Speed, Time and Distance
- 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?
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Required average speed = Total distance covered/Total time taken
= (3 x 39)/[39/26 + 39/39 +39/52]Correct Option: D
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
- 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?
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Relative speed = 21 - 15 = 6 m per minute.
Relative distance = 114 mCorrect Option: A
Relative speed = 21 - 15 = 6 m per minute.
Relative distance = 114 m
Time taken to catche = 144/6 min = 19 min
- 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. ?
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Let the certain distance be d and time t.
Now, by given condition,
d/3 = ( t + 15 ) min ( t +15 )/60 h and
d/4 = (t - 15) min = ( t - 15 )/60 hCorrect Option: B
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.
- 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 ?
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Let distance and original speed of the man be d km and s km/h.
Then,
d/s - d/s + 3 = 2/3 ....(i)
and d/(s - 2) - d/s = 2/3 ...(ii)Correct Option: D
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
- 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?
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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)Correct Option: C
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