Questions on Principal Money Divided into Parts

Duration: 12 min

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This educational video is a lecture on simple and compound interest, presented by an instructor named Yash Jain. The video begins with an introductory slide defining the topic. It then proceeds to solve three distinct word problems. The first problem involves dividing a sum of Rs. 4000 into two parts to be invested at 3% and 5% simple interest, with a total interest of Rs. 144. The instructor sets up the equation P(3/100) + (4000-P)(5/100) = 144 and solves for P, finding the two parts to be Rs. 2800 and Rs. 1200. The second problem asks to divide Rs. 3600 into two parts such that the simple interest on the first part at 5% for 3 years equals the interest on the second part at 6.25% for 4 years. The instructor sets up the equation (P*5*3)/100 = ((3600-P)*6.25*4)/100 and solves for P, finding the parts to be Rs. 2250 and Rs. 1350. The third problem involves a merchant lending Rs. 4000, with one part at 10% and the other at 12% simple interest for 3 years, resulting in a total amount of Rs. 5350. The instructor sets up the equation 4000 + (P*10*3)/100 + ((4000-P)*12*3)/100 = 5350 and solves for P, finding the amount given at 10% to be Rs. 2500. The video concludes with a 'Thank You' screen.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title slide for a lesson on 'Simple and Compound Interest'. The instructor, Yash Jain, appears in a smaller window and introduces the topic. The first problem is presented on screen: 'A sum of rupees 4000 is divided into two parts such that if one part is invested at the rate of 3% and the other part at the rate of 5% and at the end of the year rupees 144 is collected as the total interest from both portions. Find each part?'. The instructor begins to solve this by setting up variables, letting the first part be P and the second part be 4000-P.

  2. 2:00 – 5:00 02:00-05:00

    The instructor continues to solve the first problem. He writes the formula for simple interest, SI = (PRT)/100, and applies it to both parts. He writes the equation: (P * 3 * 1)/100 + ((4000 - P) * 5 * 1)/100 = 144. He simplifies this to (3P + 20000 - 5P)/100 = 144, which becomes (20000 - 2P)/100 = 144. He then multiplies both sides by 100 to get 20000 - 2P = 14400. He solves for P, finding 2P = 5600, so P = 2800. The two parts are Rs. 2800 and Rs. 1200.

  3. 5:00 – 10:00 05:00-10:00

    The video transitions to the second problem: 'A sum of Rs. 3600 is divided into two parts such that the simple interest on first part at the rate of 5% per annum for 3 years and the other part at the rate of 6.25% per annum for 4 years is equal. These parts are:'. The instructor sets up the equation: (P * 5 * 3)/100 = ((3600 - P) * 6.25 * 4)/100. He simplifies this to 15P/100 = (90000 - 25P)/100. He multiplies both sides by 100 to get 15P = 90000 - 25P. He combines like terms to get 40P = 90000, so P = 2250. The two parts are Rs. 2250 and Rs. 1350.

  4. 10:00 – 11:45 10:00-11:45

    The third problem is presented: 'A merchant gives Rs. 4000 to a man on borrow for 3 years. On this, there is an interest of 10% per annum on one part and on the remaining part there is an interest of 12% per annum. After 3 years, the merchant received Rs. 5350. Find the amount given at 10%?'. The instructor sets up the equation for the total amount: 4000 + (P * 10 * 3)/100 + ((4000 - P) * 12 * 3)/100 = 5350. He simplifies this to 4000 + 30P/100 + (144000 - 36P)/100 = 5350. He combines terms to get 4000 + (144000 - 6P)/100 = 5350. He solves for P, finding P = 2500. The video ends with a 'Thank You for Watching' screen.

The video provides a clear, step-by-step demonstration of solving three different types of simple interest problems. It begins with a basic problem of finding two parts of a sum based on total interest, then moves to a problem where the interests are equal, and finally to a problem where the total amount (principal + interest) is given. The instructor consistently uses the formula SI = (PRT)/100 and demonstrates how to set up and solve linear equations to find the unknown principal amounts. The progression builds from a simple two-variable setup to more complex scenarios, reinforcing the core concept of simple interest calculation.