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A man went to his office on cycle at the rate of 10 km/hr and reached late by 6 minutes. When he increased the speed by 2 km/hr, he reached 6 minutes before time. What is the distance between his office and his departure point ?
  • a)
    6 km
  • b)
    7 km
  • c)
    12 km
  • d)
    16 km
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A man went to his office on cycle at the rate of 10 km/hr and reached ...
Let the distance be d.
D/10 = T+ 1/10
D/12 = T - 1/10
So, D = 12 kilometers.
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Most Upvoted Answer
A man went to his office on cycle at the rate of 10 km/hr and reached ...
Given data:
Speed of the cycle = 10 km/hr
Increased speed = 2 km/hr
Late by 6 minutes at the initial speed
Reaches 6 minutes before time at the increased speed

To find: Distance between the office and departure point

Let us assume the distance between the office and departure point as 'd' km.

Calculation:
Let us calculate the time taken by the man at the initial speed and the increased speed.

At the initial speed of 10 km/hr,
Time = Distance/Speed
Time taken for the distance 'd' = d/10 hrs
As he reached late by 6 minutes, the time taken = d/10 + 6/60 hrs

At the increased speed of 12 km/hr,
Time taken for the same distance 'd' = d/12 hrs
As he reached 6 minutes before time, the time taken = d/12 - 6/60 hrs

Equating the above two equations,
d/10 + 6/60 = d/12 - 6/60
d/60 = 1/30
d = 60/30
d = 2 km

Therefore, the distance between the office and departure point is 2 km.

Hence, the correct answer is option (c) 12 km.
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A man went to his office on cycle at the rate of 10 km/hr and reached late by 6 minutes. When he increased the speed by 2 km/hr, he reached 6 minutes before time. What is the distance between his office and his departure point ?a)6 kmb)7 kmc)12 kmd)16 kmCorrect answer is option 'C'. Can you explain this answer?
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