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 km/h after every hour. If it travels 35 km in the first hour, what is the total distance travelled by the car in 12 hours?

Accepted Answer

Introduction / Context: This question belongs to the time and distance chapter and involves an arithmetic progression in speeds. The car does not move at a uniform speed; instead, its speed increases by a fixed amount every hour. You must recognise that the distance covered in each hour forms an arithmetic series because distance in each hour equals speed in that hour for a duration of one hour. Given Data / Assumptions: Distance covered in the first hour = 35 km, so speed in first hour = 35 km/h. Speed increases by 2 km/h after every hour. Total duration of travel = 12 hours. We assume the change in speed happens at the end of each full hour. Concept / Approach: For each hour, distance = speed * time. Time each hour is 1 hour. Hence distances in successive hours are equal to the speeds: 35, 37, 39, and so on, forming an arithmetic progression with first term a = 35, common difference d = 2, and number of terms n = 12. Total distance is the sum of this arithmetic series: Sum = n / 2 * [2a + (n - 1) * d]. Step-by-Step Solution: Step 1: First hour speed a = 35 km/h. Step 2: Common difference d = 2 km/h. Step 3: Number of hours n = 12, so there are 12 terms in the series. Step 4: Use formula: Sum = n / 2 * [2a + (n - 1) * d]. Step 5: Sum = 12 / 2 * [2 * 35 + (12 - 1) * 2] = 6 * [70 + 22] = 6 * 92 = 552 km. Verification / Alternative check: We can manually list the speeds for a quick check: 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57. Pairing first and last terms: 35 + 57 = 92. There are 6 such pairs, so total distance = 6 * 92 = 552 km. This matches the formula based calculation and confirms the answer. Why Other Options Are Wrong: 456 km: Too low; it corresponds to using fewer terms or a smaller common difference. 482 km: Also lower; may come from miscounting number of hours or misusing the formula. 556 km: Slightly higher, indicating a numerical mistake when adding terms. Common Pitfalls: Assuming constant speed and multiplying 35 by 12, which ignores the increasing pattern. Using n instead of n - 1 inside the bracket of the arithmetic series formula. Forgetting that each term represents distance for exactly one hour of travel. Final Answer: The total distance travelled by the car in 12 hours is 552 km.

Question 17

A person crosses a street that is 600 metres long in 5 minutes. What is his walking speed in kilometres per hour?

Accepted Answer

Introduction / Context: This is a straightforward time and distance question that checks whether you can correctly convert units between metres per minute and kilometres per hour. Such conversions are very common in aptitude tests and form the foundation for more advanced time and distance problems. Given Data / Assumptions: Length of the street = 600 metres. Time taken to cross the street = 5 minutes. We must find the walking speed in km/h. The person walks at a constant speed. Concept / Approach: Speed is defined as distance divided by time. However, because the required unit is km/h, we need to convert 600 metres to kilometres and 5 minutes to hours. Basic conversions are: 1000 metres = 1 kilometre. 60 minutes = 1 hour. After converting, we compute speed = distance / time in the desired units. Step-by-Step Solution: Step 1: Distance = 600 metres = 600 / 1000 km = 0.6 km. Step 2: Time = 5 minutes = 5 / 60 hours = 1 / 12 hour. Step 3: Speed = distance / time = 0.6 / (1 / 12). Step 4: 0.6 / (1 / 12) = 0.6 * 12 = 7.2. Step 5: Therefore the walking speed is 7.2 km/h. Verification / Alternative check: Alternatively, calculate speed in metres per minute first. Distance is 600 metres in 5 minutes, so speed = 600 / 5 = 120 metres per minute. Now convert to km/h: 120 metres per minute * 60 minutes per hour = 7200 metres per hour, or 7.2 km/h. This matches the previous method and verifies the answer. Why Other Options Are Wrong: 3.6 km/h: This is half of 7.2 and may arise from forgetting a factor of 2 during conversion. 8.4 km/h: Slightly higher than the true value and likely results from an error in division. 10 km/h: Too high and does not correspond to the given distance and time. Common Pitfalls: Confusing metres with kilometres and minutes with hours. Using 5/100 instead of 5/60 when converting minutes to hours. Inverting the speed formula and accidentally computing time / distance. Final Answer: The walking speed of the person is 7.2 km/h.

