Speed, Time and Distance
- A man walks a distance of 35 kms. He walks for some time at 4 km per hour and for some time at 5 km per hr. If he walks at 5 km per hr. instead of 4 km per hr. and 4 km per hr. instead of 5 km per hr, he will walk 2 kms more in the same span of time. Find his total time of total journey.
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Let the man walks for x hours at 4 km per hr. and y hours at 5 km per hr. and covers a distance of 35 kms.
∴ Distance = 4x + 5y = 35 ...(i)
Now, he walks at 5 km per hr.
for x hours and at 4 km per hr.
for y hours and covers a distance (35 + 2) = 37 kms
∴ Distance = 5x + 4y = 37...(ii)
By 5 × (i) – 4 × (ii) we have20x + 25y = 175 20x + 16y = 148 - - - ____________ 9y = 27
⇒ y = 3
Putting the value of (y) in equation (i), we have
4x + 5 × 3 = 35
⇒ 4x = 35 – 15 = 20
⇒ x = 5
∴ Total time taken
= x + y = 5 + 3 = 8 hours.Correct Option: C
Let the man walks for x hours at 4 km per hr. and y hours at 5 km per hr. and covers a distance of 35 kms.
∴ Distance = 4x + 5y = 35 ...(i)
Now, he walks at 5 km per hr.
for x hours and at 4 km per hr.
for y hours and covers a distance (35 + 2) = 37 kms
∴ Distance = 5x + 4y = 37...(ii)
By 5 × (i) – 4 × (ii) we have20x + 25y = 175 20x + 16y = 148 - - - ____________ 9y = 27
⇒ y = 3
Putting the value of (y) in equation (i), we have
4x + 5 × 3 = 35
⇒ 4x = 35 – 15 = 20
⇒ x = 5
∴ Total time taken
= x + y = 5 + 3 = 8 hours.
- Ram travelled one-third of a journey with a speed of 10 km per hr, the next one-third with a speed of 9 km per hr. and the rest at a speed of 8 km per hr. If he had travelled half the journey at speed of 10 km per hr. and the other half with a speed of 8 km per hr, he would have been 1 minute longer on the way. What distance did he travel?
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Let the total distance travelled be x kms.
Case I :
Speed for the first one-third distancei.e. x kms =10 km per hr. 3 ∴ Time taken = x hours 30
Similarly, time taken for the next one-third distance= x hours 27
and time taken for the last onethird distance= x hours. 24
∴ Total time taken to cover x kms.= x + x + x hours. 30 27 24
Case II :
Time taken for one-half distance at the speed of 10 km per hr.= x hrs. 20
and time taken for remaining 1/2 of distance= x hrs. at 8 km per hr. 16 Total time taken = x + x hrs. 20 16
Time taken in (Case II – Case I)= 1 minute = 1 hr. 60
∴ According to the questionx + x − x + x + x 20 16 30 27 24 = 1 60 ⇒ 108x + 135x − 72x − 80x − 90x 2160 = 1 60 ⇒ 243x − 242x = 1 2160 60 ⇒ x = 1 2160 60 ⇒ x = 2160 = 36 km. 60
Hence the required distance
= 36 km.Correct Option: A
Let the total distance travelled be x kms.
Case I :
Speed for the first one-third distancei.e. x kms =10 km per hr. 3 ∴ Time taken = x hours 30
Similarly, time taken for the next one-third distance= x hours 27
and time taken for the last onethird distance= x hours. 24
∴ Total time taken to cover x kms.= x + x + x hours. 30 27 24
Case II :
Time taken for one-half distance at the speed of 10 km per hr.= x hrs. 20
and time taken for remaining 1/2 of distance= x hrs. at 8 km per hr. 16 Total time taken = x + x hrs. 20 16
Time taken in (Case II – Case I)= 1 minute = 1 hr. 60
∴ According to the questionx + x − x + x + x 20 16 30 27 24 = 1 60 ⇒ 108x + 135x − 72x − 80x − 90x 2160 = 1 60 ⇒ 243x − 242x = 1 2160 60 ⇒ x = 1 2160 60 ⇒ x = 2160 = 36 km. 60
Hence the required distance
= 36 km.
- Two men A and B start walking simultaneously from P to Q, a distance of 21 kms, at the speed of 3 km and 4 km an hour respectively. B reaches Q, returns immediately and meets A at R. Find the distance from P to R.
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Let A and B meet after time t hours.
Distance covered by A in t hours = 3 t km.
Distance covered by B in t hours = 4 t km.
Total distance covered by A and B = (3t + 4t) km = 7 t km.
From the diagram we can see that the total distance covered by A and B is equal to twice the distance between P and Q.
∴ 7t = 2 × 21t = 2 × 21 = 6 hours 2
Distance PR = 6 × 3 = 18 km.Correct Option: D
Let A and B meet after time t hours.
Distance covered by A in t hours = 3 t km.
Distance covered by B in t hours = 4 t km.
Total distance covered by A and B = (3t + 4t) km = 7 t km.
From the diagram we can see that the total distance covered by A and B is equal to twice the distance between P and Q.
∴ 7t = 2 × 21t = 2 × 21 = 6 hours 2
Distance PR = 6 × 3 = 18 km.
- Two men A and B walk from X to Y a distance of 42 kms at 5 km and 7 km an hour respectively. B reaches Y and returns immediately and meets A at R. Find the distance from X to R.
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When B meets A at R, by then B has walked a distance (XY + YR) and A, the distance XR.
That is both of them have together walked twice the distance from X to Y, i.e., 42 kms.
Now, the ratio of speed of A and B is 5 : 7 and they walk 84 kms.
∴ Hence, the distance XR travelled byA = 5 × 84 = 35 kms. 5 + 7 Correct Option: C
When B meets A at R, by then B has walked a distance (XY + YR) and A, the distance XR.
That is both of them have together walked twice the distance from X to Y, i.e., 42 kms.
Now, the ratio of speed of A and B is 5 : 7 and they walk 84 kms.
∴ Hence, the distance XR travelled byA = 5 × 84 = 35 kms. 5 + 7
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Walking at 3 of his usual speed a man is late by 4 2 1 hours. The usual time would have been what? 2
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New speed is 3 of the usual speed 4 ∴ New time taken = 4 of the usual time 3 ∴ 4 of the usual time – Usual time = 5 3 2 ⇒ 1 of the usual time = 5 3 2 ∴ Usual time = 5 × 3 2 = 15 hours or 7.5 hrs 2 Correct Option: B
New speed is 3 of the usual speed 4 ∴ New time taken = 4 of the usual time 3 ∴ 4 of the usual time – Usual time = 5 3 2 ⇒ 1 of the usual time = 5 3 2 ∴ Usual time = 5 × 3 2 = 15 hours or 7.5 hrs 2