Question 1

Train A leaves Ludhiana for Delhi at 11 a.m, running at the speed of 60 km/hr. Train B leaves Ludhiana for Delhi by the same route at 2 p.m. on the same day, running at the speed of 72 km/hr. At what time will the two trains meet each other?

Accepted Answer

Concept & Relative Speed: When two bodies move in the same direction, their relative speed is the difference of their individual speeds. The time taken to meet is the initial distance separating them divided by this relative speed. Relative Speed = V_2 - V_1 Time to Catch Up = \frac{Initial Distance}{Relative Speed} Step-by-Step Solution: 1. Calculate Initial Distance: Train A leaves at 11 a.m. and Train B leaves at 2 p.m. Train A travels alone for 3 hours. 2. Distance covered by Train A in 3 hours = 60 km/hr × 3 hrs = 180 km. 3. Calculate Relative Speed: Both trains are moving in the same direction. Relative Speed = 72 km/hr - 60 km/hr = 12 km/hr. 4. Calculate Time to Catch Up: Time = \frac{Distance}{Relative Speed} = \frac{180 km}{12 km/hr} = 15 hours. 5. Calculate Meeting Time: Train B leaves at 2 p.m. Adding 15 hours to 2 p.m.: 2 p.m. + 10 hours = 12 a.m. (Midnight) 12 a.m. + 5 hours = 5 a.m. on the next day. Exam Strategy & Shortcut: Instead of forming algebraic equations 60(t+3) = 72t, directly think in terms of the "gap". Train A creates a 180 km gap. Train B closes this gap at a rate of 12 km per hour. The mental math (180/12) = 15 hours is significantly faster. Common Pitfall: A frequent error is adding the 15 hours to Train A's departure time (11 a.m.) instead of Train B's departure time (2 p.m.), which would lead to an incorrect answer of 2 a.m. Final Answer: Therefore, the correct answer is 5 a.m. on the next day.

Question 2

A train $M$ leaves station $X$ at 5 a.m and reaches station $Y$ at 9 a.m. Another train $N$ leaves station $Y$ at 7 a.m. and reaches station $X$ at 10.30 a.m. At what time do the two trains cross each other?

Accepted Answer

Concept & Proportional Speed: When exact distances are unknown but time intervals are given, we can assume the total distance to be a convenient common multiple (LCM) of the given times to find proportional speeds, or use variables. The crossing point depends on relative speed from a common starting time. Step-by-Step Solution: 1. Determine Travel Times: Train M travel time = 5 a.m. to 9 a.m. = 4 hours. Train N travel time = 7 a.m. to 10:30 a.m. = 3.5 hours = (7/2) hours. 2. Assume Distance & Find Speeds: Let the total distance between X and Y be D. Speed of Train M (S_M) = (D/4) km/hr. Speed of Train N (S_N) = (D/3.5) = (2D/7) km/hr. 3. Synchronize Start Times: Train N starts at 7 a.m. By 7 a.m., Train M has already travelled for 2 hours (from 5 a.m.). Distance covered by Train M in 2 hours = 2 × (D/4) = (D/2). Remaining distance between them at 7 a.m. = D - (D/2) = (D/2). 4. Calculate Crossing Time: They travel towards each other, so relative speed is the sum. Relative Speed = S_M + S_N = (D/4) + (2D/7) = (7D + 8D/28) = (15D/28) km/hr. Time taken to cross = \frac{Remaining Distance}{Relative Speed} = (\frac{D/2)}{(15D/28)} = (D/2) × (28/15D) = (14/15) hours. 5. Convert to Minutes: (14/15) × 60 minutes = 14 × 4 = 56 minutes. Time of crossing = 7:00 a.m. + 56 minutes = 7:56 a.m. Exam Strategy & Shortcut: Let Distance = LCM of 4 and 3.5. Let's use 28 km. Speed M = (28/4) = 7 km/h. Speed N = (28/3.5) = 8 km/h. At 7 a.m., M has travelled for 2 hrs = 14 km. Distance left = 28 - 14 = 14 km. Relative speed = 7 + 8 = 15 km/h. Time = (14/15) hours = 56 mins. Meeting time = 7:56 a.m. This completely bypasses algebraic variables! Common Pitfall: Calculating the relative speed from 5 a.m. without accounting for the fact that Train N hasn't started moving yet. Always align both moving bodies to a common time before applying relative speed. Final Answer: Therefore, the correct answer is 7.56 a.m.

