Question 1

The speeds of two cars are in the ratio 7 : 8. If the second car covers 200 km in 5 hours (i.e., 40 km/h), find the speed of the first car (in km/h).

Accepted Answer

Introduction / Context: Speed ratios scale actual speeds by a common factor. Once you identify the common factor from one car, you can compute the other car’s speed directly. Given Data / Assumptions: Speed ratio (Car1 : Car2) = 7 : 8. Car2 speed = 200 / 5 = 40 km/h. Concept / Approach: If 8 parts correspond to 40 km/h, then 1 part is 40/8, and 7 parts (Car1) give the required speed. Step-by-Step Solution: Common factor = 40 / 8 = 5 km/h per part.Car1 speed = 7 × 5 = 35 km/h. Verification / Alternative check: Proportion: 35 : 40 simplifies to 7 : 8, matching the ratio. Why Other Options Are Wrong: 28, 25, 32, 40 km/h do not maintain the 7 : 8 proportion with a 40 km/h second car. Common Pitfalls: Confusing the mapping of “7” and “8” to the first/second car; not simplifying to get the correct scale factor. Final Answer: 35 km/h

Question 2

The speeds of two buses are in the ratio 5 : 3. If the first bus covers 400 km in 8 hours (50 km/h), find the speed of the second bus (in km/h).

Accepted Answer

Introduction / Context: Again, ratios scale real values. With the first bus speed known from distance/time, use the 5 : 3 ratio to compute the second bus speed. Given Data / Assumptions: Speed ratio (Bus1 : Bus2) = 5 : 3. Bus1 speed = 400/8 = 50 km/h. Concept / Approach: Each “part” equals 50/5 = 10 km/h. Thus, Bus2 speed is 3 parts = 30 km/h. Step-by-Step Solution: Part size = 50 / 5 = 10 km/h.Bus2 speed = 3 × 10 = 30 km/h. Verification / Alternative check: 50 : 30 reduces to 5 : 3—consistent with the ratio. Why Other Options Are Wrong: 15, 27, 37, 25 km/h do not maintain the correct proportionality with Bus1 = 50 km/h. Common Pitfalls: Applying the ratio inversely or mixing which bus corresponds to 5 or 3 parts. Final Answer: 30 km/h

Question 3

The speeds of three cars are in the ratio 2 : 3 : 5. For the same distance, what is the ratio of their times taken?

Accepted Answer

Introduction / Context: For equal distances, time is inversely proportional to speed. Given a speed ratio, the time ratio is the inverse of those numbers. Given Data / Assumptions: Speed ratio = 2 : 3 : 5. All travel the same distance. Concept / Approach: Time ∝ 1/speed, so time ratio = 1/2 : 1/3 : 1/5. Multiply by the LCM (30) to get whole numbers. Step-by-Step Solution: 1/2 : 1/3 : 1/5 → multiply by 30.Times = 15 : 10 : 6. Verification / Alternative check: Check intuition: the fastest car (5 parts) should have the smallest time (6 parts), which it does. Why Other Options Are Wrong: They place larger time for faster cars or mismatch the inverse relationship entirely. Common Pitfalls: Forgetting to invert ratios; accidentally copying the same 2:3:5 order for times. Final Answer: 15 : 10 : 6

Question 4

Nilu covers a certain one-way distance in 6 hours at her onward speed. While returning, her speed decreases by 2 km/h and she takes 9 hours to cover the same distance back. What was her return speed (in km/h)?

Accepted Answer

Introduction / Context: When the same distance is covered at two different speeds, distance equality lets us equate speed×time for each leg. The speed drop and the longer return time together determine the original and return speeds. Given Data / Assumptions: Onward time = 6 h at speed v. Return time = 9 h at speed v − 2. Distances are equal (out and back). Concept / Approach: Set v*6 = (v − 2)*9 and solve for v, then compute the return speed v − 2. Step-by-Step Solution: 6v = 9(v − 2) ⇒ 6v = 9v − 18 ⇒ 3v = 18 ⇒ v = 6 km/h.Return speed = v − 2 = 4 km/h. Verification / Alternative check: Distances: 6×6 = 36 km outward; 9×4 = 36 km return—consistent. Why Other Options Are Wrong: 2 or 5 km/h breaks the equality 6v = 9(v − 2); 7 and 6 km/h also fail the equation. Common Pitfalls: Using 2 km/h as the absolute return speed instead of the decrement; not ensuring both legs’ distances match. Final Answer: 4 km/h

Question 5

A takes 4 hours more than B to walk a fixed distance D. If A doubles her speed, she can complete D in 2 hours less than B. How much time (in hours) does B require to walk D at his usual speed?

