Three cars leave A for B in equal time intervals. They reach B simultaneously and then leave for Point C which is 240 km 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 80 km away from C. What is the difference between the speed of the first and the third car?
A)80 kmph B)20 kmph C)60 kmph D)40 kmph

• Let V1, V2 and V3 be the speeds of the cars.
  
  ABV1 - ABV2 = ABV2 - ABV3
  240V1 - 240V2 = 1
  V3 = 2V1
  Condition I states that the cars leave in equal intervals of time and arrive at the same time. Or, the difference in the time taken between cars 1 and 2 should be equal to the time taken between cars 2 and 3.
  
  We get ABV1 - ABV2 = ABV2 - ABV3
  As the second car arrived at C an hour earlier than the first, we get a second equation
  240V1 - 240V2 = 1
  The third car covered 240 + 80 kms when the first one covered 240 – 80 kms. Therefore, 320V3 = 160V1
  This gives us V3 = 2V1
  From condition 1, we have ABV1 - ABV2 = ABV2 - ABV3
  Substituting V3 = 2V1, this gives us ABV1 - ABV2 = ABV2 - AB2V1
  or 3AB2V1 = 2ABV2 or V2 = 4V13
  Solving 240V1 - 240V2 = 1 , we get 60V1 = 1 or V1 = 60 kmph
  => V2 = 80 kmph and V3 = 120 kmph
  
  Hence, the answer is 60 kmph
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