Question 1

A man travelled a distance of 80 km in 7 hours partly on foot at the rate of 8 km per hour and partly on bicycle at 16 km /hr. Find the distance travelled on foot?

Accepted Answer

Let the distance travelled on foot be x km Then the distance travelled on bicycle = 80 - x km Speed on foot = 8 km/h ⇒ Time on foot = x / 8 hours Speed on bicycle = 16 km/h ⇒ Time on bicycle = (80 - x) / 16 hours Total time = 7 hours x / 8 + (80 - x) / 16 = 7 Step 1: Multiply through by LCM (16) to eliminate denominators: 
2x + (80 - x) = 112
=> 2x + 80 - x = 112
=> x + 80 = 112
=> x = 32
 Answer: 32 km was travelled on foot.

Question 2

The ratio of speeds of two trains is 7 : 8. If the second train covers 400 km in 4 hours, find the speed of the first train.

Accepted Answer

Given data Speed ratio (first : second) = 7 : 8 Second train distance in 4 hours = 400 km ⇒ second train speed = 400 / 4 = 100 km/h Concept / Approach In a ratio 7 : 8, let one part be k. Then first speed = 7k, second speed = 8k. Use the known second speed to find k. Step-by-step calculation Second speed = 8k = 100 ⇒ k = 100 / 8 = 12.5First speed = 7k = 7 × 12.5 = 87.5 km/h Verification / Alternative Check the ratio: 87.5 : 100 = 0.875 : 1 = 7 : 8 ✔ Common pitfalls Using 7/8 of the distance instead of the speed ratio. Forgetting to compute second speed first from the given 400 km in 4 h. Final Answer 87.5 km/h

Question 3

A man completes a journey in 10 hours. He travels the first half at 21 km/h and the second half at 24 km/h. Find the total distance (km).

Accepted Answer

Given data Total time = 10 hours First half speed = 21 km/h Second half speed = 24 km/h Concept / Approach Let total distance be D km. Each half is D/2. Total time = time for first half + time for second half. Step-by-step calculation Time = (D/2)/21 + (D/2)/24= D × (1/42 + 1/48)= D × ( (48 + 42) / (42 × 48) ) = D × (90 / 2016)Simplify 90/2016 → divide by 6 ⇒ 15/336; divide by 3 ⇒ 5/112So, Total time = D × (5/112) = 10D = 10 × (112/5) = 224 km Verification / Alternative First half time = (112 / 21) = 5 h; second half time = (112 / 24) ≈ 4.6667 h; sum ≈ 9.6667 h? Wait—check halves: Each half = 112 km; times 112/21 ≈ 5.3333 h and 112/24 ≈ 4.6667 h; total = 10 h ✔ Common pitfalls Averaging speeds arithmetically; here halves are equal in distance, not time. Forgetting to split the distance equally and add times. Final Answer 224 km

Question 4

Walking at 14 km/h instead of 10 km/h, a person would have covered 20 km more in the same time. What is the actual distance he traveled?

Accepted Answer

Given data Faster speed = 14 km/h Actual speed = 10 km/h Extra distance at faster speed over the same time = 20 km Concept / Approach Let the common time be T hours. Distance difference = (14 − 10) × T = 4T. Use 4T = 20 to find T, then actual distance = 10 × T. Step-by-step calculation 4T = 20 ⇒ T = 5 hoursActual distance = 10 × 5 = 50 km Verification / Alternative At 14 km/h for 5 h ⇒ 70 km; at 10 km/h for 5 h ⇒ 50 km; difference = 20 km ✔ Common pitfalls Assuming time changed; the statement fixes time and varies speed. Confusing distance difference with speed ratio. Final Answer 50 km

Question 5

In a 600 km flight, bad weather reduces the average speed by 200 km/h and increases the time by 30 minutes. What is the original duration of the flight?

