Question 1

A car covers the distance from Sonagarh to Raipur in 35 min at an average speed of 69 km/h. If its average speed increases by 36 km/h for the same route, how long (in minutes) will the journey take?

Accepted Answer

Introduction / Context: This is a same-distance, different-speed time conversion. Once we compute the distance from the first scenario, we can reuse it to find the time with a higher speed. Conversions between hours and minutes must be handled carefully for accuracy. Given Data / Assumptions: Initial average speed v1 = 69 km/h. Initial time t1 = 35 min = 35/60 h. New speed v2 = 69 + 36 = 105 km/h. Same distance each time. Concept / Approach: Distance D = v1 * t1. New time t2 = D / v2. Convert t2 (hours) to minutes by multiplying by 60. Step-by-Step Solution: D = 69 * (35/60) = 69 * 0.5833… = 40.25 km.t2 (h) = 40.25 / 105 ≈ 0.383333… h.t2 (min) = 0.383333… * 60 = 23 minutes (exact). Verification / Alternative check: At 105 km/h, per minute distance is 1.75 km. Covering 40.25 km takes 40.25 / 1.75 = 23 minutes—consistent. Why Other Options Are Wrong: 24, 27, 29 minutes correspond to different (lower) speeds than 105 km/h for the 40.25 km route. Common Pitfalls: Mixing hours and minutes; assuming percentage change applies to time directly (it applies inversely via speed). Final Answer: 23 min

Question 2

A student walks from home to school at 2.5 km/h and arrives 6 minutes late. The next day, increasing speed by 1 km/h (to 3.5 km/h), the student arrives 6 minutes early. What is the distance from home to school (in km)?

Accepted Answer

Introduction / Context: Here two commuting scenarios bracket the scheduled time: one is 6 minutes late, the other is 6 minutes early. For a fixed distance, time is inversely proportional to speed. Setting up equations with the same scheduled time lets us solve the distance cleanly. Given Data / Assumptions: Speed day 1 = 2.5 km/h (late by 6 min). Speed day 2 = 3.5 km/h (early by 6 min). Let scheduled time be T hours, distance be D km. Concept / Approach: Equations: D/2.5 = T + 0.1 and D/3.5 = T − 0.1, since 6 minutes = 0.1 h. Subtract the second from the first to eliminate T and solve for D. Step-by-Step Solution: D/2.5 − D/3.5 = 0.2.(1/2.5) = 0.4; (1/3.5) ≈ 0.285714; difference = 0.1142857.D * 0.1142857 = 0.2 ⇒ D ≈ 1.75 km = 7/4 km. Verification / Alternative check: At 2.5 km/h: time = 1.75/2.5 = 0.7 h (42 min). At 3.5 km/h: 1.75/3.5 = 0.5 h (30 min). Midpoint is 36 min; late/early by ±6 min as stated. Why Other Options Are Wrong: 5/4, 9/4, 11/4 km do not produce a symmetrical ±6 minute deviation around a single schedule for the given speeds. Common Pitfalls: Using minutes without converting to hours; averaging speeds rather than setting up time equations. Final Answer: 7/4 km

Question 3

A policeman spots a thief 200 m ahead. The thief runs at 10 km/h and the policeman chases at 11 km/h. After 6 minutes of running, what is the distance between them (in metres)?

Accepted Answer

Introduction / Context: Pursuit problems use relative speed along a straight line. The gap closes at (chaser speed − runner speed). Once we find how much is closed in the given time, we subtract from the initial lead to get the remaining separation. Given Data / Assumptions: Initial lead L = 200 m. Speeds: thief = 10 km/h, policeman = 11 km/h. Time t = 6 minutes = 360 s. Concept / Approach: Relative speed = (11 − 10) km/h = 1 km/h. Convert 1 km/h to m/s: 1000/3600 ≈ 0.277777… m/s. Distance closed in time t equals relative_speed * t. Step-by-Step Solution: Relative speed = 0.277777… m/s.Closed distance = 0.277777… * 360 ≈ 100 m.Remaining gap = 200 − 100 = 100 m. Verification / Alternative check: In 6 minutes, thief covers (10 km/h)*(0.1 h)=1 km; policeman covers (11 km/h)*(0.1 h)=1.1 km; difference 0.1 km = 100 m—same result. Why Other Options Are Wrong: 180/150/125 m do not match the closed distance computed from the correct relative speed. Common Pitfalls: Adding speeds when both move in the same direction; mixing metres and kilometres without proper conversion. Final Answer: 100 m

Question 4

A truck covers 368 km in 8 h (constant speed). A car travels 16 km farther than the truck at a speed 18 km/h faster than the truck. How many hours will the car take to cover its distance?

