PRACTICE QUESTIONS ON TIME AND DISTANCE

Let half of the distance travelled be d.
Time taken at the speed of 4 km/hr = (d/4)
Time taken at the speed of 5 km/hr = (d/5)
Therefore, total time = (d/4) + (d/5) = 9 hrs.
⇒ (5d + 4d)/20 = 9
⇒ 9d = 180
⇒ d = 20.
Therefore total distance = d x 2 = 40 kms

Circumference of the circular path = 2Π50 = 100Π m
This means 100Π m is the distance covered in 60 minutes
Therefore distance covered in 15 minutes = 1/4 x 100Π m = 25Π m
Now distance covered in one revolution of the wheel = 2Π(1/2) m = Π m
Therefore number of revolutions in 15 minutes = 25Π/Π = 25

Let the total distance be d km. We have speed = 8km/hr and 10km/hr.
We know, time (t) = distance (d)/speed (s).
⇒ t = d/8 (for the first day)
⇒ t = d/10 (for the second day)
Difference between the time taken on first day and second day is 2.5 + 5 = 7.5 mins.
Therefore, (d/8) – (d/10) = 7.5/60. (7.5 mins converted to hours)
⇒ (10d – 8d)/80 = 75/600
⇒ d = (75 x 80)/1200.
⇒ d = 5 kms.
Therefore, the distance between his school and house 5 km.

Let the distance travelled at the rate of 12 kmph be x km
Time taken to travel this distance = x/12 .......(1)
Then the distance travelled at the rate of 10 km/hr will be 75 – x
Time taken to travel this distance = (75 – x)/10 .......(2)
Total time taken to cover the total distance = 7 hrs
⇒ (x/12) + (75 – x)/10 = 7 [sum of (1) and (2)]
⇒ 10x + 900 – 12x = 840
⇒ –2x = –60
⇒ x = 30.
Therefore, the distance travelled at the rate of 12 km/hr = 30km.

Time taken to travel 20 kms at the speed of 5 kmph.
⇒ t = 20/5 = 4 hrs.
Actual time of reaching = 4 hrs – 40 mins = 3 hrs 20 mins ......(1)
Time taken to travel 20 kms at the speed of 8 kmph.
⇒ t = 20/8 = 2.5 hrs. ......(2)
Difference between (1) and (2) = 3 hrs 20 mins – 2 hrs 30 mins
200 mins – 150 mins = 50 mins.
Sanu would have reached 50 mins earlier.

Average speed is represented by total distance divided by the time taken.
Total distance = 10 kms + 12 kms = 22 kms
Total Time = (10/12) + (12/10)
⇒ (10/12) + (12/10) = (100 + 144)/120 = 244/120 hrs
Average speed = 22/(244/120) = 10.8 kmph.
Average speed for the entire trip is 10.8 kmph.

We know that time = distance/speed
Time taken by the athlete for the first half = 15/10
And time taken by the athlete for the second half = 15/20
Total time = 15/10 + 15/20 = (45/20) x 60 = 135 mins
Time taken to cover the total distance is 135 mins.

Total distance travelled in 1 ½ hour
= 50 + (&frac; x 32) km
= 66 km
∴ Average speed = Total distance/Total time
= 66/(3/2) = 44 kmph.

Time = Distance/Speed
Time taken for each 6 km can be given by
6/25, 6/50, 6/75 and 6/150
Total time = (6/25) + (6/50) + (6/75) + (6/150) = (36 + 18 + 12 + 6)/150 = 72/150
Average speed = Distance/time = (24/72) x 150 = 50 kmph

Let the speeds of A and B be 3 kmph and 4 kmph.
And the distance be a common multiple e.g. 12 km
Time taken by A to cover the distance = 4 hrs
And the time taken by B = 3 hrs
The time difference between the two = 1 hour
But as per the question the difference in time = 30 min
∴ the distance is 6 km and the time taken by A is 2 hours

Let the distance to his office be D km
∴ the time taken at 10 kmph would be D/10 and the time taken at 15 kmph would be D/15
As per the question the time difference between the two = 2 hrs (1 p.m. – 11 a.m.)
⇒ D/10 – D/15 = 2 hrs
D = 60 km
∴ he takes 60/10 = 6 hrs when he travels at 10 kmph i.e. he starts at 7 a.m.
From the question to reach at 12 noon, he has 5 hours
∴ speed = 60/5 = 12 kmph

Let the man travel D km on foot
∴ he travels 80 - D km on bicycle
Time taken while travelling on foot = D/8
And time taken while travelling on bicycle = (80 – D)/16
From the question
D/8 + (80 – D)/16 = 7 hrs
⇒ 2D + 80 – D = 7 x 16 = 112
D = 32 km

Total distance = 300 km
40% of distance = 120 km
Time taken = 2 hours
Speed at which he travelled = 120/2 = 60 kmph
Distance left = 180 km
Time left = 12.30 – 10.30 = 2 hours
Speed required = 180/2 = 90 kmph
Increase in speed = 90 – 60 = 30 kmph

At the speed of 10 kmph time taken to travel 5 km = 30 minutes Number of rests he takes = 4 (after 1 km, 2 km, 3km and 4 km)
Total resting time = 4 x 5 = 20 minutes
Total time = 30 + 20 = 50 minutes

Let his original speed be S kmph
Time taken = 84/S
If his speed is 5 kmph more, time is 5 hrs less
∴ 84/S – 5 = 84/(S + 5)
(S + 5)(84 – 5S) = 84S
5S² + 25S – 420 = 0
S = 7
Simpler method
84 has the following factors - 2, 3, 4, 6, 7, 12, 14, 21, 28, 42
Of the above factors 7 and 12 are at a difference of 5 (note the use of 5 kmph in speed in the question)
Through simple trial and error we can find that to travel 84 km he takes 12 hours at 7 kmph
and at 12 kmph he takes 7 hours both of which are at a difference of 5.

Let as assume the distance to be common multiple of 30 and 45 for e.g. 90 km
At 30 kmph he takes 3 hours and at 45 kmph he takes 2 hours
The difference between the two times is 1 hour
Since as per the question the difference in time is 2 hours the distance is double of that assumed i.e. 180 km