Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 4)
4.
A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
Answer: Option
Explanation:
Let speed of the car be x kmph.
| Then, speed of the train = | 150 | x | = |  |
3 | x |  kmph. |
| 100 | 2 |
 |
75 | - | 75 | = | 125 |
| x | (3/2)x | 10 x 60 |
 |
75 | - | 50 | = | 5 |
| x | x | 24 |
| x = |
|
25 x24 | |
= 120 kmph. |
| 5 |
Discussion:
342 comments Page 4 of 35.
Reddy said:
1 decade ago
Hi friend that is not delay thats the time wasted by the train due to its stoppings if the train don't stop any where it will come first than car.
Divs said:
1 decade ago
Raj would you please tell me how come the speed be 25?
Hemant said:
1 decade ago
Explain the simple way, (75/x)-(75/(3/2)x)=12.5/60
Atul said:
1 decade ago
(75/x)denotes time taken by car.
this means time taken by the car to cover the distance of 75 km with speed x.
(75/(3/2)x)denotes time taken by train to cover the 75 km distance with speed 3/2x.
in the problem it is said that train rest for 12.5 min at station
and both reach 75 km at same time.
There fore difference between the time is (75/x)-(75/(3/2)x)=12.5/60 in hrs.
this means time taken by the car to cover the distance of 75 km with speed x.
(75/(3/2)x)denotes time taken by train to cover the 75 km distance with speed 3/2x.
in the problem it is said that train rest for 12.5 min at station
and both reach 75 km at same time.
There fore difference between the time is (75/x)-(75/(3/2)x)=12.5/60 in hrs.
Rameez said:
1 decade ago
Let t' is the exact time of the train without stopping...
Let t is the exact time of the car
12.5 minutes is the wastage of time because of train stopping
Since in given problem it is said that both car & train arrives at same time....
Therefore t'+(125/600)=t
75/(3x/2)+125/600=75/x
75/x-75/(3x/2)=125/600
By solving it we get x=120Kmph
Let t is the exact time of the car
12.5 minutes is the wastage of time because of train stopping
Since in given problem it is said that both car & train arrives at same time....
Therefore t'+(125/600)=t
75/(3x/2)+125/600=75/x
75/x-75/(3x/2)=125/600
By solving it we get x=120Kmph
Hemant Sharma said:
1 decade ago
Hi, its right, I'm little confuse that why calculate difference b/w both but car lost time 12.5.so the difference b/w boths time is equal to the lost time | i.e 12.5 is equal to the 125/10 and change into hrs so 125/(60*10) .
Prawinn said:
1 decade ago
Simple train and car reach the place in a same time if car takes 8 hours to reach the place train also take 8 hours that 8 hour be t+somewaste of time the diff b/w both train and car is 0.
So, t1-(t+x) = 0
=> t1-t = x .
So, t1-(t+x) = 0
=> t1-t = x .
Kiru said:
1 decade ago
Thanks senhil I understand easily.
Sreekanth said:
1 decade ago
Think... Y 3/2x came with an example...
if u think car has x speed, then train ll have 2x.
then substitute x with simple value..
x:2x
if x=10 then
10:20
it mean train has double value of car speed.
but x:3/2x
then 10:15x
In which train has 50% faster than car.
50% of car is 5. so 10+5=15.
if u think car has x speed, then train ll have 2x.
then substitute x with simple value..
x:2x
if x=10 then
10:20
it mean train has double value of car speed.
but x:3/2x
then 10:15x
In which train has 50% faster than car.
50% of car is 5. so 10+5=15.
Arun said:
1 decade ago
Rameez your explanation is great.
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