Accepted Answer

Introduction / Context: Comparative speed problems often require first extracting one vehicle’s speed, then applying adjustments to distance and speed for the second vehicle to compute time. Keeping units consistent simplifies the calculation. Given Data / Assumptions: Truck: distance 368 km, time 8 h ⇒ speed = 46 km/h. Car: speed = 46 + 18 = 64 km/h. Car distance = 368 + 16 = 384 km. Concept / Approach: Time = distance / speed. Compute truck speed first, then plug in the car’s parameters to find time needed for 384 km at 64 km/h. Step-by-Step Solution: Truck speed = 368 / 8 = 46 km/h.Car speed = 64 km/h; car distance = 384 km.Car time = 384 / 64 = 6 h. Verification / Alternative check: At 64 km/h, 6 hours covers 384 km exactly (64 * 6 = 384). Why Other Options Are Wrong: 5 h would cover 320 km; 7 h gives 448 km; 8 h 512 km—none match 384 km at 64 km/h. Common Pitfalls: Forgetting to add both the distance and speed increments; mixing up which value applies to which vehicle. Final Answer: 6 h

Question 5

Shantanu rides a motorcycle for 715 km at constant speed v. If he had ridden 10 km/h faster, he would have saved 2 hours over the same distance. What is his original speed v (in km/h)?

Accepted Answer

Introduction / Context: This is a classic “time saved with higher speed” equation. Over a fixed distance, increasing speed reduces time according to t = D/v. Setting up the difference in times equal to the saved amount yields a quadratic in v that can be solved exactly. Given Data / Assumptions: Distance D = 715 km. Original speed v km/h; faster speed v + 10 km/h. Time saved = 2 h. Concept / Approach: Equation: 715/v − 715/(v+10) = 2. Simplify to solve for v. Positive, realistic highway values are expected. Step-by-Step Solution: 715[(1/v) − (1/(v+10))] = 2.715 * (10) / (v(v+10)) = 2 ⇒ 7150 = 2v(v+10).v^2 + 10v = 3575 ⇒ v^2 + 10v − 3575 = 0.Discriminant = 100 + 14300 = 14400; sqrt = 120.v = (−10 + 120)/2 = 55 km/h (ignoring negative root). Verification / Alternative check: Times: at 55 km/h → 13 h; at 65 km/h → 11 h; saving = 2 h as required. Why Other Options Are Wrong: 60 or 65 km/h do not yield a 2-hour difference for 715 km; 36 km/h is too low and unrealistic here. Common Pitfalls: Arithmetic slips in clearing denominators; accepting the negative quadratic root. Final Answer: 55 km/h

Question 6

The speed ratio of a train to a car is 16 : 15. A bus covers 480 km in 8 h (so its speed is 60 km/h), and the bus speed is 3/4 of the train’s speed. How many kilometres will the car cover in 6 h?

Accepted Answer

Introduction / Context: We are given relative speed links among three vehicles. From the bus, we infer the train’s speed, then the car’s speed from the given ratio, and finally compute the distance for the car in 6 hours. Careful chaining avoids unnecessary algebra. Given Data / Assumptions: Bus distance/time = 480 km / 8 h ⇒ bus speed = 60 km/h. Bus speed = (3/4) * train speed ⇒ train speed = 60 * (4/3) = 80 km/h. Train : Car = 16 : 15 ⇒ car speed = 80 * (15/16) = 75 km/h. Concept / Approach: Once car speed is known, distance in 6 hours is straightforward: distance = speed * time. Step-by-Step Solution: Train speed = 80 km/h.Car speed = 80 * 15/16 = 75 km/h.Car distance in 6 h = 75 * 6 = 450 km. Verification / Alternative check: Check proportions: 75 : 80 simplifies to 15 : 16, matching the given ratio. Why Other Options Are Wrong: 480 km would require 80 km/h; 360 km corresponds to 60 km/h—not the car’s speed. Common Pitfalls: Taking 3/4 factor the wrong way (making the train slower than the bus) or mixing up the 16:15 ratio direction. Final Answer: 450 km

Question 7

Sumit drives from home to a resort at 45 km/h. On the return trip along the same route, he averages 40 km/h and takes 1 hour longer than the outbound time. How many kilometres is each one-way trip?