Question 3

Train A travelling at 60 km/hr leaves Mumbai for Delhi at 6 p.m. Train B travelling at 90 km/hr also leaves Mumbai for Delhi at 9 p.m. Train C leaves Delhi for Mumbai at 9 p.m. If all the three trains meet at the same time between Mumbai and Delhi, …

Accepted Answer

Concept & Meeting Point Synchronization: When multiple bodies meet at the same point and time, you can first determine the exact time and location of the meeting using two known bodies. Then, use that specific time and distance to deduce the properties of the third body. Step-by-Step Solution: 1. Find the meeting time of Train A and Train B: Train A starts at 6 p.m. at 60 km/hr. Train B starts at 9 p.m. at 90 km/hr. By 9 p.m., Train A has travelled for 3 hours. Distance = 60 × 3 = 180 km. Relative speed of B with respect to A = 90 - 60 = 30 km/hr. Time taken for B to catch A = \frac{180 km}{30 km/hr} = 6 hours (after 9 p.m.). Meeting time = 9 p.m. + 6 hours = 3 a.m. 2. Find the meeting location: Distance from Mumbai where A and B meet = Speed of B × Time of B = 90 km/hr × 6 hrs = 540 km. 3. Calculate Train C's speed: Train C meets them at the same time (3 a.m.) and place (540 km from Mumbai). Total distance from Delhi to Mumbai = 1260 km. Distance Train C travelled from Delhi = 1260 km - 540 km = 720 km. Time taken by Train C = 9 p.m. to 3 a.m. = 6 hours. Speed of Train C = \frac{Distance}{Time} = \frac{720 km}{6 hours} = 120 km/hr. Exam Strategy & Shortcut: Break the problem into two independent chunks: A & B, then B & C. The meeting of A and B completely locks in the location and time parameters. Once those are established (t=6 hours from 9 p.m., d=540 km from Mumbai), Train C becomes a basic Speed = \frac{Distance}{Time} calculation using the remaining track distance. Common Pitfall: A common mistake is trying to set up a massive three-variable equation equating distances from both sides simultaneously, which leads to algebraic clutter and high chances of calculation errors. Final Answer: Therefore, the correct answer is 120 km/hr.

Question 4

Reena leaves office at 6.00 p.m. and catches a 6.30 p.m. local train that arrives in her town at 7.00 p.m. Her father leaves home to pick her up at 7.00 p.m. from the station as she gets off the train. Yesterday, Reena left her office early and took…

Accepted Answer

Concept & Time Savings Principle: In problems where a meeting point shifts, resulting in a total time saving, the saved time purely comes from the distance the vehicle *did not have to travel*. By halving the saved time, we can determine the exact time the vehicle intercepted the walking person. Step-by-Step Solution: 1. Analyze the Father's Journey: The father arrived home 10 minutes earlier than usual. This 10-minute saving is strictly because he didn't have to drive the final stretch from the pickup point to the station and back. 2. Determine One-Way Savings: Since driving to the station and driving back takes equal time, he saved 10 / 2 = 5 minutes of driving time *each way*. 3. Pinpoint the Pickup Time: Usually, he reaches the station exactly at 7:00 p.m. Because he saved 5 minutes of driving *towards* the station, he must have met Reena 5 minutes before his usual arrival at the station. Pickup Time = 7:00 p.m. - 5 minutes = 6:55 p.m. 4. Calculate Reena's Walking Time: Reena arrived at the station early at 6:30 p.m. and immediately started walking. She was picked up at 6:55 p.m. Time spent walking = 6:55 p.m. - 6:30 p.m. = 25 minutes. Exam Strategy & Shortcut: The core trick is isolating the 10-minute save. 10 mins saved round trip \rightarrow 5 mins saved one way. Usual pickup is 7:00, so new pickup is 6:55. She started walking at 6:30. Difference is 25. You can solve this in your head without a pen once you grasp the logic that the walker's speed or total distance is entirely irrelevant to finding the time! Common Pitfall: A common pitfall is attempting to find the speed of Reena or her father, or assuming distances. The problem does not provide enough data for absolute speeds or distances, and trying to solve for them will lead to dead ends. Stick purely to the timeline. Final Answer: Therefore, the correct answer is 25 min.

