Bunuel
The distance between two cities, A and B, is 900 miles. A bus starts from A for B at 7:00 am. Another bus starts from B for A at 8:30 am. What is the speed of the first bus?
(1) The speed of the second bus is 4/7 of the speed of the first bus.
(2) The two buses cross each other at 7:00 pm on the same day.
Here's my solution. But it took me around 3 mins with this. Hoping for a better one to be posted by some experts.
From the question:First, let d1 be the distance covered by bus A in 1.5 hours (bw 7-8:30 AM)
Also, \(d1 = 1.5*S_a\)
Total time the two buses travel together till they meet= T
Let \(S_a\) be the speed of bus A and \(S_b\) be the speed of bus B.
Using the concept of relative speed, we can say that the distance of 900- d1 is travelled
over a time period of T with a speed of \(S_a-S_b\)
we get \((900 - d1)/T = S_a - S_b\)
From (1)We have, \(S_b=(4/7)*S_a\)
---> \((900 - d1)/T = -(3/7)*S_a\)
This equation has two variables T and \(S_a\), hence (1) is not sufficient
From (2)T = 10.5 hours ( from 8:30 AM till 7 PM)
Again, we get an equation with two variables \(S_a\) and \(S_b\)
therefore, (2) is not sufficient
But if we combine (1) and (2), we get an equation with one variable, and hence we can find \(S_a\)
Therefore answer is C.