Accepted Answer

Introduction / Context: For a fixed distance D, time equals D/v. If one leg takes 1 hour more than the other, equate the difference of the two times to 1 hour to solve for D. This avoids needing the absolute times individually. Given Data / Assumptions: Outbound speed = 45 km/h. Return speed = 40 km/h. Return time − Outbound time = 1 h. Distance each way = D (km). Concept / Approach: Set D/40 − D/45 = 1 ⇒ D * ( (1/40) − (1/45) ) = 1. Solve for D by taking common denominators. Step-by-Step Solution: (1/40) − (1/45) = (45 − 40) / 1800 = 5/1800 = 1/360.D * (1/360) = 1 ⇒ D = 360 km. Verification / Alternative check: Outbound time = 360/45 = 8 h; return time = 360/40 = 9 h; indeed, 1 hour longer. Why Other Options Are Wrong: 250 or 375 km give a time gap different from exactly 1 hour with the given speeds. Common Pitfalls: Using average speeds or adding the speeds; the key is the time difference at fixed distance. Final Answer: 360 km

Question 8

A car runs at 65 km/h when serviced and 40 km/h when not serviced. After servicing, the car covers a certain distance in 5 h (distance = 325 km). Approximately how long will it take to cover the same distance when not serviced at 40 km/h?

Accepted Answer

Introduction / Context: Speed–time–distance conversions are direct when one of the trip legs provides the total distance. Here, we first compute the route length from the serviced run, then reuse that length to find the required time at the lower speed. The question seeks an approximate time, so rounding to the nearest option is appropriate. Given Data / Assumptions: Serviced speed = 65 km/h; time = 5 h. Distance D = 65 * 5 = 325 km. Not serviced speed = 40 km/h. Concept / Approach: Time = distance / speed. Compute 325 / 40 hours, then compare to the nearest choice as “approximate”. Step-by-Step Solution: D = 325 km.t = 325 / 40 = 8.125 h.Approximate to the nearest listed value: 8 h. Verification / Alternative check: At 40 km/h for 8.125 h you cover exactly 325 km; rounding to 8 h is within the intent of “approximate”. Why Other Options Are Wrong: 7 h gives 280 km; 10 h gives 400 km—both far from 325 km. Common Pitfalls: Mistaking “approximate” for percentage rounding; ensure you use the same distance for both conditions. Final Answer: 8 h

Question 9

The travel-rate ratio of A to B is 2 : 3, so A takes 20 minutes more than B to reach the same destination. If A had walked at double speed, how long would A take to cover the distance?

Accepted Answer

Introduction / Context: Time to cover a fixed distance is inversely proportional to speed. With a speed ratio 2:3 (A slower than B), and a 20-minute time difference, we can back out a scale for distance/speed and then compute A’s time at doubled speed. This avoids finding the absolute distance explicitly. Given Data / Assumptions: A:B speed ratio = 2:3 ⇒ let speeds be 2k and 3k. Time difference = 20 min = 1/3 h. Distance for both is the same. Concept / Approach: Times: t_A = D/(2k), t_B = D/(3k). Difference t_A − t_B = D/(6k) = 1/3 ⇒ D/k = 2. Doubling A’s speed to 4k makes new time t_A’ = D/(4k) = (D/k)/4. Step-by-Step Solution: D/k = 2.t_A’ = 2/4 = 1/2 h.Convert to minutes: 1/2 h = 30 min. Verification / Alternative check: Original times: D/(2k) = 1 h and D/(3k) = 2/3 h (difference 1/3 h = 20 min), consistent with D/k = 2. Why Other Options Are Wrong: 35, 20, 45 min do not equal the implied 1/2 hour at doubled speed. Common Pitfalls: Applying the 2:3 ratio to time instead of speed (they are inversely related). Final Answer: 30 min

Question 10

A car travels 75 km at 25 km/h, then 25 km at 5 km/h, and finally 50 km at 25 km/h along the same route. What is the overall average speed for the entire 150 km journey?

Accepted Answer

Introduction / Context: Average speed equals total distance divided by total time, never the simple average of the listed speeds unless times or distances are equal in a specific way. Here, segment distances and speeds are given explicitly, making the computation direct by summing segment times and distances. Given Data / Assumptions: Segment 1: 75 km at 25 km/h. Segment 2: 25 km at 5 km/h. Segment 3: 50 km at 25 km/h. Concept / Approach: Compute each segment’s time (distance/speed), add them to get total time. Sum distances to get total distance. Then average speed = total distance / total time. Step-by-Step Solution: t1 = 75/25 = 3 h.t2 = 25/5 = 5 h.t3 = 50/25 = 2 h.Total time = 3 + 5 + 2 = 10 h.Total distance = 75 + 25 + 50 = 150 km.Average speed = 150 / 10 = 15 km/h. Verification / Alternative check: Weighted average by time produces the same value; the very slow 5 km/h segment dominates time. Why Other Options Are Wrong: 25 or 40 km/h ignore the long slow segment; 12.5 km/h would require more than 10 h total time. Common Pitfalls: Taking the simple mean of 25, 5, 25, which is not applicable across unequal times. Final Answer: 15 km/h

