Time) to find the principal. The video concludes with a thank you message.
Chapters
0:00 – 2:00 00:00-02:00
The video opens with a title slide for a lesson on 'Simple and Compound Interest'. The slide features a graph with an upward-trending orange line and a row of gold coins, symbolizing financial growth. The instructor, Yash Jain, appears in a small window on the right, introducing the topic. The background is a light blue gradient with a decorative pattern of stars and triangles. The text on the slide includes 'Simple Interest and Compound Interest', 'Basic To Advance', and 'by YASH JAIN'. The instructor is seen speaking and gesturing, setting the stage for the lesson.
2:00 – 5:00 02:00-05:00
The video transitions to a problem-solving segment. The first problem is displayed on a yellow background with a pattern of small shapes: 'A sum of money is given on simple interest with a specific rate of interest for 2 years. If the rate of interest will be 3% more, then extra amount of Rs. 300 will be received. Find the principal?'. The instructor begins to solve this by writing the formula for the amount (A = P + SI) and then defines the initial amount (A1 = P + SI1) and the new amount (A2 = P + SI2). He then calculates the difference (A2 - A1), which simplifies to the difference in interest (SI2 - SI1). He writes the formula for the difference in interest as (P * (R+3) * 2)/100 - (P * R * 2)/100 = 300.
5:00 – 8:11 05:00-08:11
The instructor continues to solve the first problem. He simplifies the equation (P * (R+3) * 2)/100 - (P * R * 2)/100 = 300 to 2P(R+3) - 2PR = 30000. This further simplifies to 2PR + 6P - 2PR = 30000, which reduces to 6P = 30000. He then solves for P, finding P = 5000. He then introduces a shortcut formula: Principal = (Extra Interest * 100) / (Extra Rate * Time). He applies this to the second problem: 'A sum of money invested on a rate of simple interest for 2 years. If this money is invested on 3% more interest then the interest received by the same amount is Rs.72 more than before. Find Principal?'. He uses the formula: (72 * 100) / (3 * 2) = 7200 / 6 = 1200. The video ends with a 'THANK YOU FOR WATCHING' screen.
The video provides a clear, step-by-step tutorial on solving simple interest problems where the rate of interest changes. It effectively uses a combination of algebraic derivation and a direct formula to find the principal. The instructor's methodical approach, from setting up the problem with the standard formula to simplifying the equation and finally applying a shortcut, makes the concepts accessible. The progression from a detailed algebraic solution to a quick formula demonstrates a practical approach to problem-solving, which is beneficial for students preparing for competitive exams.