Question 18

A policeman notices a thief at a distance of 200 metres and both start running in the same direction, the thief at 10 km/h and the policeman at 11 km/h. What will be the distance between them after 6 minutes?

Accepted Answer

Introduction / Context: This problem is a relative speed question involving a chase in one dimension. The policeman and the thief are moving in the same direction with different speeds, and we are asked to find how the gap between them changes after a certain time. Understanding relative speed is crucial for many time and distance and chase problems. Given Data / Assumptions: Initial distance between policeman and thief = 200 metres. Speed of thief = 10 km/h. Speed of policeman = 11 km/h. Both run in the same direction. We need the distance between them after 6 minutes. Concept / Approach: When two objects move in the same direction, their relative speed is the difference of their speeds. Since the policeman runs faster, he is closing the gap with the thief at a rate equal to (11 - 10) km/h = 1 km/h. We convert time into hours and then compute how much of the initial gap is closed in that time. The remaining distance is the new separation. Step-by-Step Solution: Step 1: Relative speed of policeman with respect to thief = 11 - 10 = 1 km/h. Step 2: Convert 6 minutes into hours: 6 minutes = 6 / 60 = 0.1 hour. Step 3: Distance by which the gap reduces in 0.1 hour = Relative speed * Time = 1 * 0.1 = 0.1 km. Step 4: Convert 0.1 km to metres: 0.1 * 1000 = 100 metres. Step 5: Initial gap = 200 metres, so distance between them after 6 minutes = 200 - 100 = 100 metres. Verification / Alternative check: We can find actual distances covered in 6 minutes. In 0.1 hour, thief covers 10 * 0.1 = 1 km = 1000 metres. Policeman covers 11 * 0.1 = 1.1 km = 1100 metres. The difference in distances covered in that time is 1100 - 1000 = 100 metres, which matches the computed remaining gap of 100 metres, confirming the answer. Why Other Options Are Wrong: 150 m: This would imply the policeman closed only 50 metres, which contradicts relative speed of 1 km/h over 0.1 hour. 190 m: Implies the gap reduced by just 10 metres, clearly incorrect. 200 m: Means no change in distance, which is impossible when speeds are different. Common Pitfalls: Adding the speeds instead of subtracting them for motion in the same direction. Forgetting to convert minutes into hours before using km/h. Interpreting 200 metres as 200 km and making unit errors. Final Answer: The distance between the policeman and the thief after 6 minutes is 100 m.

Question 19

Two trains A and B start simultaneously from stations P and Q in opposite directions and meet once. After meeting, train A reaches its destination in 16 hours and train B reaches its destination in 9 hours. If train A runs at 120 km/h, what is the s…

Accepted Answer

Introduction / Context: This time and distance problem involves two trains travelling towards each other and then continuing to their destinations after they meet. The times taken by each train to reach its destination after the meeting point, combined with one known speed, allow us to find the other speed. This is a standard relative motion question over a fixed distance between two stations. Given Data / Assumptions: Train A travels from P to Q at 120 km/h. Train B travels from Q to P at an unknown speed v km/h. They start at the same time and meet once. After meeting, train A takes 16 hours to reach its destination. After meeting, train B takes 9 hours to reach its destination. Concept / Approach: Let distance from P to meeting point be x km, and from meeting point to Q be y km. For train A: Time after meeting to reach Q = 16 hours, so y = 120 * 16. For train B: Time after meeting to reach P = 9 hours, so x = v * 9. Before meeting, times taken by both trains are equal. Distance covered by A before meeting is x = 120 * t, and by B is y = v * t. Thus we can equate x and y with these expressions and solve for v. Step-by-Step Solution: Step 1: From meeting to Q, train A travels y = 120 * 16 = 1920 km. Step 2: From meeting to P, train B travels x = v * 9 km. Step 3: Before meeting, train A travels x = 120 * t, so x = 120t. Step 4: Before meeting, train B travels y = v * t. Step 5: From Step 1, y = 1920, so 1920 = v * t. Also x = v * 9 = 120t. Step 6: From v * 9 = 120t, we get v = 120t / 9. Step 7: Substitute into 1920 = v * t to get 1920 = (120t / 9) * t = 120t^2 / 9. Step 8: Solve for t: t^2 = 1920 * 9 / 120 = 144, so t = 12 hours. Step 9: Then v = 120 * 12 / 9 = 160 km/h. Verification / Alternative check: We can verify consistency using distances. Before meeting, A covers 120 * 12 = 1440 km, and afterwards covers 1920 km, so total distance P to Q is 1440 + 1920 = 3360 km. Train B covers 160 * 12 = 1920 km before meeting and 160 * 9 = 1440 km after meeting, also totalling 3360 km. The distances match, confirming the result. Why Other Options Are Wrong: 90 km/h: Too low; it fails to satisfy the distance relations with the given times. 67.5 km/h: Even lower and clearly inconsistent with total distance calculations. None of these: Incorrect because 160 km/h is a valid option and satisfies all conditions. Common Pitfalls: Incorrectly assuming distances after meeting are equal instead of using time and speed. Forgetting that both trains travel for the same time before meeting. Algebraic errors when solving the simultaneous equations for t and v. Final Answer: The speed of the second train B is 160 km/h.