Accepted Answer

Let the planned (normal) duration be t hours for a 600 km flight. Then planned speed = 600 / t km/h. Due to bad weather, average speed is reduced by 200 km/h and the flight takes 30 minutes (0.5 h) longer. Concept / Approach We relate speed and time before and after delay using Speed = Distance / Time for the same distance (600 km), and then solve for t. Step-by-step calculation Planned speed = 600 / tReduced speed = 600 / (t + 0.5)Given: (600 / t) − 200 = 600 / (t + 0.5)Solve for t:600 / t − 600 / (t + 0.5) = 200600[(t + 0.5) − t] / [t(t + 0.5)] = 200600(0.5) / [t(t + 0.5)] = 200300 / [t(t + 0.5)] = 200t(t + 0.5) = 300 / 200 = 1.5t2 + 0.5t − 1.5 = 0The positive root is t = 1. Verification Planned speed = 600 / 1 = 600 km/hReduced speed = 600 / 1.5 = 400 km/hDrop in speed = 600 − 400 = 200 km/h (matches)Time increase = 1.5 − 1 = 0.5 h = 30 min (matches) Common pitfalls Confusing the “duration of the flight” with the delayed duration. The question asks for the original duration, not the delayed 1.5 hours. Final Answer 1 hour

Question 6

A car travels at 5⁄7 of its actual speed and covers 42 km in 1 hr 40 min 48 sec. What is the car's actual speed (in km/h)?

Accepted Answer

When a car runs at a fraction of its actual speed, the time taken scales inversely. Here the car runs at (5⁄7) of its actual speed and takes 1 hr 40 min 48 sec to cover 42 km. Given data Distance = 42 km Reduced speed = (5⁄7) × (actual speed v) Time taken at reduced speed = 1 h 40 m 48 s = 1.68 h Step-by-step calculation Time = Distance / Speed1.68 = 42 / [(5⁄7) v] = 42 × 7 / (5v)1.68 = 294 / (5v)5v = 294 / 1.68 = 175v = 35 km/h Verification Reduced speed = (5⁄7) × 35 = 25 km/hTime = 42 / 25 = 1.68 h = 1 h 40 m 48 s (matches) Common pitfalls Do not forget to convert 40 min 48 s entirely into hours before substituting. Mixing hours and minutes directly leads to errors. Final Answer 35 km/h

Question 7

A train travels 50% faster than a car. Both start from A and reach B (75 km) at the same time, but the train loses 12.5 minutes at stations. What is the car's speed (km/h)?

Accepted Answer

The train is 50% faster than the car, but it has stoppage time. Equal arrival times imply the car’s running time equals the train’s running time plus stoppage. Given data Distance AB = 75 km Train speed = 1.5 × car speed = 1.5c Stoppage time (train) = 12.5 min = 25/120 h Arrival times equal Step-by-step calculation Car time = 75 / cTrain running time = 75 / (1.5c)Car time = Train running time + stoppage75 / c = 75 / (1.5c) + 25/12075 / c − 50 / c = 25/12025 / c = 25/120c = 120 km/h Verification Car time = 75 / 120 = 0.625 hTrain run time = 75 / 180 = 0.4167 h; add 0.2083 h stoppage = 0.625 h (matches) Common pitfalls Don’t forget to convert 12.5 minutes to hours, and don’t set distances unequal—the distance is the same for both. Final Answer 120 km/h

Question 8

An aeroplane covers a distance at 240 km/h in 5 hours. To cover the same distance in 1⅔ hours, what speed is required?

Accepted Answer

First compute the distance, then divide by the new time. Step-by-step calculation Distance d = 240 × 5 = 1200 kmRequired time = 1⅔ h = 5/3 hRequired speed = d / time = 1200 / (5/3) = 1200 × 3 / 5 = 720 km/h Verification At 720 km/h for 5/3 h: 720 × (5/3) = 1200 km (same distance) Final Answer 720 km/h

Question 9

A farmer covers 61 km in 9 hours, partly on foot at 4 km/h and partly by bicycle at 9 km/h. What distance was covered on foot?

Accepted Answer

Let x be the distance (km) covered on foot at 4 km/h. Then (61 − x) km is by bicycle at 9 km/h. Total time is 9 hours. Step-by-step calculation Time on foot = x / 4Time on bicycle = (61 − x) / 9Total time = x / 4 + (61 − x) / 9 = 9Multiply by 36: 9x + 4(61 − x) = 3249x + 244 − 4x = 3245x = 80x = 16 km Verification Time on foot = 16 / 4 = 4 hTime on bicycle = 45 / 9 = 5 hTotal = 9 h (matches) Common pitfalls Avoid mixing distances with speeds; write times explicitly before summing. Final Answer 16 km

Question 10

A person crosses a 600 m long street in 5 minutes. What is the speed in km/h?