Accepted Answer

Introduction / Context: This is a time comparison with speed changes. If A normally takes 4 hours longer than B, but halving A’s time (by doubling speed) makes A finish 2 hours faster than B, you can form equations in terms of the unknown times. Given Data / Assumptions: Let B’s time = t hours. A’s time = t + 4 hours. If A doubles speed, her time halves: (t + 4)/2 hours, which is 2 hours less than B’s time t. Concept / Approach: Equation: (t + 4)/2 = t − 2. Solve for t to get B’s time directly without needing distances or speeds explicitly. Step-by-Step Solution: (t + 4)/2 = t − 2t + 4 = 2t − 4 ⇒ 8 = tHence B’s time = 8 hours. Verification / Alternative check: A’s usual time = 12 h. If A doubles speed, A’s time becomes 6 h, which is indeed 2 hours less than B’s 8 h. Why Other Options Are Wrong: 4, 6, 7, 9 hours do not satisfy (t + 4)/2 = t − 2. Common Pitfalls: Forgetting that doubling speed halves time; setting up the inequality rather than the strict equation. Final Answer: 8 h

Question 6

Sohail reaches his office 5 minutes late when walking at 16 km/h, and 10 minutes early when walking at 20 km/h. Find the one-way distance from his home to his office (in km).

Accepted Answer

Introduction / Context: When departure time is fixed, early/late arrival relative to schedule gives two equations for the same distance at different speeds. The total early–late gap is 15 minutes (0.25 h) between the two scenarios. Given Data / Assumptions: D/16 = T + 5/60. D/20 = T − 10/60. D is the distance; T is the scheduled time (unknown). Concept / Approach: Subtract the equations to eliminate T: D/16 − D/20 = 15/60 = 1/4 hour. Solve for D. Step-by-Step Solution: D(1/16 − 1/20) = 1/4 ⇒ D( (5 − 4)/80 ) = 1/4D/80 = 1/4 ⇒ D = 20 km. Verification / Alternative check: Check times: 20/16 = 1.25 h (5 min late if T = 1.1667 h); 20/20 = 1.0 h (10 min early)—consistent. Why Other Options Are Wrong: 16, 18, 22, 24 km do not satisfy the 15-minute difference produced by 16 vs 20 km/h. Common Pitfalls: Adding instead of subtracting the equations; mixing minutes with hours. Final Answer: 20 km

Question 7

Walking at 3/4 of his usual speed, Sachin covers a fixed distance in 2 hours more than at his usual speed. Find the time (in hours) Sachin takes to cover the distance at his usual speed.

Accepted Answer

Introduction / Context: Time varies inversely with speed for a fixed distance. Reducing speed to 3/4 of usual increases time by a factor of 4/3. The difference between the stretched time and the original time is given as 2 hours. Given Data / Assumptions: Usual time = t hours. Reduced speed = 3/4 of usual ⇒ time becomes (4/3)t. Time increase = (4/3)t − t = 2 hours. Concept / Approach: Set (4/3)t − t = 2 and solve for t to find the usual-time value directly. Step-by-Step Solution: (4/3)t − t = (1/3)t = 2 ⇒ t = 6 hours. Verification / Alternative check: At reduced speed, time = (4/3)*6 = 8 hours, which is 2 hours more than 6 hours—matches the statement. Why Other Options Are Wrong: 4, 5, 7, 9 hours would not make the time difference exactly 2 hours under a 3/4 speed reduction. Common Pitfalls: Using 3/4 on time instead of on speed; forgetting that time scales as the reciprocal of speed. Final Answer: 6 h

Question 8

A car covers 200 km in 2 h 40 min, while a jeep covers the same 200 km in 2 h. What is the ratio of their speeds (car : jeep)?

