if a student runs at 1.5 times of his usual speeed he reaches 20 minutes early if he runs 0.5 times of hus usual speed how late will he reach his school
Answers
Solution :-
Let us Assume that, Distance between his home to his school is x km and usual speed of the boy is y km/h.
Than,
→ usual Time = (Distance/Speed) = (x/y) Hours.
Now,
it has been said that, if the student runs at 1.5 times of his usual speeed he reaches 20 minutes early.
So,
→ New Speed = 1.5 times of usual speed = 1.5y = (3/2)y km/h.
→ New Time = (x/y) - 20 Min. = (x/y) - (20/60) = {(x/y) - (1/3)} Hours.
And,
→ Distance from home to school = Speed * Time
→ x = (3y/2) * {(x/y) - (1/3)}
→ x = (3x/2) - (y/2)
→ (y/2) = (3x/2) - x
→ (y/2) = (x/2)
→ y = x .
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Now, we have to find he he runs 0.5 times of hus usual speed how late will he reach his school ?
So,
→ New Speed = 0.5 times of y = (1/2) of y = (y/2) km/h .
→ Distance from home to school = x km
→ Time taken by boy this time = Distance/speed = x/(y/2) = (2x/y) Hours.
Now, putting y = x , we get,
→ (2x/y) = (2y/y) = 2 Hours.
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And,
→ Our usual Times was = (x/y) = 1 Hours. { As x = y }.
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Hence,
→ Extra Time student took because of decrease in speed = 2 - 1 = 1 Hour. (Ans.)