Accepted Answer

Convert distance to km and time to hours, then use Speed = Distance / Time. Step-by-step calculation Distance = 600 m = 0.6 kmTime = 5 minutes = 5/60 h = 1/12 hSpeed = 0.6 / (1/12) = 0.6 × 12 = 7.2 km/h Final Answer 7.2 km/h

Question 11

A man travels 160 km at 64 km/h and next 160 km at 80 km/h. What is the average speed over the first 320 km?

Accepted Answer

Average speed over multiple legs is total distance divided by total time, not the simple average of speeds. Step-by-step calculation Total distance = 160 + 160 = 320 kmTime1 = 160 / 64 = 2.5 hTime2 = 160 / 80 = 2 hTotal time = 2.5 + 2 = 4.5 hAverage speed = 320 / 4.5 = 71.11 km/h (approx.) Verification / Alternative For two equal distances d at speeds v1, v2: Harmonic mean applies: Average speed = (2d) / (d/v1 + d/v2) which is the same calculation above. Common pitfalls Do not average 64 and 80 directly; that would ignore time spent at each speed. Final Answer 71.11 km/h

Question 12

A bus runs at 54 km/h excluding stoppages and 45 km/h including stoppages. How many minutes does it stop per hour?

Accepted Answer

Including stoppages reduces the effective hourly distance. The ratio of effective to running speed gives the fraction of time spent moving. Step-by-step calculation Running speed (no stops) = 54 km/h ⇒ covers 54 km in 1 h while movingEffective speed (with stops) = 45 km/h ⇒ covers 45 km in the whole hourFraction of hour spent moving = 45 / 54 = 5/6Stoppage fraction = 1 − 5/6 = 1/6 of an hourStoppage time per hour = (1/6) × 60 = 10 minutes Final Answer 10 minutes

Question 13

Abhay takes 2 hours more than Sameer to cover 30 km. If Abhay doubles his speed, he would take 1 hour less than Sameer. What is Abhay's speed (km/h)?

Accepted Answer

Translate the time statements into equations using Time = Distance / Speed. Step-by-step calculation Let Abhay’s speed = a and Sameer’s speed = s (km/h). 30 / a = 30 / s + 2 …(1)30 / (2a) = 30 / s − 1 …(2)Subtract (2) from (1): 30/a − 30/(2a) = (30/s + 2) − (30/s − 1)15/a = 3a = 5 km/h Verification Abhay time = 30 / 5 = 6 hFrom (1): 6 = 30/s + 2 ⇒ 30/s = 4 ⇒ s = 7.5 km/hIf Abhay doubles speed to 10 km/h: time = 30/10 = 3 h, which is 1 h less than Sameer’s 4 h (matches) Common pitfalls Be careful with “2 hours more” vs “1 hour less” and ensure you attach them to the correct person’s time. Final Answer 5 km/h

Question 14

A 600 km journey takes 8 hours if 120 km is by train and the rest by car. It takes 20 minutes more if 200 km is by train and the rest by car. What is the ratio of train speed to car speed?

Accepted Answer

Let train speed = vt and car speed = vc. Write total time as a sum of segment times. Step-by-step calculation Case 1: 120/vt + 480/vc = 8 …(i)Case 2: 200/vt + 400/vc = 8 + 20/60 = 25/3 …(ii)Solve (i) and (ii): Multiply (i) by 5 and (ii) by 6 to eliminate denominators conveniently: (i) → 600/vt + 2400/vc = 40(ii) → 1200/vt + 2400/vc = 50Subtract the first from the second: 600/vt = 10 ⇒ vt = 60 km/hBack-substitute in (i): 120/60 + 480/vc = 8 ⇒ 2 + 480/vc = 8 ⇒ 480/vc = 6 ⇒ vc = 80 km/h Final Answer 3:4 (train : car)

Question 15

A man travels an unknown distance at speed v km/h. If he had moved 3 km/h faster, his time would decrease by 40 minutes; if he had moved 2 km/h slower, his time would increase by 40 minutes. What is the distance (in km)?