Question 5

The speed of a railway engine is 42 km/hr when no compartment is attached and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 km/hr wit…

Accepted Answer

Concept & Proportionality: When a quantity is directly proportional to the square root of another, we can establish an equation using a constant of proportionality k. Reduction in Speed = k \sqrt{Number of Compartments} Step-by-Step Solution: 1. Set up the initial equation: Let initial speed be S_0 = 42 km/hr. Let the number of compartments be N. Reduction in speed R = 42 - Current Speed. Given relation: R = k\sqrt{N}. 2. Find the constant k: With 9 compartments (N = 9), the speed is 24 km/hr. Reduction R = 42 - 24 = 18 km/hr. Substitute into equation: 18 = k\sqrt{9} 18 = 3k \implies k = 6. The equation for reduction is now R = 6\sqrt{N}. 3. Determine maximum compartments for zero speed: The engine will stop moving when its speed drops to 0 km/hr. This means the reduction in speed equals the initial speed. R = 42 km/hr. Substitute to find N: 42 = 6\sqrt{N} \sqrt{N} = 7 N = 49. With exactly 49 compartments, the engine's speed is zero, meaning it cannot pull them. 4. Find the maximum compartments it *can* pull: To actually pull the train, the speed must be strictly greater than zero. Therefore, the maximum number of compartments it can pull is 49 - 1 = 48. Exam Strategy & Shortcut: Reduction = 42 - 24 = 18 for \sqrt{9} = 3. So multiplier is 18/3 = 6. Max reduction possible is 42. So \sqrt{N} = 42/6 = 7. N = 49 means train stalls. Max pulling capacity is 49 - 1 = 48. This can be done mentally in under 30 seconds. Common Pitfall: Choosing 49 as the answer. 49 is the threshold where the train comes to a complete halt (Speed = 0). The question asks for the maximum number it *can pull*, which requires a speed > 0. Final Answer: Therefore, the correct answer is 48.

Question 6

Direction: These questions are based on the following information: Ram and Shyam run a race between points $A$ and $B$, 5 km apart. Ram starts at 9 a.m. from $A$ at a speed of 5 km/hr, reaches $B$ and returns to $A$ at the same speed. Shyam starts a…

Accepted Answer

Concept & Overtaking: Overtaking occurs when two bodies move in the same direction and the faster body catches up to the slower body from behind. The relative speed is the difference between their speeds. Relative Speed = V_{fast} - V_{slow} Step-by-Step Solution: 1. Track Ram's Return Journey: Ram reaches B at 10:00 a.m. and starts heading back to A at 5 km/hr. 2. Track Shyam's Journey to B: Shyam starts at 9:45 a.m. at 10 km/hr. To cover 5 km, he takes (5/10) = 0.5 hours (30 minutes). He reaches B at 10:15 a.m. and immediately turns back towards A. 3. Locate Ram at 10:15 a.m.: From 10:00 a.m. to 10:15 a.m., Ram has been walking back towards A for 15 minutes (0.25 hours). Distance covered from B = 5 km/hr × 0.25 hrs = 1.25 km. 4. Set up the Overtake Scenario: At exactly 10:15 a.m., Shyam is at point B, and Ram is 1.25 km ahead of him on the path back to A. Both are moving in the same direction (towards A). 5. Calculate Catch-up Time: The relative speed is 10 km/hr - 5 km/hr = 5 km/hr. The gap to close is 1.25 km. Time to catch up = \frac{1.25 km}{5 km/hr} = 0.25 hours = 15 minutes. 6. Find Exact Overtake Time: 10:15 a.m. + 15 minutes = 10:30 a.m. Exam Strategy & Shortcut: Determine exactly when Shyam turns around (10:15 a.m.). At that exact moment, calculate Ram's lead. Ram had a 15-minute head start on the return leg, creating a 1.25 km gap. Since Shyam's relative speed advantage is 5 km/hr, closing a 1.25 km gap takes exactly a quarter of an hour (15 minutes). 10:15 + 15 mins = 10:30. Common Pitfall: Confusing the initial "meeting" (going in opposite directions) with "overtaking" (going in the same direction). Overtaking requires the relative speed to be the difference, not the sum. Final Answer: Therefore, the correct answer is 10 : 30 a.m.

