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15 Apr 2014, 15:26
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C
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Difficulty:
15%
(low)
Question Stats:
79% (01:39) correct
21%
(02:07)
wrong
based on 258
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At 12:00 hour a truck left the city A to city B, at a constant speed of 45 km/h. One hour later, a car left the city B to the city A, along the same road, at a constant speed of 100 km/h. If the distance between A and B is 190 km, at what time will the truck and the car meet each other?
A. 12:40
B. 13:00
C. 14:00
D. 14:40
E. 16:00
A. 12:40
B. 13:00
C. 14:00
D. 14:40
E. 16:00
tauchmeister
It's best to draw a diagram in these questions to help you see the situation.
A ____45______T (1:00)_____________145____________ B
The truck starts from A at 12:00 and reaches a point, T, 45 km away in 1 hour i.e. at 1:00 because the speed of the truck is 45 km per hour.
Since total distance between A and B is 190 km and the truck has already covered 45, the distance between T and B is 190 - 45 = 145 km.
Now, at 1:00, truck starts from T and Car starts from B and they travel towards each other at a relative speed of 45 + 100 = 145 km/hr (since they are moving in opposite directions, relative speed is the sum of their speeds)
To cover a distance of 145 km at a relative speed of 145 km/hr, they will take 145/145 = 1 hour.
So they will meet 1 hour after 1:00 i.e. at 2:00.
Answer (C)
15 Apr 2014, 22:21
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At 1300 Hrs, truck has already travelled 45 Kms & car has started from City B
Distance remaining = 190 - 45 = 145
Setting up the equation
Lets say truck travels x kms & meets car
\(\frac{x}{45} = \frac{145-x}{100}\)
29x = 145 * 9
x = 45
To travel 45Kms, truck would again require 1 hr
So they would meet at 1400 Hrs
Answer = C
Distance remaining = 190 - 45 = 145
Setting up the equation
Lets say truck travels x kms & meets car
\(\frac{x}{45} = \frac{145-x}{100}\)
29x = 145 * 9
x = 45
To travel 45Kms, truck would again require 1 hr
So they would meet at 1400 Hrs
Answer = C
