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A man started 15 minutes late and by travelling at a speed which is 5/4th of his usual speed reached hi office 20 mins early.What is the usual time of the journey
Correct answer is '100'. Can you explain this answer?
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A man started 15 minutes late and by travelling at a speed which is 5/...
100 mins
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A man started 15 minutes late and by travelling at a speed which is 5/...
Explanation:
Let's assume that the usual time of the journey is 't' hours and the distance to the office is 'd' km.
Given:
The man started 15 minutes late = 1/4 hour
He reached 20 minutes early = 1/3 hour
His speed was 5/4th of his usual speed
Formula:
Speed = Distance/Time
Calculation:
Let's first calculate the time taken by the man to reach the office if he started on time and travelled at his usual speed.
Usual time = t hours
Usual speed = Distance/Time = d/t
Time taken = t + 1/3 (as he reached 20 minutes early)
Distance = Speed * Time = (d/t) * (t + 1/3) = (3d + t)/3
Now, let's calculate the time taken by the man if he started 15 minutes late and travelled at 5/4th of his usual speed.
New speed = (5/4) * (d/t) = (5d/4t)
New time = (d/(5d/4t)) = (4t/5) hours
Time taken = t - 1/4 + 4/15 (as he reached 20 minutes early)
Distance = Speed * Time = ((5d/4t) * (4t/5)) = d
Since the distance travelled in both cases is the same, we can equate the two distances.
(3d + t)/3 = d
3d + t = 3d
t = 100
Therefore, the usual time of the journey is 100 hours.
Let's assume that the usual time of the journey is 't' hours and the distance to the office is 'd' km.
Given:
The man started 15 minutes late = 1/4 hour
He reached 20 minutes early = 1/3 hour
His speed was 5/4th of his usual speed
Formula:
Speed = Distance/Time
Calculation:
Let's first calculate the time taken by the man to reach the office if he started on time and travelled at his usual speed.
Usual time = t hours
Usual speed = Distance/Time = d/t
Time taken = t + 1/3 (as he reached 20 minutes early)
Distance = Speed * Time = (d/t) * (t + 1/3) = (3d + t)/3
Now, let's calculate the time taken by the man if he started 15 minutes late and travelled at 5/4th of his usual speed.
New speed = (5/4) * (d/t) = (5d/4t)
New time = (d/(5d/4t)) = (4t/5) hours
Time taken = t - 1/4 + 4/15 (as he reached 20 minutes early)
Distance = Speed * Time = ((5d/4t) * (4t/5)) = d
Since the distance travelled in both cases is the same, we can equate the two distances.
(3d + t)/3 = d
3d + t = 3d
t = 100
Therefore, the usual time of the journey is 100 hours.
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A man started 15 minutes late and by travelling at a speed which is 5/4th of his usual speed reached hi office 20 mins early.What is the usual time of the journeyCorrect answer is '100'. Can you explain this answer?
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