Question 11

A car runs at 50 km/h when not serviced and at 60 km/h when serviced. After servicing, it covers a certain distance in 6 h (distance = 360 km). How long will it take to cover the same distance when not serviced at 50 km/h?

Accepted Answer

Introduction / Context: The stem wording is commonly intended to compare the time for the same fixed distance under two speed conditions (serviced vs not serviced). We first find the route length using the serviced run, then compute the time at the lower “not serviced” speed for that same distance. Given Data / Assumptions: Serviced speed = 60 km/h; serviced time = 6 h ⇒ distance D = 360 km. Not serviced speed = 50 km/h. We seek time at 50 km/h for D = 360 km. Concept / Approach: Use t = D / v with D fixed. Keep units in hours throughout to avoid conversion errors. Step-by-Step Solution: D = 60 * 6 = 360 km.t (not serviced) = 360 / 50 = 7.2 h. Verification / Alternative check: At 60 km/h, 6 h is 360 km; at 50 km/h, 7.2 h also yields 360 km—consistent. Why Other Options Are Wrong: 8 or 8.2 h overestimate; 6.5 h underestimates time for 50 km/h. Common Pitfalls: Misreading the stem to ask the time “when serviced” again; the meaningful comparison is serviced vs not serviced for the same distance. Final Answer: 7.2 h

Question 12

Stations A and B are 1000 km apart. A train leaves A for B at 40 km/h, and at the same moment another train leaves B for A at 60 km/h. How far from A will they meet?

Accepted Answer

Introduction / Context: For two objects moving toward each other, the meeting point divides the total distance in the ratio of their speeds (or equivalently, each covers distance = speed * common time). Using either method gives the same result cleanly and quickly. Given Data / Assumptions: Total separation = 1000 km. Speeds: A→B = 40 km/h, B→A = 60 km/h. Concept / Approach: Since they start together, time to meet is the same for both. Solve by proportional division: the distance from A will be (speed from A / sum of speeds) * total distance. Step-by-Step Solution: Sum of speeds = 40 + 60 = 100 km/h.Fraction from A = 40/100 = 0.4 of 1000 km.Distance from A = 0.4 * 1000 = 400 km. Verification / Alternative check: Common meeting time = 1000/100 = 10 h; A’s train covers 40 * 10 = 400 km, B’s covers 600 km—sums to 1000 km. Why Other Options Are Wrong: 350/300/525 km do not match the 40:60 ratio or the equal-time product. Common Pitfalls: Using difference of speeds (that is for relative speed in time calculation) to split distances; here use ratio or equal time logic. Final Answer: 400 km

Question 13

A train runs 3 h at 40 km/h and then 4.5 h at 60 km/h, which totals 3/5 of its full trip distance. It wants to cover the remaining distance in 4 h. What average speed is needed for the remaining part (in km/h)?

Accepted Answer

Introduction / Context: Segmented trips require careful accumulation of distance and then scaling to the whole. Once we find how much distance remains, the required speed for a given remaining time is just distance/time. The key is converting the “3/5 of total” into an absolute distance via the already covered kilometres. Given Data / Assumptions: Stage 1: 3 h @ 40 km/h ⇒ 120 km. Stage 2: 4.5 h @ 60 km/h ⇒ 270 km. Covered so far = 390 km = 3/5 of total. Remaining time allowed = 4 h. Concept / Approach: If 390 km is 3/5, then total distance = 390 * (5/3) = 650 km. Remaining distance = 650 − 390 = 260 km. Required speed = 260/4. Step-by-Step Solution: Total distance = 390 * 5/3 = 650 km.Remaining distance = 650 − 390 = 260 km.Required average speed = 260 / 4 = 65 km/h. Verification / Alternative check: At 65 km/h for 4 h, the train adds 260 km, reaching the computed total of 650 km. Why Other Options Are Wrong: 45/35 km/h would be too slow to cover 260 km in 4 h; 80 km/h is unnecessarily high. Common Pitfalls: Confusing 3/5 of total time with 3/5 of total distance—here it explicitly refers to distance. Final Answer: 65 km/h

Question 14

Two cars A and B leave the same point for the same destination. Speeds are 30 km/h (A) and 45 km/h (B). Car B reaches 2 hours earlier than car A. What is the distance from start to destination (in km)?

