Raju goes to school at the rate of 2% km/hr and
reaches 6 minutes late. If he travels at the speed
of 3 km/hr , he is 10 minutes early. What is the
distance to the school?
Answers
Answer:
in which class you in....
Answer:
This is the problem we face in our day to day lives. We want to be early,
to office so that we do not miss an important meeting or
to school so that we do not miss our favourite class
or be at the movie theatre so that we do not miss the first few minutes of the suspense thriller for which you have been waiting for almost an year.
This student, for that matter anybody, has only two options under such circumstances
Start early or
Speed up, otherwise.
Probably our student under discussion choose to experiment with various speeds to reach his school and probably is making note of time it takes correspondingly.
The information we have is:
travels at 2.5 kmph, reaches late by 6 minutes ( 1/10 hours)
travels at 2.5 kmph, reaches early by 10 minutes ( 1/6 hours)
Even though we need to calculate the distance between his home and school, but let's look at the speeds under discussion.
With 2.5 kmph he is late, but with 4.5 kmph he arrives early. This means the optimal speed for which he reaches school on time lies somewhere between 2.5 kmph and 4.5 kmph. Let this optimal speed be s . With d as the distance between his home and school and t as time taken for which he arrive at school on time. This gives the known relation,
d=s∗t
With 2.5 kmph, time taken to cover this distance would be t+1/10 .
With 4.5 kmph, time taken to cover this distance would be t−1/6.
This actually means that with two different speed and different time durations he covers the same distance. i.e
2.5∗(t+1/10)=4.5∗(t−1/6)
5∗(t+1/10)=9∗(t−1/6)
5t+1/2=9t−3/2
2=4t
time taken to cover the distance with optimal speed t is 1/2 hours.
Using any of the two equations given and the time obtained previously we can find the distance.
d=2.5∗(t+1/10)=4.5∗(t−1/6)
This gives us the distance as 1.5 km.