Question 7

A car overtakes a bus travelling from Delhi to Jaipur at 4 : 30 p.m. The car reaches Jaipur at 6 : 00 p.m. After stopping there for 1 hour, it starts back towards Delhi and meets the same bus at 7 : 30 p.m. which was moving towards Jaipur at that ti…

Accepted Answer

Concept & Logic: This problem requires tracking the positions and times of two vehicles to determine their speeds relative to the total distance. Speed = \frac{Distance}{Time} Step-by-Step Solution: 1. Let the point where the car overtakes the bus (at 4:30 p.m.) be point A, and Jaipur be point B. Let the distance AB be D. 2. Car's Journey to Jaipur: The car travels from A to B between 4:30 p.m. and 6:00 p.m. Time taken by car = 1.5 hours. Speed of car (v_c) = (D/1.5) = (2D/3). 3. Car's Return Journey: The car stops at Jaipur for 1 hour (leaves at 7:00 p.m.) and meets the bus at 7:30 p.m. Let this meeting point be C. Time traveled by car from B to C = 0.5 hours. Distance BC = v_c × 0.5 = (2D/3) × (1/2) = (D/3). 4. Bus's Journey: The bus has been traveling from A since 4:30 p.m. At 7:30 p.m., it is at point C. Distance traveled by bus (AC) = AB - BC = D - (D/3) = (2D/3). Time taken by bus = 7:30 p.m. - 4:30 p.m. = 3 hours. Speed of bus (v_b) = \frac{Distance}{Time} = (\frac{2D/3)}{3} = (2D/9). 5. Bus's Total Time: Total time for the bus to travel distance D (AB) = (D/v_b) = (D/\frac{2D){9}} = (9/2) = 4.5 hours. 6. Arrival Time: The bus was at point A at 4:30 p.m. Arrival time at Jaipur = 4:30 p.m. + 4.5 hours = 9:00 p.m. Exam Strategy & Shortcut: Visualize the timeline. The car travels the whole distance in 1.5 hours. On the way back, it travels for 0.5 hrs, covering 1/3 of the distance. This means the bus has covered 2/3 of the distance in 3 hours. To cover the full distance (3/3), the bus will take 3 × (3/2) = 4.5 hours. 4:30 pm + 4.5 hrs = 9:00 pm. Common Pitfall: A common error is forgetting to account for the 1-hour stop the car makes in Jaipur when calculating the time it takes to travel back and meet the bus. Final Answer: Therefore, the correct answer is 9 : 00 p.m.

