A student rides a bicycle at $8$ km/hr and reaches the school $2.5$ minutes late.The next day he increases the speed $10$ kms/hr and reaches school $5$ minutes early.How far is the school from his house?

Using the distance formula with the given speeds and the total time difference, distance equals the product of the two speeds and the time difference divided by the speed difference. The time difference between being 2.5 minutes late and 5 minutes early is 7.5 minutes, which we convert to hours by dividing by 60 to get 0.125 hours. Plugging in the values, we calculate 8 multiplied by 10 multiplied by 0.125 divided by 2, which simplifies to 10 divided by 2, resulting in 5 kms. The school is 5 kms away.

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