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? (Ratio)

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

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.

$∴ R a t i o o f s p e e d s o f c_{1} : c_{2} = 1 : 2 A g a i n, a t f i r s t m e e t i n g o f a l l t h r e e c a r s a s t i m e s a r e i n a r i t h m e t i c p r o g r e s s i o n, s p e e d i s i n h a r m o n i c m e a n . S o, c_{2}' s s p e e d = \frac{2 \times 1 \times 2}{1 + 2} = \frac{4}{3} ∴ R a t i o o f s p e e d = 1 : \frac{4}{3} : 2 = 3 : 4 : 6 T h e n o f t i m e = \frac{1}{3} : \frac{1}{4} \frac{1}{6} = 4 : 3 : 2 ∴ D i f f e r e n c e b e t w e e n t h e s p e e d o f c_{1} a n d c_{3} = \frac{240}{2} - \frac{240}{4} = 120 - 60 = 60 k m p h .$