Accepted Answer

Introduction / Context: For equal distances, speed is distance/time. Convert mixed hours to pure hours, compute both speeds, and simplify their ratio. Given Data / Assumptions: Car: 200 km in 2 h 40 min = 2 + 40/60 = 2.6667 h. Jeep: 200 km in 2 h. Concept / Approach: Compute speeds, then ratio (car : jeep) = (car speed) : (jeep speed) and reduce to simplest form. Step-by-Step Solution: Car speed = 200 / 2.6667 ≈ 75 km/h.Jeep speed = 200 / 2 = 100 km/h.Ratio = 75 : 100 = 3 : 4. Verification / Alternative check: Cross-multiply: 3×100 = 4×75 = 300—consistent. Why Other Options Are Wrong: Other ratios do not equal 75:100 when simplified. Common Pitfalls: Not converting 40 minutes to 2/3 hour; swapping the order (jeep : car) instead of (car : jeep). Final Answer: 3 : 4

Question 9

Amit walks at 4 km/h. Four hours later, Brijesh starts from the same point on a bicycle at 20 km/h. At what distance from the start will Brijesh catch Amit?

Accepted Answer

Introduction / Context: This is a pursuit (catch-up) problem on a straight path. The early head start distance equals the product of Amit’s speed and the lead time. The catch-up time then depends on relative speed. Given Data / Assumptions: Amit speed = 4 km/h; Brijesh speed = 20 km/h. Lead start = 4 hours for Amit. Concept / Approach: Head start distance = 4 km/h × 4 h = 16 km. Relative speed = 20 − 4 = 16 km/h. Time to catch after Brijesh starts = head start / relative speed = 1 hour. Position from start = Amit’s total time × 4 km/h (or Brijesh’s 1 hour × 20 km/h). Step-by-Step Solution: Head start = 16 km.Relative speed = 16 km/h.Catch-up time after Brijesh starts = 16/16 = 1 hour.Distance from start at meeting = 20 km (by Brijesh), equals 5 h × 4 km/h for Amit = 20 km. Verification / Alternative check: Both computations give 20 km, confirming the meeting point. Why Other Options Are Wrong: 15, 18, 24, 13 km do not satisfy head-start and relative-speed logic. Common Pitfalls: Forgetting to use relative speed; miscalculating the 4-hour head start distance. Final Answer: 20 km

Question 10

A train runs at 50 km/h when moving (no stoppage) but averages 40 km/h including stops. On average, how many minutes per hour does the train stop?

Accepted Answer

Introduction / Context: Overall average speed over a clock hour equals distance covered in that hour. If the running speed is higher, the train must have stopped for some fraction of that hour. Solve for the moving-time fraction, then convert the remainder to minutes. Given Data / Assumptions: Running speed = 50 km/h. Overall speed = 40 km/h across 1 clock-hour. Concept / Approach: Let moving time per hour be t hours. Then distance in one clock-hour = 50t. Given overall average 40 km/h, 50t = 40 ⇒ t = 0.8 h. Step-by-Step Solution: t = 40/50 = 0.8 h = 48 min moving.Stoppage per hour = 60 − 48 = 12 minutes. Verification / Alternative check: In 48 minutes at 50 km/h, distance is 40 km, matching the overall 40 km/h. Why Other Options Are Wrong: 6, 10, 14, 18 minutes correspond to incorrect moving fractions. Common Pitfalls: Confusing arithmetic mean of speeds with time-weighted average; not solving for t first. Final Answer: 12 min

Question 11

A bus runs at 54 km/h when moving (no stoppage) but averages 45 km/h including stops. On average, how many minutes per hour does the bus stop?

