Walking with 6/7th of his usual speed , Prakash was late by 12 minutes . What is the usual time taken by Prakash to cover the distance ?

Let's denote the usual speed of Prakash as S (in km/h for example), and the usual time he takes to cover the distance as T (in hours for example).

When Prakash walks with 6/7th of his usual speed, his speed becomes 6/7 * S.

We know that speed = distance / time. Therefore, time = distance / speed.

So, the time it takes for Prakash to cover the distance at this slower speed is distance / (6/7 * S).

We also know that this slower speed made him 12 minutes late, so the time it took him at this slower speed is T + 12/60 (we divide by 60 to convert minutes into hours).

We can now set up the equation:

distance / (6/7 * S) = T + 12/60

We can simplify this equation by multiplying each side by (6/7 * S):

distance = (6/7 * S) * (T + 12/60)

We know that the distance is also equal to S * T (from the speed = distance / time formula), so we can set up the equation:

S * T = (6/7 * S) * (T + 12/60)

We can cancel out S from both sides:

T = (6/7) * (T + 12/60)

Solving for T, we get:

T = 6/7 * T + 72/420

Subtracting 6/7 * T from both sides, we get:

1/7 * T = 72/420

Multiplying both sides by 7, we get:

T = 72/60 = 1.2 hours

So, the usual time taken by Prakash to cover the distance is 1.2 hours.