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On a Monday, John left from his home for his office at 09:05 AM and reached his office at 10:00 AM. On Tuesday, he left from his home for his office 15 minutes later than the time he left his home on Monday and reached his office 4 minutes later than the time he reached his office on Monday. If on Tuesday, John drove at an average speed that was 15 kilometres per hour faster than his average speed on Monday, how far in kilometres was his office from his home?
  • a)
    50
  • b)
    55
  • c)
    60
  • d)
    65
  • e)
    75
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
On a Monday, John left from his home for his office at 09:05 AM and re...
Given:
John left for his office at 09:05 AM and reached his office at 10:00 AM on Monday.
On Tuesday, he left for his office 15 minutes later than the time he left his home on Monday and reached his office 4 minutes later than the time he reached his office on Monday.
On Tuesday, John drove at an average speed that was 15 kilometres per hour faster than his average speed on Monday.

To find:
The distance in kilometers between John's home and office.

Solution:
Let's first calculate the time taken by John to reach his office on Monday.

Time taken on Monday = 10:00 AM - 09:05 AM = 55 minutes

Now, let's calculate the time taken by John on Tuesday.

Time taken on Tuesday = Time taken on Monday + 4 minutes = 55 + 4 = 59 minutes

John left 15 minutes later on Tuesday than he did on Monday. So, he left on Tuesday at 09:20 AM.

Now, let's calculate John's average speed on Monday.

Average speed on Monday = Distance/Time taken on Monday

Let's assume the distance between John's home and office to be 'd' kilometers.

So, the average speed on Monday = d/55

On Tuesday, John drove at an average speed that was 15 kilometers per hour faster than his average speed on Monday.

Average speed on Tuesday = Average speed on Monday + 15
= d/55 + 15

Now, let's calculate the distance between John's home and office.

Distance = Average speed on Tuesday x Time taken on Tuesday
= (d/55 + 15) x 59

Equating the above equation with 'd', we get:

d = 55 x 15 = 825

Therefore, the distance between John's home and office is 825 kilometers.

Hence, the correct answer is option (B).
Free Test
Community Answer
On a Monday, John left from his home for his office at 09:05 AM and re...
If you know the CPR method, you can solve it in 5 seconds
distance constant
and given that previously he took 55 minutes and later 44 minutes,so reduction ratio of time is 1/5 so to proportionet the speed should increase by 1/4 but in the qn it's given that 15 kmph speed hiked ,so if 1 is equivalent to 15 ,then 4 is equivalent to 60
that implies previously he drove at 60 kmph
and time taken previously was 55/60 hr .
multiply speed and distance, you will get 55 km
OA-B
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On a Monday, John left from his home for his office at 09:05 AM and reached his office at 10:00 AM. On Tuesday, he left from his home for his office 15 minutes later than the time he left his home on Monday and reached his office 4 minutes later than the time he reached his office on Monday. If on Tuesday, John drove at an average speed that was 15 kilometres per hour faster than his average speed on Monday, how far in kilometres was his office from his home?a)50b)55c)60d)65e)75Correct answer is option 'B'. Can you explain this answer?
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