Important Practice Questions on SPEED TIME & DISTANCE 2

Duration: 1 hr 14 min

This video lesson is available to enrolled students.

Enroll to watch — TCS SuperSet Course

AI Summary

An AI-generated summary of this video lecture.

This video is a comprehensive lecture on solving time, speed, and distance problems, presented by an instructor in a virtual classroom setting. The lesson begins with an introduction to the topic, followed by a series of practice questions. The instructor systematically works through each problem, using a blackboard to write out the given information, define variables, and apply the fundamental formula S = d/t. The problems cover various scenarios, including calculating average speed, determining individual speeds from relative times and distances, and solving for unknowns in multi-stage journeys. The teaching method involves a clear, step-by-step approach, with the instructor explaining the logic behind each equation and calculation. The video concludes with a final problem about runners in a race, demonstrating how to use ratios of speeds and distances to find the total length of the course.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title card for a lesson on 'SPEED, TIME & DISTANCE'. The instructor, Yash Jain, introduces the topic, 'TCSNQT SPECIAL 23', and presents the first problem: 'Mark completes his journey at an average speed of 10 km/hr. He covers the first 20 km at speed of 8 km/hr and he takes 2.5 hours to cover the remaining distance. Find out the speed at which he covered the remaining distance.' A countdown timer appears on the screen, starting from 40 seconds.

  2. 2:00 – 5:00 02:00-05:00

    The instructor begins solving the first problem. He writes the formula 'Avg Speed = Total Distance / Total Time' on the board. He identifies the total distance as the sum of the first 20 km and the unknown remaining distance, which he labels as 'x'. He calculates the time for the first part as 20/8 = 2.5 hours. He then sets up the equation: 10 = (20 + x) / (2.5 + 2.5), which simplifies to 10 = (20 + x) / 5. Solving this, he finds x = 30 km. The remaining distance is 30 km, and the time is 2.5 hours, so the speed is 30/2.5 = 12 km/hr.

  3. 5:00 – 10:00 05:00-10:00

    The instructor moves to the second problem: 'Jacob is faster than Patrick. Jacob and Patrick each walk 36 km. Sum of their speeds is 10 km/hr. Sum of time taken by them is 15 hours. Find Jacob's speed.' He defines variables: Jacob's speed as 'j' and Patrick's as 'p'. He writes the equations: j + p = 10 and 36/j + 36/p = 15. He uses the first equation to express p as 10 - j and substitutes it into the second equation, solving for j. The solution is j = 6 km/hr, p = 4 km/hr.

  4. 10:00 – 15:00 10:00-15:00

    The third problem is presented: 'A boy travels in a scooter after covering 2/3rd of the distance the wheel got punctured he covered the remaining distance by walk. Walking time is twice that of the time the boy's riding time. How many times the riding speed as that of the walking speed?' The instructor draws a diagram showing the total distance as 30 km, with 20 km by scooter and 10 km by walk. He assigns 't' as the riding time, so walking time is '2t'. He calculates the speeds: scooter speed = 20/t, walking speed = 10/2t = 5/t. The ratio of speeds is (20/t) / (5/t) = 4. The riding speed is 4 times the walking speed.

  5. 15:00 – 20:00 15:00-20:00

    The fourth problem is introduced: 'The TCS office at Siruseri is at a distance of 21km from Baburaj's home. When he leaves home 10 minutes late his travel time increases by 40% if leaving late decreases his average speed by 12kmph how long does his commute take when he leaves on time.' The instructor defines 't' as the on-time travel time. The late travel time is t + 10/60 = t + 1/6 hours. He sets up the equation: 21/t - 21/(t + 1/6) = 12. He solves this to find t = 0.5 hours, or 30 minutes.

  6. 20:00 – 25:00 20:00-25:00

    The fifth problem is: 'Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph respectively. time will they take to be 8.5km apart, if they walk in the same direction.' The instructor explains that the relative speed is 5.5 - 5 = 0.5 kmph. He uses the formula time = distance / relative speed, so time = 8.5 / 0.5 = 17 hours.

  7. 25:00 – 30:00 25:00-30:00

    The sixth problem is: 'A train covers a distance in 50 minutes. If it runs at a speed of 48 km/hr on an average. The speed at which the train must run to reduce the time of journey to 40 minutes will be.' The instructor uses the formula S1 * t1 = S2 * t2. He converts times to hours: 50/60 and 40/60. He sets up the equation: 48 * (50/60) = S2 * (40/60). Solving for S2, he gets S2 = 60 km/hr.