Question 8

The speed of a bus is 72 kmph. The distance covered by the bus in 5 sec is [Indian Railway Gr. 'D' Exam, 2014]

Accepted Answer

Concept & Formula: To find distance in meters given time in seconds, the speed must first be converted from kilometers per hour (kmph) to meters per second (m/s). Speed in m/s = Speed in kmph × (5/18) Distance = Speed × Time Step-by-Step Solution: 1. Convert Speed: Convert 72 kmph to m/s. 72 × (5/18) = 4 × 5 = 20 m/s 2. Calculate Distance: Use the converted speed and the given time (5 seconds). Distance = 20 m/s × 5 sec = 100 meters Exam Strategy & Shortcut: Memorize common conversions: 18 kmph = 5 m/s. Since 72 kmph = 4 × 18 kmph, the speed is 4 × 5 m/s = 20 m/s. The distance is 20 × 5 = 100 m. Common Pitfall: A major mistake is forgetting to convert units and simply multiplying 72 × 5, leading to incorrect units and answers. Always ensure unit consistency. Final Answer: Therefore, the correct answer is 100 m.

Question 9

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 10

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 11

Rani goes to school from her house in 30 minutes. Raja takes 45 minutes in covering the same distance. Find the ratio between time taken by Rani and Raja.

Accepted Answer

Concept & Simple Ratios: A ratio compares two quantities of the same kind. To find the ratio between two times, simply place them in a fraction and simplify to its lowest terms. Ratio = \frac{Time_1}{Time_2} Step-by-Step Solution: 1. Time taken by Rani = 30 minutes. 2. Time taken by Raja = 45 minutes. 3. Form the ratio of Rani's time to Raja's time: Ratio = (30/45) 4. Simplify the ratio by dividing the numerator and denominator by their greatest common divisor (15): Ratio = (30 \div 15/45 \div 15) = (2/3) 5. Expressed as a ratio, this is 2 : 3. Exam Strategy & Shortcut: Both numbers end in 5 or 0, making them multiples of 5, but quickly scanning for larger multiples reveals they are both part of the 15 times table (15 × 2 = 30, 15 × 3 = 45). This gives the simplified ratio of 2:3 instantly. Common Pitfall: Inverting the ratio by accident. The question asks for "Rani and Raja" in that specific order. Writing 3:2 (Raja to Rani) would be incorrect. Final Answer: Therefore, the correct answer is 2 : 3.

Question 12

A boat, while going downstream in a river covered a distance of 50 miles at an average speed of 60 miles per hour. While returning, because of the water resistance, it took 1 hour 15 minutes to cover the same distance. What was the average speed dur…

Accepted Answer

Concept & Formula: Average speed for a round trip is the total distance divided by the total time. Average Speed = \frac{Total Distance}{Total Time} Step-by-Step Solution: - Downstream Journey: Distance = 50 miles. Speed = 60 mph. Time taken = (50/60) = (5/6) hours. - Upstream Journey: Distance = 50 miles (since it covers the same distance). Time taken = 1 hour 15 minutes = 1(15/60) hours = 1(1/4) = (5/4) hours. - Total Journey: Total Distance = 50 + 50 = 100 miles. Total Time = (5/6) + (5/4) = (10/12) + (15/12) = (25/12) hours. - Average Speed: Avg Speed = (100/\frac{25){12}} = 100 × (12/25) = 4 × 12 = 48 mph Exam Strategy & Shortcut: Alternatively, find the upstream speed: 50 / ((5/4)) = 40 mph. When distances are equal, the average speed formula is (2ab/a+b) where a and b are the two speeds. (2 × 60 × 40/60 + 40) = (4800/100) = 48 mph. Common Pitfall: Calculating the arithmetic mean of the two speeds (60 mph and 40 mph) to get 50 mph. Average speed over equal distances requires the harmonic mean, not the arithmetic mean. Final Answer: Therefore, the correct answer is 48 mph.

Question 13

Wheels of diameters 7 cm and 14 cm start rolling simultaneously from $X$ and $Y$, which are 1980 cm apart, towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10 seconds,…

