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?
60 kmph
20 kmph
40 kmph
80 kmph
Answers
Answer:
60 km/Hr
Step-by-step explanation:
Let say Speed of Car1 = X km/Hr
Let say Speed of Car2 = Y km/Hr
Let say speed of car3 = Z km/Hr
Let say distance between AB = D km/Hr
Time taken by Car 1 = D/X
Time taken by Car 2 = D/Y
Time taken by Car 3 = D/Z
D/Z - D/Y = D/Y - D/X
=> 2/Y = 1/X + 1/Z - eq 1
Time taken by car 2 for 240 km = 240/Y
Time taken by Car 1 for 240 km = 240/X
240/X = 240/Y + 1
=> 1/X - 1/240= 1/Y - eq 2
Putting 1/Y in eq 1
2/X - 1/120 = 1/X + 1/Z
=> 1/X = 1/Z + 1/120 - eq 3
When 1st & Third car meet
Distance traveled by third car = 240 + 80 = 320 km
Distance traveled by First Car = 240 - 80 = 160 km
Time taken by both car is equal
320/Z = 160/X
=> Z = 2X
putting in eq 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 car = 60 km/Hr
Answer:
1st Option is correct that is Difference between the speed of the Car1 and Car3 is 60 km/hr
Step-by-step explanation:
Given:
Car1 , Car2 , Car3 leaves in equal interval of time from A to B.
Distance between B and C = 240 km.
Time gap between Car 1 and Car2 = 1 hr.
After returning from C , Car1 and Car3 meet at 80 km from C.
To find: Difference between the speed of the Car1 and Car3.
Let Speed of Car1 = x km/hr ,
Speed of Car2 = y km/hr
Speed of Car3 = z km/hr
we need to find value of z - x ( as car3 return from C while car1 coming to C ⇒ Speed of car3 is greater than car1 )
Let say distance between A and B = d km
using, Speed Distance & time formula
Time taken by Car1 to reach B = d/x
Time taken by Car2 to reach B = d/y
Time taken by Car3 to reach B = d/z
They reaching simultaneously,
So
d/z - d/y = d/y - d/x
d/y + d/y = d/z + d/x
2/y = 1/x + 1/x .................................(1)
Time taken by car2 to reach C = 240/y
Time taken by car1 to reach C = 240/x
So,
240/x = 240/y + 1
1/x = 1/y + 1/240
1/x - 1/240 = 1/y .......................(2)
putting value of 1/y in (1), we get
2( 1/x - 1/240 ) = 1/x + 1/z
2/x - 1/120 = 1/x + 1/z
2/x - 1/x - 1/z = 1/120
1/x - 1/x = 1/120 ............................(3)
Lastly when car1 and car3 meets say at D
Distance traveled by car1 from B meeting point D = 240 - 80 = 160 km
Distance traveled by car 3 from B meeting point D = 240 + 80 = 320 km
Time taken by car1 to D = 160/x
Time taken by car3 to D = 160/z
We know that time taken both cars are equal. So,
320/z = 160/x
2x = z
Putting value of z in (3), we get
1/x - 1/2x = 1/120
1/2x = 1/120
2x = 120
x = 60
⇒ z = 2x = 2(60) = 120
Thus, z - x = 120 - 60 = 60 km/hr
Therefore, 1st Option is correct that is Difference between the speed of the Car1 and Car3 is 60 km/hr