Accepted Answer

Introduction / Context: As with trains, the difference between running speed and overall average indicates stoppage time. Compute moving time t per hour, then convert the idle portion to minutes. Given Data / Assumptions: Running speed = 54 km/h. Overall average = 45 km/h across 1 clock-hour. Concept / Approach: Let moving time = t hours. Then 54t = 45 ⇒ t = 45/54 = 5/6 h = 50 min. Stoppage = 10 min per hour. Step-by-Step Solution: 54t = 45 ⇒ t = 5/6 h = 50 min.Stop time/hour = 60 − 50 = 10 min. Verification / Alternative check: Distance in 50 minutes at 54 km/h is 45 km; over 60 minutes this averages 45 km/h. Why Other Options Are Wrong: 4, 6, 8, 12 minutes correspond to wrong t values. Common Pitfalls: Arithmetic mean mistakes; not converting fractional hours to minutes properly. Final Answer: 10 min

Question 12

A bus covers a certain journey in 16 hours, traveling half the distance at 40 km/h and the other half at 60 km/h. Find the total length of the journey (in km).

Accepted Answer

Introduction / Context: When equal distances are traveled at two different speeds, average speed for the whole trip is the harmonic mean of the two speeds. Multiplying that average by total time gives the total distance. Given Data / Assumptions: Half the distance at 40 km/h; half at 60 km/h. Total time = 16 h. Concept / Approach: Average speed for equal-distance halves: V = 2 / (1/40 + 1/60) = 48 km/h. Then Distance = V × Time. Step-by-Step Solution: 1/40 + 1/60 = 5/120 = 1/24.V = 2 / (1/24) = 48 km/h.Distance = 48 × 16 = 768 km. Verification / Alternative check: Let total distance be 2x. Time = x/40 + x/60 = x/24. Given total time 16 = x/24 ⇒ x = 384 ⇒ distance = 768 km. Why Other Options Are Wrong: Other values arise from using the arithmetic mean of speeds or misworking the time equation. Common Pitfalls: Averaging speeds directly for equal distances; not distinguishing equal “distance” vs equal “time.” Final Answer: 768 km

Question 13

A man starts 20 minutes late but travels at 1.5 times his usual speed and reaches right on time. What is his usual travel time (in minutes) to reach the office?

Accepted Answer

Introduction / Context: If you start late but increase speed, you can still arrive on schedule. Doubling logic: time is inversely proportional to speed. Here, increasing speed to 1.5× scales time by 2/3. Given Data / Assumptions: Usual time = T hours. Faster time = (2/3)T hours (because speed is 1.5×). He starts 20 minutes = 1/3 hour late and still arrives on time. Concept / Approach: Because he arrives exactly on time, (2/3)T must equal T − 1/3. Solve for T, then convert to minutes if needed. Step-by-Step Solution: (2/3)T = T − 1/3 ⇒ (1/3)T = 1/3 ⇒ T = 1 hour. Verification / Alternative check: At usual speed he would need 60 min; starting 20 min late would cause a 20 min delay, but at 1.5× speed his travel time is 40 min, exactly compensating. Why Other Options Are Wrong: 30, 40, 50, 80 minutes fail (2/3)T = T − 20 min. Common Pitfalls: Using 1/1.5 = 3/2 incorrectly; mixing minutes/hours while solving. Final Answer: 1 h

Question 14

X and Y start walking toward each other at 8:00 am from points 17.5 km apart. Their speeds are 3 km/h and 4 km/h, respectively. At what time do they meet?

Accepted Answer

Introduction / Context: When two people move toward each other on a line, the closing speed is the sum of their speeds. Divide the initial separation by the closing speed to get the meeting time from the start. Given Data / Assumptions: Distance apart = 17.5 km. Speeds: 3 km/h and 4 km/h. Start time = 8:00 am. Concept / Approach: Relative (closing) speed = 3 + 4 = 7 km/h. Time to meet = distance / relative speed. Step-by-Step Solution: Time = 17.5 / 7 = 2.5 hours = 2 h 30 min.Meeting time = 8:00 am + 2 h 30 min = 10:30 am. Verification / Alternative check: Positions after 2.5 h: X covers 7.5 km, Y covers 10 km; total 17.5 km—consistent. Why Other Options Are Wrong: Evening times are impossible here; 11:30 am implies 3.5 hours instead of 2.5. Common Pitfalls: Using the difference of speeds instead of the sum for opposite directions. Final Answer: 10 : 30 am

Question 15

You travel three equal legs of 39 km each at 26 km/h, 39 km/h, and 52 km/h, respectively. What is your average speed for the entire journey?

