Question

Town P and Q are 265 km apart. Car A started from P at 60 km/hr at 8.00 am towards Q. Car B started from Q at 40 km/hr at 10.00 am. B stopped for half an hour. A stopped for 28 minutes at town S which is at a distance of 210 km from P. Find the time at which B crossed A.

Answer 1

A went 210 km in 3.5 hr (11:30)
A stopped till 11:58
A then went on.

So, A's position x in terms of time elapsed from 8:00 is

x =
60t for 0 <= t <= 7/2
210 for 3.5 <= t <= 3 29/30
210+60(t-3 29/30) for t > 3 29/30
A reached Q at 11:58+0:55 = 12:53

B's distance y is thus
y = 40(t-2)

y=210 at t=7.25, or 15:15
B never passed A, since it started later and was going slower.

I suspect a typo. Maybe the cars are backwards?

Answer 2

A, Starts from P, S=60kmph, At 11.30 AM (Rest of 28mins.)
B, Starts from Q, S=40kmph, At 11.00 AM (Rest of 30mins.)

At 8AM, distance between is 265km
At 10AM, distance between is 145km as A travels 120km in 2hrs
At 11AM, distance between is 45km as A&B travel 100km
At 11.30AM, distance b/w is 15km at A travels 30km

Now d=15,s=40(bec. A is resting)
t=22.5 minutes
Thus, 11.30+22.5 min=11.52.30AM