Time and Distance problems
- 1. A bus travels at the rate of 54 km/h without stoppages and it travels at 45 km/h with stoppages. How many minutes does the bus stop on an average per hour?
Options- A. 8 min
- B. 10 min
- C. 12 min
- D. 4 min Discuss
Correct Answer: 10 min
Explanation:
Due to stoppages, bus covers 9 km less per hour.
Time taken to cover 9 km =(9/54) x 60 = 10 min
Hence, the train stops on an average 10 min per hour.
- 2. A bus covers a certain distance in 16 h. It covers half the distance at 40 km/h and the rest at 60 km/h. Find the length of the journey.?
Options- A. 520 km
- B. 448 km
- C. 384 km
- D. 768 km Discuss
Correct Answer: 768 km
Explanation:
Let the total distance covered = L = Length of journey
According to the question,
(L/2 x 1/40) + (L/2 x 1/60) = 16
? L/80 + L/120 = 16
? (3L + 2L)/240 = 16
? 5L = 16 x 240
? L = (16 x 240)/5 = 16 x 48 = 768 km
- 3. A man started 20 min late and travelling at a speed of 1 1/ 2 times of his usual speed reaches his office in time. The time taken by the man to reach his office at his usual speed is?
Options- A. 40 min
- B. 1 h 20 min
- C. 1 h
- D. 30 min Discuss
Correct Answer: 1 h
Explanation:
Let usual speed of a man is V and time is t h.
By the formula, V1t1 = V2t2
Vt = 11/2 V(t - 20/60)
? t = 3/2(t - 1/3)
? 3/2 t - t = 1/2
? t = 1 h
Hence, the time taken by the man to reach his office at his usual speed is 1 h.
- 4. X and Y start walking toward each other at 8 : 00 am at the speed of 3 km/h and 4 km/h. respectively. They were initially 17.5 km apart, at what time do they meet?
Options- A. 10 : 30 am
- B. 10 : 30 pm
- C. 11 : 30 am
- D. 11 : 30 pm Discuss
Correct Answer: 10 : 30 am
Explanation:
Let they meet after T h.
then, according to the question, 3T + 4T = 17.5
? 7T = 17.5
? T = 17.5/7 = 2.5 h
So, they meet 2.5 h after 8 : 00 am.
It means they meet at 10 : 30 am.
- 5. If you travel 39 km at a speed of 26 km/h, another 39 km at a speed of 39 km/h and again 39 km at a speed of 52 km/h, what is your average speed for the entire journey?
Options- A. 39 km/h
- B. 37 km/h
- C. 33.33 km/h
- D. 36 km/h Discuss
Correct Answer: 36 km/h
Explanation:
Required average speed = Total distance covered/Total time taken
= (3 x 39)/[39/26 + 39/39 +39/52]
= 3 x 13/(1/2 + 1/3 + 1/4)
= 3 x 13 x 12/13 = 36 km/h
- 6. Dalbir Singh, a polieman, is 114 m behind a thief. Dalbir Singh runs 21 m and the thief 15 m in a minute. In what time will Dalbir singh catch the thief?
Options- A. 19 min
- B. 16 min
- C. 21 min
- D. 23 min Discuss
Correct Answer: 19 min
Explanation:
Relative speed = 21 - 15 = 6 m per minute.
Relative distance = 114 m
Time taken to catche = 144/6 min = 19 min
- 7. A person travels a certain distance at 3 km/h and reaches 15 min late. If he travels at 4 km/h, he reaches 15 min earlier. The distance he has to travel is.?
Options- A. 4.5 km
- B. 6 km
- C. 7.2 km
- D. 12 km Discuss
Correct Answer: 6 km
Explanation:
Let the certain distance be d and time t.
Now, by given condition,
d/3 = ( t + 15 ) min ( t +15 )/60 h
? 20d = t + 15
? t = 20d - 15 ......(i)
And from another condition,
d/4 = (t - 15) min = ( t - 15 )/60 h
? 15d = t - 15
? t = 15d + 15 ...(ii)
From Eqs. (i) and (ii), we get
20d - 15 = 15d + 15
? 5d = 30
? d = 6 km.
- 8. A man covers a certain distance on scooter. Had he moved 3 km/h faster, he would have taken 40 min less. If he had moved 2 km/ slower, he would have taken 40 min more. The distance (in
km) is?
Options- A. 42.5
- B. 36
- C. 37.5
- D. 40 Discuss
Correct Answer: 40
Explanation:
Let distance and original speed of the man be d km and s km/h.
Then,
d/s - d/s + 3 = 2/3
? [d(s + 3 - s)]/[s(s + 3)] = 2/3
? 9d = 2s(s + 3) ...(i)
And d/(s - 2) - d/s = 2/3
? [d(s - s + 2)]/[s(s - 2)] = 2/3
? 3d = s(s - 2) ...(ii)
From Eqs.(i) and (ii), we get
3s(s - 2) = 2s(s + 3)
? 3s2 - 6s = 2s2 + 6s
? s2 = 12s
? s = 12
From Eq.(ii), we get
3d = 12(12 - 2)
? d = 40 km
- 9. A man riding a bicycle form his house at 10km/h and reaches his office late by 6 min. He increases his speed by 2km/h and reaches 6 min before. How far is the office form his house?
Options- A. 6 km
- B. 7 km
- C. 12 km
- D. 16 km Discuss
Correct Answer: 12 km
Explanation:
Here, a = 10 km/h
b = ( 10 + 2 ) = 12 km/h,
t1 = 6 min = 1/10 h
and t2= 1/10 h.
? Required distance = ab( t1 + t2 )/(b - a)
= 10 x 12 x ( 1/10 + 1/10 ) /(12 - 10) = (10 x 12 x 1)/(5 x 2) = 12km
- 10. A walks at a uniform rate of 2 km/h and 2 h after his start, B cycles after him at the uniform rate of 5 km/h. How far from the starting point will B catch A?
Options- A. 62/3 km
- B. 61/3 km
- C. 65/7 km
- D. 63/7 km Discuss
Correct Answer: 62/3 km
Explanation:
Let B catches A after T h.
Then, distance travelled by A in (T + 2) h = Distance travelled by B in T h
According to the question,
2 (T + 2 ) = 5T
? 2T + 4 = 5T
? 3T = 4
? T = 4/3 h
Distance travelled by B in 4/3 h
= (4/3) x 5 = 20/3 = 62/3 km
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