Accepted Answer

Introduction / Context: For equal distances traveled at different speeds, the average speed is not the arithmetic mean. It is computed via total distance divided by total time, effectively a harmonic-mean aggregate across segments. Given Data / Assumptions: Segment distances: 39 km, 39 km, 39 km (equal). Speeds: 26, 39, and 52 km/h. Concept / Approach: Total distance = 117 km. Total time = 39/26 + 39/39 + 39/52 hours. Average speed = total distance / total time = 3 / (1/26 + 1/39 + 1/52) for equal d segments. Step-by-Step Solution: Times: 39/26 = 1.5 h; 39/39 = 1 h; 39/52 = 0.75 h.Total time = 1.5 + 1 + 0.75 = 3.25 h.Average speed = 117 / 3.25 = 36 km/h. Verification / Alternative check: Harmonic expression: V = 3 / (1/26 + 1/39 + 1/52) = 36 km/h. Why Other Options Are Wrong: 33.33 and 35 are near-misses; 39 is the arithmetic mean, not applicable for equal-distance segments. Common Pitfalls: Averaging speeds directly; forgetting equal-distance condition. Final Answer: 36 km/h

Question 16

Dalbir Singh, a policeman, is 114 m behind a thief. Dalbir runs 21 m per minute while the thief runs 15 m per minute. In how many minutes will Dalbir catch the thief?

Accepted Answer

Introduction / Context: This is a simple linear pursuit problem. The closing speed equals the difference of their speeds when both move in the same direction. Time to catch equals the gap divided by closing speed. Given Data / Assumptions: Initial gap = 114 m. Policeman speed = 21 m/min; thief speed = 15 m/min. Concept / Approach: Relative speed (closing) = 21 − 15 = 6 m/min. Time = gap / relative speed in minutes. Step-by-Step Solution: Time = 114 / 6 = 19 minutes. Verification / Alternative check: Distance each covers in 19 minutes: policeman 399 m, thief 285 m; difference equals 114 m—the initial gap—so they meet at that time. Why Other Options Are Wrong: 16, 18, 21, 23 minutes do not match the exact ratio 114 : 6. Common Pitfalls: Adding speeds instead of taking the difference; using seconds instead of minutes inadvertently. Final Answer: 19 min

Question 17

A person travels a fixed distance at 3 km/h and arrives 15 minutes late; at 4 km/h the person arrives 15 minutes early. What is the distance (in km)?

Accepted Answer

Introduction / Context: With the same departure time, different walking speeds give arrival times relative to a schedule. Setting up two equations in distance and scheduled time allows elimination of the schedule variable to solve for distance directly. Given Data / Assumptions: D/3 = T + 0.25 h (15 min late). D/4 = T − 0.25 h (15 min early). Concept / Approach: Subtract the equations: D/3 − D/4 = 0.5 h. Solve for D using the difference of reciprocals (1/3 − 1/4) = 1/12. Step-by-Step Solution: D(1/12) = 0.5 ⇒ D = 6 km. Verification / Alternative check: Check with T = 1.25 h: at 3 km/h time = 2 h (0.75 h late); adjust T properly (e.g., T = 1.25 h gives 6/3 = 2 h = 45 min late). Alternatively, use the symmetric 30-minute spread: the 0.5 h difference matches (1/3 − 1/4)×D only when D = 6. Why Other Options Are Wrong: 4.5, 5, 7.2, 12 km do not satisfy D(1/3 − 1/4) = 0.5. Common Pitfalls: Adding the two equations or mixing up early and late signs; forgetting to convert minutes to hours. Final Answer: 6 km

Question 18

A person covers a distance on a scooter. If he had gone 3 km/h faster, he would have taken 40 minutes less; if he had gone 2 km/h slower, he would have taken 40 minutes more. What is the distance (in km)?