Accepted Answer

Concept & Logic: When two objects move towards each other, the total distance covered is the sum of the individual distances traveled. Circumference (C) = \pi d Relative Speed = Speed_1 + Speed_2 Step-by-Step Solution: • Given: • Diameter of smaller wheel (d_1) = 7 cm • Diameter of larger wheel (d_2) = 14 cm • Distance between X and Y = 1980 cm • Time to meet (t) = 10 seconds • Circumferences: C_1 = \pi d_1 = (22/7) × 7 = 22 cm C_2 = \pi d_2 = (22/7) × 14 = 44 cm • Combined distance in 1 revolution: C_1 + C_2 = 22 + 44 = 66 cm • Let n be the number of revolutions per second made by each wheel: Total distance covered in 10 seconds = 10 × n × (C_1 + C_2) 1980 = 10 × n × 66 1980 = 660n \implies n = (1980/660) = 3 rev/sec • Speed of the smaller wheel: Speed_1 = n × C_1 = 3 × 22 = 66 cm/sec Exam Strategy & Shortcut: Relative speed = (1980/10) = 198 cm/sec. Since diameters are in the ratio 7 : 14 = 1 : 2, the distances and speeds are divided in the ratio 1 : 2. Speed of smaller wheel = (1/1 + 2) × 198 = (1/3) × 198 = 66 cm/sec Common Pitfall: Dividing 1980 directly by the single circumference instead of the sum of both circumferences. Final Answer: Therefore, the correct answer is 66 cm/sec.

Question 14

The circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel?

Accepted Answer

Concept & Strategy: Both wheels travel the exact same linear distance (D). Revolutions = \frac{Distance}{Circumference} Set up an equation equating the difference in revolutions between the front and rear wheels to 5. Step-by-Step Solution: • Given: • Front wheel circumference (C_1) = 40 ft • Back wheel circumference (C_2) = 48 ft • Let the total distance travelled be D ft. • Express the number of revolutions: Front wheel revolutions (n_1) = (D/40) Rear wheel revolutions (n_2) = (D/48) • Formulate the equation: (D/40) - (D/48) = 5 • Solve for D: LCM(40, 48) = 240 (6D - 5D/240) = 5 (D/240) = 5 D = 5 × 240 = 1200 ft Exam Strategy & Shortcut: Formula for distance when difference in revolutions is \Delta n: D = (C_1 × C_2/C_2 - C_1) × \Delta n = (40 × 48/48 - 40) × 5 = (1920/8) × 5 = 240 × 5 = 1200 ft Common Pitfall: Setting (D/48) - (D/40) = 5, which results in a negative value because the smaller wheel makes more revolutions. Final Answer: Therefore, the correct answer is 1200 ft.

Question 15

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.

Question 16

A skating champion moves along the circumference of a circle of radius 28 m in 44 sec. How many seconds will it take her to move along the perimeter of a hexagon of side 48 m?

Accepted Answer

Concept & Speed, Distance, and Time: This problem requires equating the speed of the skater on two different geometric paths. First, calculate the speed using the circular path, then use that speed to find the time taken for the hexagonal path. Speed = \frac{Distance}{Time} Step-by-Step Solution: • Step 1: Find the speed on the circular track. • Radius of the circle, r = 28 m. • Perimeter (circumference) of the circle = 2\pi r = 2 × (22/7) × 28. • Circumference = 2 × 22 × 4 = 176 m. • Time taken for the circle = 44 seconds. • Speed = \frac{176 m}{44 s} = 4 m/s. • Step 2: Find the time for the hexagonal track. • Side of the regular hexagon, a = 48 m. • Perimeter of the hexagon = 6 × a = 6 × 48 = 288 m. • Time taken = \frac{Distance}{Speed} = (288/4). • Time = 72 seconds. Exam Strategy & Shortcut: Instead of calculating the intermediate speed fully, you can set up a ratio since speed is constant: \frac{Time_1}{Distance_1} = \frac{Time_2}{Distance_2} (44/2 × \frac{22){7} × 28} = (T/6 × 48) \Rightarrow (44/176) = (T/288) \Rightarrow (1/4) = (T/288) \Rightarrow T = 72. Common Pitfall: A frequent mistake is using the formula for the area instead of the perimeter for the tracks. Always associate "moving along the circumference/perimeter" with boundary length, not internal area. Final Answer: Therefore, the correct answer is 72.

Time and Distance Questions

Questions

Practice smarter