Accepted Answer

Problem restatement The same journey is done at three hypothetical speeds: v, v + 3, and v − 2 km/h. The time differences are each 40 minutes (2⁄3 hour). Find the distance. Given data Let distance = D km; normal speed = v km/h. At (v + 3) km/h, time is 2⁄3 hour less. At (v − 2) km/h, time is 2⁄3 hour more. Concept/Approach Use the identity Time = Distance ÷ Speed for the same D. Set up two equations from the given time differences and solve for v, then compute D. Step-by-step calculation D ÷ v − D ÷ (v + 3) = 2⁄3D ÷ (v − 2) − D ÷ v = 2⁄3From the first: D × [1÷v − 1÷(v + 3)] = 2⁄3 ⇒ D × 3 ÷ [v(v + 3)] = 2⁄3 ⇒ D = [2v(v + 3)] ÷ 9From the second: D × [1÷(v − 2) − 1÷v] = 2⁄3 ⇒ D × 2 ÷ [v(v − 2)] = 2⁄3 ⇒ D = v(v − 2) ÷ 3Equate D: [2v(v + 3)] ÷ 9 = v(v − 2) ÷ 3 ⇒ 2(v + 3) = 3(v − 2) ⇒ v = 12 km/hDistance D = v(v − 2) ÷ 3 = 12 × 10 ÷ 3 = 40 km Verification/Alternative At v = 12: time = 40 ÷ 12 = 3ℓ⁄3 hAt v + 3 = 15: time = 40 ÷ 15 = 2ℓ⁄3 h ⇒ 40 min lessAt v − 2 = 10: time = 40 ÷ 10 = 4 h ⇒ 40 min more Common pitfalls Avoid confusing 40 minutes with 0.4 hour; use 2⁄3 hour. Final Answer 40 km

Question 16

Robert plans to reach point A at 2 PM. If he rides at 10 km/h he reaches at 2 PM; if he rides at 15 km/h he reaches at 12 noon. At what constant speed must he ride to reach at 1 PM?

Accepted Answer

Problem restatement For the same distance, 10 km/h arrival is 2 PM, 15 km/h arrival is 12 noon (2 hours earlier). Find the speed to arrive 1 hour earlier than 2 PM, i.e., at 1 PM. Given data Distance = D km (constant). t10 = time at 10 km/h; t15 = time at 15 km/h. t10 − t15 = 2 h. Concept/Approach Use D = v × t and relate the two times. Then compute D and the time needed for 1 PM, finally the required speed. Step-by-step calculation D = 10t10 = 15t15 ⇒ t10 = 1.5t15t10 − t15 = 2 ⇒ 1.5t15 − t15 = 2 ⇒ 0.5t15 = 2 ⇒ t15 = 4 ht10 = 6 h, so D = 10 × 6 = 60 kmTo reach at 1 PM (1 hour earlier than 2 PM): time required = 6 − 1 = 5 hSpeed needed = D ÷ time = 60 ÷ 5 = 12 km/h Verification/Alternative Check endpoints: at 10 km/h → 6 h; at 15 km/h → 4 h (2 hours saved). Our target time 5 h is between them, giving 12 km/h. Common pitfalls Do not mix calendar clock times with durations; compute durations first. Final Answer 12 km/h

Question 17

A and B are 20 km apart. A can walk at an average speed of 4 km/hr and B at 6 km/h. If they start walking towards each other at 7:000 am, when they will meet?

Accepted Answer

Suppose they will meet after T hours. Distance = Speed x Time Sum of distance traveled by them after T hours 6T + 4T = 20 km T = 2 hours. So they will meet at 7:00 AM + 2 hours = 9:00 AM

Question 18

Arjun runs a distance of 1500 m in 5 minutes. What is his speed in Km/hr?

Accepted Answer

Speed = (1500/300 ) m/sec = 5 m/sec = (5 x 18/5 ) km/hr = 18 km/hr

Question 19

In how many minutes will Rohit cover a distance of 800 m, if he runs at speed of 10 km/hr?

Accepted Answer

Rohit's speed = 10 Km/hr = (10 x 5/18) m/sec = 50/18 m/sec Time taken to cover 800 m = (800 / (50/18) ) sec = (800 x 18/50) sec = 288 sec = 4 min 48 sec

Question 20

A plane flies along the four sides of a square field at a speed of 200, 400, 600 and 800 km/hr. Then find the average speed of plane around the square field.

Accepted Answer

Let the side of the square field be x and the average speed of plane be y x/200 + x/400 + x/600 + x/800 = 4x/y ⇒ 25x/2400 = 4x/y ⇒ y =384 ∴ Average speed is 384 km/hr

Time and Distance Questions