Accepted Answer

Introduction / Context: Two what-if changes to speed produce symmetric changes in time around the actual (unknown) scenario. Setting up time equations for the same distance at (v + 3), v, and (v − 2) allows solving for both v and distance D. Given Data / Assumptions: Let actual speed = v km/h; distance = D km. D/(v + 3) = D/v − 2/3 h. D/(v − 2) = D/v + 2/3 h. Concept / Approach: Use the pair of equations to eliminate D or v. Solving simultaneously yields concrete values that satisfy both constraints. Step-by-Step Solution: From the system, the consistent solution is v = 12 km/h and D = 40 km.Check faster case: time at 12 vs 15 km/h differs by (D/12 − D/15) = 40(1/12 − 1/15) = 40(1/60) = 2/3 h = 40 min.Check slower case: time at 10 vs 12 km/h differs by 40(1/10 − 1/12) = 40(1/60) = 2/3 h = 40 min. Verification / Alternative check: Both constraints are satisfied exactly for D = 40 km, confirming the solution. Why Other Options Are Wrong: 36, 37.5, 42.5, 45 km fail at least one of the two 40-minute differences. Common Pitfalls: Assuming linearity in speed vs time change; not expressing both scenarios with the same D. Final Answer: 40

Question 19

A man cycles from his house to office at 10 km/h and reaches 6 minutes late. At 12 km/h, he reaches 6 minutes early. How far is his office from his house (in km)?

Accepted Answer

Introduction / Context: Two speeds around a schedule produce symmetric early/late differences. As with similar problems, build two equations in distance and scheduled time, subtract to eliminate the schedule, and solve for distance directly. Given Data / Assumptions: D/10 = T + 0.1 h (6 minutes late). D/12 = T − 0.1 h (6 minutes early). Concept / Approach: Subtract equations: D/10 − D/12 = 0.2 h (12 minutes total spread). Use 1/10 − 1/12 = 1/60 to compute D. Step-by-Step Solution: D(1/60) = 0.2 ⇒ D = 12 km. Verification / Alternative check: At 10 km/h, time = 1.2 h = 72 min; if schedule is 66 min, that is 6 min late. At 12 km/h, time = 60 min, which is 6 min early. Both conditions match. Why Other Options Are Wrong: 6, 7, 9, 16 km do not satisfy the 12-minute spread with speeds 10 and 12 km/h. Common Pitfalls: Using arithmetic mean of speeds; forgetting to convert 6 minutes to 0.1 hour. Final Answer: 12 km

Question 20

A walks at a uniform speed of 2 km/h. Two hours later, B starts cycling behind A at 5 km/h on the same straight route. At what distance from the starting point will B catch A?

Accepted Answer

Introduction / Context: This is a classic pursuit problem on a straight line with a head start. We compare a walker (2 km/h) who starts first with a cyclist (5 km/h) who starts later. The key is to use the initial lead and relative speed to find the catch-up time and then the meeting position from the start. Given Data / Assumptions: A's speed = 2 km/h (walking). B's speed = 5 km/h (cycling). B starts 2 hours after A. Straight route; speeds are constant. Concept / Approach: When one chases another along the same line, the catch-up time equals head start distance divided by relative speed. Meeting distance from the start equals the leader's speed multiplied by total time the leader has been traveling at meeting. Step-by-Step Solution: A's head start distance = 2 km/h * 2 h = 4 km.Relative speed (B w.r.t. A) = 5 − 2 = 3 km/h.Catch-up time after B starts = 4 / 3 h = 1 h 20 min.Total time A has moved at meet = 2 + 4/3 = 10/3 h.Meeting distance from start = 2 * (10/3) = 20/3 km = 6 2/3 km. Verification / Alternative check: Position of B at meet: B travels (4/3) h at 5 km/h = 20/3 km, same as A's position, confirming the meeting point. Why Other Options Are Wrong: 6 1/3 km, 6 3/7 km, 6 5/7 km come from incorrect relative-speed or head-start handling. Common Pitfalls: Using total time 4/3 h to compute distance for A without adding A's initial 2-hour lead time. Final Answer: 6 2/3 km

Time and Distance Questions