A boy walking at a speed of 20kmph reaches his school 15min late. Next time at a speed of 25kmph reaches his school 10m early find the usual time to reach the school?

Sure, let's solve this step by step:

Step 1: Convert the speeds from km/h to km/min. 1 km/h = 1/60 km/min So, 20 km/h = 20/60 km/min = 1/3 km/min And, 25 km/h = 25/60 km/min = 5/12 km/min

Step 2: Let's denote the usual time to reach the school as 't' (in minutes) and the distance to the school as 'd' (in km).

Step 3: According to the problem, when the boy walks at a speed of 20 km/h, he is 15 minutes late. So, the equation for the distance will be: d = speed * time = (1/3) * (t + 15)

Step 4: The problem also states that when the boy walks at a speed of 25 km/h, he is 10 minutes early. So, the equation for the distance will be: d = speed * time = (5/12) * (t - 10)

Step 5: Since the distance to school remains the same, we can equate the two equations: (1/3) * (t + 15) = (5/12) * (t - 10)

Step 6: Solve this equation for 't'. First, clear the fractions by multiplying the entire equation by 12: 4*(t + 15) = 5*(t - 10)

Step 7: Expand both sides: 4t + 60 = 5t - 50

Step 8: Rearrange the equation to find 't': t = 110 minutes

So, the usual time for the boy to reach school is 110 minutes.