Question 20

Two horses start trotting towards each other from points A and B, which are 50 km apart. They meet after 1 hour. After meeting, the first horse reaches B exactly 5/6 hour before the second horse reaches A. What is the speed of the slower horse?

Accepted Answer

Introduction / Context: This is a relative speed problem involving two objects starting from opposite ends and then continuing to their destinations after they meet. The key idea is that the distances from the meeting point to each end are related to the distances travelled before the meeting, and the difference in the times taken after the meeting gives a relation between their speeds. Given Data / Assumptions: Distance between A and B = 50 km. The two horses meet after 1 hour of trotting towards each other. Horse 1 reaches B exactly 5/6 hour before Horse 2 reaches A. We must find the speed of the slower horse. Concept / Approach: Let speeds of horses be v1 and v2 km/h, with v1 > v2. They meet after 1 hour, so: v1 + v2 = 50. After meeting, the remaining distance for Horse 1 to B equals the distance Horse 2 travelled before meeting, which is v2. So time taken by Horse 1 after meeting is t1 = v2 / v1. The remaining distance for Horse 2 to A equals v1, so its time after meeting is t2 = v1 / v2. The given condition is: t2 - t1 = 5/6 hour. This gives a relation in terms of the ratio r = v1 / v2. Step-by-Step Solution: Step 1: v1 + v2 = 50 because they meet after 1 hour. Step 2: Let r = v1 / v2, so v1 = r * v2. Step 3: Time after meeting for Horse 1, t1 = v2 / v1 = 1 / r. Step 4: Time after meeting for Horse 2, t2 = v1 / v2 = r. Step 5: Given t2 - t1 = 5/6, so r - 1 / r = 5 / 6. Step 6: Multiply both sides by r: r^2 - 1 = (5 / 6) r, so 6r^2 - 6 = 5r. Step 7: Rearranging: 6r^2 - 5r - 6 = 0. Solving gives r = 3/2 (taking the positive root). Step 8: Thus v1 = (3/2) v2. Using v1 + v2 = 50: (3/2)v2 + v2 = 50, or (5/2)v2 = 50, so v2 = 20 km/h. Verification / Alternative check: With v2 = 20 km/h and v1 = 30 km/h, the total distance in 1 hour is 30 + 20 = 50 km, matching the separation. After the meeting, Horse 1 has remaining distance 20 km and takes 20 / 30 = 2/3 hour. Horse 2 has remaining distance 30 km and takes 30 / 20 = 3/2 hour. The difference is 3/2 - 2/3 = (9 - 4) / 6 = 5/6 hour, exactly as given. Why Other Options Are Wrong: 30 km/h: This is the speed of the faster horse, not the slower horse. 15 km/h: Too low; it does not satisfy v1 + v2 = 50 with a reasonable positive partner speed. 25 km/h: Leads to inconsistent meeting and time difference conditions. Common Pitfalls: Assuming the times after meeting are equal instead of using the given 5/6 hour difference. Mixing up which horse is faster and which is slower when interpreting the conditions. Not converting the ratio equation correctly when solving r - 1 / r = 5 / 6. Final Answer: The speed of the slower horse is 20 km/h.

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