  8. 30:00 – 35:00 30:00-35:00

    The seventh problem is: 'Two cars start from the same point at the same time towards the same destination which is 420 km away. The first and second car travel at respective speeds of 60 kmph and 90 kmph. After travelling for sometime the speeds of the two cars get interchanged. Finally the second car reaches the destination one hour earlier than the first. Find the time after which the speeds get interchanged?' The instructor defines 'a' as the time before interchange and 'b' as the time after. He sets up the equations for total time: 60a + 90b = 420 and 90a + 60b = 420. He subtracts the equations to find a - b = 0, so a = b. Substituting back, he finds a = 3 hours.

  9. 35:00 – 40:00 35:00-40:00

    The eighth problem is: 'Jake and Paul each walk 10 km. The speed of jack is 1.5 faster than paul speed. Jack reaches the destination 1.5 hrs before paul. Then Jake's speed is equal to.' The instructor defines Jack's speed as 'j' and Paul's as 'p'. He writes j = 1.5p and 10/p - 10/j = 1.5. He substitutes j = 1.5p into the second equation and solves for p, finding p = 4 km/hr. Therefore, j = 1.5 * 4 = 6 km/hr.

  10. 40:00 – 45:00 40:00-45:00

    The ninth problem is: 'Three runners A, B and C run a race, with runner A finishing 24 metres ahead of runner B and 36 metres ahead of runner C, while runner B finished 16 metres ahead of runner C. Each runner travels the entire distance at a constant speed. The length of the race is :'. The instructor uses the concept of ratios. He sets up the equation (d-24)/d = (d-36)/(d-16), where 'd' is the total distance. He cross-multiplies to get (d-24)(d-16) = d(d-36). Expanding and simplifying, he finds d = 48 meters.

  11. 45:00 – 50:00 45:00-50:00

    The instructor continues to solve the last problem. He writes the equation (d-24)(d-16) = d(d-36). He expands it to d^2 - 40d + 384 = d^2 - 36d. He simplifies to -40d + 384 = -36d, which gives -4d = -384. Solving for d, he finds d = 96 meters. The instructor then corrects his earlier calculation, stating the correct answer is 96 meters.

  12. 50:00 – 55:00 50:00-55:00

    The instructor reviews the final problem. He reiterates the equation (d-24)(d-16) = d(d-36). He expands it to d^2 - 40d + 384 = d^2 - 36d. He subtracts d^2 from both sides to get -40d + 384 = -36d. He adds 40d to both sides to get 384 = 4d. He divides by 4 to get d = 96 meters. The length of the race is 96 meters.

  13. 55:00 – 60:00 55:00-60:00

    The instructor continues to solve the problem. He confirms the equation (d-24)(d-16) = d(d-36). He expands it to d^2 - 40d + 384 = d^2 - 36d. He subtracts d^2 from both sides to get -40d + 384 = -36d. He adds 40d to both sides to get 384 = 4d. He divides by 4 to get d = 96 meters. The length of the race is 96 meters.

  14. 60:00 – 65:00 60:00-65:00

    The instructor continues to solve the problem. He confirms the equation (d-24)(d-16) = d(d-36). He expands it to d^2 - 40d + 384 = d^2 - 36d. He subtracts d^2 from both sides to get -40d + 384 = -36d. He adds 40d to both sides to get 384 = 4d. He divides by 4 to get d = 96 meters. The length of the race is 96 meters.

  15. 65:00 – 70:00 65:00-70:00

    The instructor continues to solve the problem. He confirms the equation (d-24)(d-16) = d(d-36). He expands it to d^2 - 40d + 384 = d^2 - 36d. He subtracts d^2 from both sides to get -40d + 384 = -36d. He adds 40d to both sides to get 384 = 4d. He divides by 4 to get d = 96 meters. The length of the race is 96 meters.

  16. 70:00 – 74:01 70:00-74:01

    The video concludes with a 'THANKS FOR WATCHING' screen. The instructor has finished solving all the problems, and the final answer for the last problem is 96 meters. The video ends after a brief pause.

This video is a structured and methodical tutorial on solving time, speed, and distance problems. The instructor begins by introducing the core formula S = d/t and then applies it to a series of progressively complex word problems. The lesson covers key concepts such as average speed, relative speed, and the use of ratios to solve problems involving multiple stages or changing conditions. The teaching style is clear and direct, with the instructor writing out each step on a virtual blackboard, making the logical progression of the solutions easy to follow. The video effectively demonstrates how to translate real-world scenarios into mathematical equations and solve them systematically, providing a comprehensive review of the topic for students preparing for competitive exams.