Accepted Answer

Introduction / Context: When two travellers cover the same distance at different speeds, the faster one arrives earlier. The difference in their times equals the given time gap, which leads to a simple equation in the unknown distance D that we can solve directly. Given Data / Assumptions: Speeds: 30 km/h (A), 45 km/h (B). Arrival time difference = 2 h (B earlier). Same route and distance. Concept / Approach: Times: t_A = D/30, t_B = D/45. Difference t_A − t_B = 2. Solve for D using common denominators. Step-by-Step Solution: D/30 − D/45 = 2.(1/30 − 1/45) = (3 − 2)/90 = 1/90.D * (1/90) = 2 ⇒ D = 180 km. Verification / Alternative check: Times: 180/30 = 6 h (A), 180/45 = 4 h (B); difference = 2 h as given. Why Other Options Are Wrong: 90 km gives 1 h gap; 270 km gives 3 h gap; 360 km gives 4 h gap—none match 2 h. Common Pitfalls: Reversing which time is larger; forgetting to use the same distance in both fractions. Final Answer: 180 km

Question 15

Two ladies start simultaneously from Cities A and B, 24 km apart, walking toward each other. The first lady walks 2 km/h faster than the second and reaches B exactly 1 hour before the second reaches A. What are their speeds (km per hour)?

Accepted Answer

Introduction / Context: Both travellers start together and traverse the entire 24 km, one going from A to B and the other from B to A. Let the slower speed be v km/h and the faster be v + 2. The faster arrives 1 hour earlier, giving a time-difference equation that can be solved for v directly. Given Data / Assumptions: Total one-way distance for each = 24 km. Faster speed = v + 2; slower speed = v. Time difference between arrivals = 1 h. Concept / Approach: Arrival times: T_fast = 24/(v+2), T_slow = 24/v. Since the faster arrives earlier by 1 h: T_slow − T_fast = 1. Solve for v by clearing denominators. Step-by-Step Solution: 24/v − 24/(v+2) = 1.24[(v+2 − v)/(v(v+2))] = 1 ⇒ 24 * 2 / (v(v+2)) = 1.v(v+2) = 48 ⇒ v^2 + 2v − 48 = 0 ⇒ (v − 6)(v + 8) = 0.v = 6 km/h (positive root) ⇒ faster = 8 km/h. Verification / Alternative check: Times: 24/6 = 4 h; 24/8 = 3 h; difference = 1 h as required. Why Other Options Are Wrong: Other pairs do not differ by exactly 2 km/h or do not yield a 1-hour arrival gap over 24 km. Common Pitfalls: Using relative speed meeting-time equations instead of whole-journey arrival times; mixing roles of the two speeds. Final Answer: 8 km, 6 km

Question 16

The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms. what was the total distance travelled in 12 hours?

Accepted Answer

Total distance travelled in 12 hours =(35+37+39+.....upto 12 terms) This is an A.P with first term, a=35, number of terms, n= 12,d=2. Required distance = 12/2[2 x 35+{12-1) x 2] =6(70+23) = 552 kms.

Question 17

A person crosses a 600 m long street in 5 minutes, What is his speed in km per hour?

Accepted Answer

Speed = 600 5 × 60 m / s e c = 2 m/sec = 2 x (18/5) km/hr = 7.2 km/hr

Question 18

A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minute…

Accepted Answer

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr Distance covered in 6 minutes = (1/60) x 6 km = 1/10 km = 100 m Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

Question 19

Two trains A and B start simultaneously in the opposite direction from two points P and Q and arrive at their destinations 16 and 9 hours respectively after their meeting each other. At what speed does the second train B travel if the first train tr…

Accepted Answer

s 1 s 2 = t 2 t 1 ⇒ 120 s 2 = 9 16 = 3 4 ⇒ s 2 = 160 k m / h

Question 20

Two horses start trotting towards each other, one from A to B and another from B to A. They cross each other after one hour and the first horse reaches B, 5/6 hour before the second horse reaches A. If the distance between A and B is 50 km. what is …

Accepted Answer

If the speed of the faster horse be f s and that of slower horse be s s then f s + s s = 50 1 = 50 and 50 s s - 50 f s = 5 6 Now, you can go through options. The speed of slower horse is 20km/h Since, 20+30=50 and 50 20 - 50 30 = 5 6

Time and Distance Questions