Can you please tell me how to solve this tough "Speed-distance-time" question

here is the question:


A boy goes from his house to his school daily by a motorbike. He drives the motorbike at three different speeds ‘a’, ‘2.5a’ and ‘4a’ km/hr for three different durations during the journey. He definitely knows the time intervals for which he should drive his motorbike at each of these speeds so as to reach the school on time.

On one particular day he starts 20 minutes late from his house and hence the extra time for which he drives his motorbike at ‘2.5a’ km/hr is 6 minutes. Find the extra time for which he drives at ‘4a’ km/hr to reach the school on time.



options are:-

a)11/3 minutes
b)15/4 minutes
c)13/3 minutes
d)cannot be determined

i will post the answer tomorrow..but i don't know how to solve it. can i apply ratio and proportion technique here if yes then how?

He drives an extra 6 minutes at 2.5a, meaning he went an extra 15a km at the middle speed.

So he must have lost 15a km in the combination of the slow and fast speeds, to be equal. And furthermore, the time lost was 20 minutes, plus the 6 minutes which were donated to the middle speed. So 26 minutes total, to go 15a km. Let s be the minutes at slow, and f be the minutes at fast.

s + f = 26
1as + 4af = 15a, so s + 4f = 15

You can now solve for f.

If time= distance devided by speed
if we know he went 2.5a km/hr for 6 min then

2.5a 4a
___ =____
6 ?

4/2.5=1.6
6*1.6+9.6

extra time....
9.6-6=3.6

3.6 final answer