A student walks at a speed of 5 km h starting from his house and reaches his office 6 minutes
later. Next day he increases his speed by 1 kmh and reaches office 6 minutes early. How far is
the office from his house?
Answers
Answer:
A student walks to school at the rate of 2.5 km/h and reaches 6 minutes late. The next day, he increases his speed by 2 km/h and then reaches school 10 minutes early. What is the distance of the school from his home?
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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 correspon
In this question Distance is constant .only speed and time varies. So in this type we use basic identity of formula of time and Distance keeping one identity which is distance constant.
S=D/T
S is indirectly proportion to T ( keep D constant)
Given data -
S1 = 2.5km /h
S2= 4.5km/h ( 2km increase speed so 2.5+2=4.5)
S1/S2= 2.5/4.5= 5/9
Therefore
T2/T1= 9/5 ( because S is indirectly proportion to T))
The reverse of speed is time taken
So T2 -T1 = 9–5 = 4unit
These 4unit = 16mint ( 16mint =6mint+10mint, time gap)
1unit = 4mint
So Time taken by T2 = 9unit = 9×4= 36mint ( convert 36mint into hour =36/60 hr )
So, Distance =speed ×time
Distance = 2.5Km/h × 36/60 hour = 1.5Km
Explanation:
Hope this unit method help you . With this method you can finish question in 5sec. Best of luck . Keep learning keep growing