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Three bikes 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 bike arrives at C an hour after the second bike. The third bike, having reached C, immediately turns back and heads towards B. The first and the third bike meet a point that is 80 km away from C. What is the difference between the speed of the first and the third bike?

Three bikes 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 bike arrives at C an hour after the second bike. The third bike, having reached C, immediately turns back and heads towards B. The first and the third bike meet a point that is 80 km away from C. What is the difference between the speed of the first and the third bike? Correct Answer 60 km/hr

Given:

Distance between B and C = 240 km

Formula:

Speed = distance/time

Calculations:

Let say Speed of bike1    = X km/Hr

Let say Speed of bike2 =   Y  km/Hr

Let say speed of bike3   =   Z km/Hr

Let say distance between  AB = D km/Hr

Time taken by Bike 1 = D/X

Time taken by Bike 2 = D/Y

Time taken by Bike 3 = D/Z

⇒ D/Z – D/Y = D/Y – D/X

⇒ 2/Y = 1/X + 1/Z       ----(1)

Time taken by bike 2 for 240 km =  240/Y

Time taken by Bike 1 for 240 km = 240/X

240/X =  240/Y + 1

⇒ 1/X – 1/240 = 1/Y      ----(2)

Putting 1/Y in (1)

⇒ 2/X  - 1/120 = 1/X + 1/Z

⇒ 1/X = 1/Z + 1/120       ----(3)

When 1st & Third bike meet

Distance  traveled by third bike = 240 + 80 = 320 km

Distance traveled by First Bike = 240 - 80 = 160 km

Time taken by both bike is equal

⇒ 320/Z  = 160/X

⇒ Z = 2X

putting in (3)

⇒ 1/X = 1/2X  + 1/120

multiplying by 120X

⇒ 120 = 60 + X

⇒ X = 60

⇒ Z = 2X = 120

⇒ Z - X = 120 - 60 = 60 km/Hr

∴ Difference between the speed of the first and the third bike is 60 km/Hr

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