Three cars leave A and B in equal time intervals. They reach B simultaneously and then leave for Point C which is 240km away from B. The first car arrives at C an hour after the second car. The third car, having reached C, immediately turns back and heads towards B. The first and the third car meet a point that is 80km away from C. What is the difference between the speed of the first and the third car?

Let, first, second and third cars are c₁, c₂ be c₃ respectively.

Distance covered by c₁ : c₃ = (240 - 80) : (240 + 80)

=160 : 320 = 1 : 2

Here, we know speed is directly proportional to the distance covered by these cars as time is constant.

$$\begin{gathered} \therefore Ratio of speeds of c_1 : c_2 = 1 : 2 \\ Again, at first meeting of all three cars as times are in arithmetic progression, \\ speed is in harmonic mean. \\ So, c_2\text{'}s speed = \frac{2\times 1\times 2}{1+2}=\frac{4}{3} \\ \therefore Ratio of speed = 1 : \frac{4}{3} : 2= 3: 4 : 6 \\ Then of time = \frac{1}{3} : \frac{1}{4} \frac{1}{6} = 4 : 3 : 2 \\ \therefore Difference between the speed of c_1 and c_3=\frac{240}{2}-\frac{240}{4}=120-60=60 kmph. \end{gathered}$$