Question 1

A man travels for 5 hours 15 minutes. If he covers the first half of the journey at 60km/h and rest at 45km/h. Find the total distance travelled by him. [SSC—CHSL (10 + 2) Exam, 2015]

Accepted Answer

Concept & Formula: When a journey is split into two equal halves (distances are the same) covered at different speeds S_1 and S_2, the relationship between total distance D and total time T can be evaluated by summing the time taken for each half. Total Time = \frac{Distance_1}{Speed_1} + \frac{Distance_2}{Speed_2} Step-by-Step Solution: 1. Convert Time: The total time is 5 hours and 15 minutes. 5 hours 15 minutes = 5 + (15/60) = 5 + (1/4) = (21/4) hours 2. Define Distance: Let the total distance be 2d. Thus, the first half is d and the second half is d. 3. Set Up Equation: The sum of times for both halves equals the total time. (d/60) + (d/45) = (21/4) 4. Solve for d: LCM of 60 and 45 is 180. (3d + 4d/180) = (21/4) (7d/180) = (21/4) d = (21 × 180/4 × 7) d = 3 × 45 = 135 km 5. Calculate Total Distance: Total Distance = 2d = 2 × 135 = 270 km Exam Strategy & Shortcut: Use the average speed formula for equal distances: Average Speed = (2 × S_1 × S_2/S_1 + S_2). Average Speed = (2 × 60 × 45/60 + 45) = (5400/105) = (360/7) km/h. Total Distance = Average Speed × Total Time = (360/7) × (21/4) = 90 × 3 = 270 km. Common Pitfall: A common mistake is forgetting to multiply d by 2 at the end if you assumed the total distance was 2d. If you assume the total distance is D, the equation is (D/2/60) + (D/2/45) = (21/4), which helps avoid this. Final Answer: Therefore, the correct answer is 270 km.

Question 2

A student goes to school at the rate of $2\frac{1}{2}$ km/h and reaches 6 min late. If he travels at the speed of 3km/h he is 10 min early. What is the distance to the school ?

Accepted Answer

Concept & Time Difference Strategy: When a person travels the same distance at two different speeds, the difference in the time taken is equal to the total time difference between arriving late and arriving early. We can equate the time difference using the formula: Time = (Distance/Speed) Step-by-Step Solution: 1. Let the distance to the school be D km. 2. The initial speed is 2(1/2) km/h = 2.5 km/h. The time taken is D / 2.5 hours. 3. The new speed is 3 km/h. The time taken is D / 3 hours. 4. Total time difference = 6 minutes (late) + 10 minutes (early) = 16 minutes. 5. Convert the time difference to hours: 16 / 60 hours. 6. Set up the equation: (D/2.5) - (D/3) = (16/60) 7. Solve for D: (3D - 2.5D/7.5) = (16/60) (0.5D/7.5) = (16/60) (D/15) = (16/60) D = 15 × (16/60) = (16/4) = 4 km Exam Strategy & Shortcut: Use the direct formula for finding distance when time difference (\Delta t) is given: Distance = (S_1 × S_2/S_1 - S_2) × \Delta t Distance = (2.5 × 3/3 - 2.5) × (16/60) Distance = (7.5/0.5) × (16/60) = 15 × (16/60) = 4 km This bypasses algebraic setups entirely. Common Pitfall: Forgetting to convert the time difference from minutes to hours before applying it in the equation alongside speeds given in km/h. Final Answer: Therefore, the correct answer is 4 km.

Question 3

In order to reach his office on time, Mr. Roy goes through the middle passage of a round fort which he takes 14 minutes to pass through. However, on a certain day, due to repairs, the straight road being blocked, he had to take the roundabout way as…

Accepted Answer

Concept & Circular Motion: The middle passage of a round fort represents its diameter. The roundabout way represents its semicircular arc. The time taken is directly proportional to the distance traveled if the speed remains constant. Distance_{diameter} = 2r Distance_{semicircle} = \pi r Step-by-Step Solution: • Let the radius of the round fort be r. • Distance through the middle passage (diameter) = 2r. • Time taken for distance 2r = 14 minutes. • Distance through the roundabout way (semicircle) = \pi r. • Let the time taken for the roundabout way be t. • Since speed is constant, the ratio of times equals the ratio of distances: (t/14) = (\pi r/2r) • t = 14 × (\pi/2) = 7\pi • Using \pi \approx (22/7): t = 7 × (22/7) = 22 minutes. • The time he was late = Time taken for roundabout way - Usual time • Late by = 22 - 14 = 8 minutes. Exam Strategy & Shortcut: The ratio of the length of a semicircle to its diameter is (\pi r/2r) = (\pi/2) \approx (22/7/2) = (11/7). Multiply the usual time by this ratio to get the new time: 14 × (11/7) = 22 mins. Difference = 22 - 14 = 8 mins. Common Pitfall: A common error is confusing the "roundabout way" with the full circumference. Since he crosses the fort to reach his office on the other side, he only travels half the circumference (a semicircle). Final Answer: Therefore, the correct answer is 8 min.

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