Question 1

A 180-metre long train crosses another 270-metre long train running in the opposite direction in 10.8 seconds. If the speed of the first train is 60 kmph, what is the speed of the second train in kmph? (Bank P.O., 2010)

Accepted Answer

Concept & Converting Combined Distance: To find an unknown speed in an opposite-direction crossing, first determine the total combined speed in m/s, convert it to km/hr, and then subtract the known speed. Relative\ Speed = (L_1 + L_2/Time) Step-by-Step Solution: 1. Given: L_1 = 180 m, L_2 = 270 m. Total distance = 180 + 270 = 450 m. Time = 10.8 s. S_1 = 60 kmph. Let S_2 be the speed of the second train in kmph. 2. Calculate the relative speed in m/s: Relative\ Speed = (450/10.8) = (4500/108)\ m/s 3. Convert this relative speed to km/hr: (4500/108) × (18/5) = (900/108) × 18 = (900/6) = 150\ km/hr 4. Since they move in opposite directions, the relative speed is the sum of their individual speeds: S_1 + S_2 = 150 60 + S_2 = 150 5. Solve for S_2: S_2 = 150 - 60 = 90\ km/hr Exam Strategy & Shortcut: Distance = 450m. Time = 10.8s. Speed in km/hr = (450/10.8) × (18/5) = 150 km/hr. Since opposite direction means speeds are added, the second train's speed is 150 - 60 = 90 km/hr. Common Pitfall: Struggling with the decimal division (450/10.8). Always clear decimals by multiplying numerator and denominator by 10 before simplifying, and apply the 18/5 conversion early to cancel out awkward denominators like 108 (since 18 × 6 = 108). Final Answer: Therefore, the correct answer is 90.

Question 2

Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is :

Accepted Answer

Concept & Relative Speed: When two trains move in opposite directions, their relative speed is the sum of their individual speeds. They must cover a distance equal to the sum of their lengths to cross each other. Relative Speed = \frac{Total Length}{Time} Step-by-Step Solution: • Total distance to cover = Sum of lengths = 100 + 100 = 200 meters. • Time taken to cross = 8 seconds. • Relative speed in m/s = (200/8) = 25 m/s. • Convert relative speed to km/hr: 25 × (18/5) = 90 km/hr. • Let the speed of the slower train be x km/hr. The faster train is 2x km/hr. • Relative speed = x + 2x = 3x. • Equating speeds: 3x = 90 \Rightarrow x = 30 km/hr. • Speed of the faster train = 2x = 60 km/hr. Exam Strategy & Shortcut: Calculate relative speed directly in m/s, assign ratios (1:2 means 3 parts total), find the value of one part (25 / 3 m/s), and then calculate the faster train's speed before converting: 2 × (25/3) × (18/5) = 60 km/hr. Common Pitfall: Forgetting to convert the relative speed from m/s to km/hr before finding the individual speeds, leading to incorrect choices. Final Answer: Therefore, the correct answer is 60 km/hr.

Question 3

A 150 m long train crosses a milestone in 15 seconds and a train of same length coming from the opposite direction in 12 seconds. The speed of the othe train is (R.R.B., 2006)

Accepted Answer

Concept & Distance: A train crossing a stationary object (like a milestone) covers a distance equal to its own length. When crossing another moving train, the distance is the sum of both lengths. Speed = \frac{Length}{Time} Step-by-Step Solution: • Length of the first train = 150 m. Time to cross milestone = 15 s. • Speed of the first train = (150/15) = 10 m/s. • Length of the second train = 150 m (since it has the same length). • Total distance when crossing each other = 150 + 150 = 300 m. • Time taken to cross each other = 12 s. • Relative speed = (300/12) = 25 m/s. • Since they move in opposite directions, Relative Speed = Speed 1 + Speed 2. • 10 + Speed 2 = 25 \Rightarrow Speed 2 = 15 m/s. • Convert to kmph: 15 × (18/5) = 54 kmph. Exam Strategy & Shortcut: You can calculate relative speed 25 m/s and subtract the first train's speed (10 m/s) to get 15 m/s. Multiply by 3.6 for a rapid m/s to km/h conversion: 15 × 3.6 = 54 kmph. Common Pitfall: Assuming the second train crosses a point in 12 seconds instead of crossing the first train. Always read whether it crosses a point or an object with length. Final Answer: Therefore, the correct answer is 54 kmph.

Question 4

A train which is moving at an average speed of 40km/h reaches its destination on time. When its average speed reduces to 35 km/h, then it reaches its destination 15 minutes late. The distance travelled by the train, is [CLAT Exam, 2016]

Accepted Answer

Concept & Formula: When a fixed distance is covered at two different speeds resulting in a time difference \Delta t, the distance D can be found using the formula: D = (S_1 × S_2/S_1 - S_2) × \Delta t Step-by-Step Solution: 1. Given Data: - Speed 1 (S_1) = 40 km/h. - Speed 2 (S_2) = 35 km/h. - Time difference (\Delta t) = 15 minutes = (15/60) = (1/4) hours. 2. Apply Distance Formula: - D = (40 × 35/40 - 35) × (1/4) - D = (1400/5) × (1/4) - D = 280 × (1/4) = 70 km. Exam Strategy & Shortcut: Memorize the product-of-speeds formula for constant distance problems. It bypasses setting up complex algebraic fractions entirely and yields the answer in seconds. Common Pitfall: Forgetting to convert the time difference from minutes to hours ((15/60)) before applying it to speeds given in km/h. Final Answer: Therefore, the correct answer is 70km.

Question 5

A train passes two bridges of lengths 500 m and 250 m in 100 seconds and 60 seconds respectively. The length of the train is [SSC—CHSL (10+2) Exam, 2015]

Accepted Answer

Concept & Logic: The speed of a train remains constant whether it crosses one bridge or another. The speed is equal to the total distance (train length + bridge length) divided by the time taken. Speed = \frac{L + Bridge_1}{t_1} = \frac{L + Bridge_2}{t_2} Step-by-Step Solution: 1. Set Up Equation: - Let train length be L. - (L + 500/100) = (L + 250/60) 2. Simplify and Solve: - Cross-multiply: 60(L + 500) = 100(L + 250) - Divide by 20 to simplify: 3(L + 500) = 5(L + 250) - 3L + 1500 = 5L + 1250 - 2L = 250 - L = 125 m. Exam Strategy & Shortcut: Reduce the time ratio first (100 : 60 = 5 : 3). Since speeds are equal, the distance ratios must also balance, making quick work of the linear equation. Common Pitfall: Forgetting to add the train's length (L) to both bridge lengths when setting up the speed equivalence, treating the bridges as standalone distances. Final Answer: Therefore, the correct answer is 125 m.

Problems on Trains Questions

Questions

Practice smarter