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 | ![]() |
100 | 2 |
![]() |
75 | - | 75 | = | 125 |
x | (3/2)x | 10 x 60 |
![]() |
75 | - | 50 | = | 5 |
x | x | 24 |
![]() |
![]() |
25 x24 | ![]() |
= 120 kmph. |
5 |
Discussion:
342 comments Page 30 of 35.
Anandhu said:
6 years ago
Can anybody tell me that, In the question, it states that the car and the train start at the same time from A and reach at the same time a B. Which means they took the same time to reach point B from A, so how can we take the difference of time taken by them?
Sandhya said:
6 years ago
Can you explain 125/10*60 how it got or is there any simple method for answering this question.
Jasmine said:
6 years ago
Distance between A and B = 75kms.
Let 'x' be the speed of the car.
Given that the train can travel 50% faster than the car i.e.,
Car speed = x.
Train speed = x+x*50/100 = 3/2*x.
Distance = speed*time.
Tc(Time of the car) = 75/x.
Tt(Time of the train) = 75/(3/2*x)+12.5 min(while stopping at the station).
=>75/(3/2*x)+12.5/60(into hours).
=>75/x=75/(3/2*x)+12.5/60.
By solving x = 120kmph.
Let 'x' be the speed of the car.
Given that the train can travel 50% faster than the car i.e.,
Car speed = x.
Train speed = x+x*50/100 = 3/2*x.
Distance = speed*time.
Tc(Time of the car) = 75/x.
Tt(Time of the train) = 75/(3/2*x)+12.5 min(while stopping at the station).
=>75/(3/2*x)+12.5/60(into hours).
=>75/x=75/(3/2*x)+12.5/60.
By solving x = 120kmph.
Anas said:
6 years ago
Both trains start at the same time and reach the destination at the same time. So the time taken by both trains should be the same right? So the difference between the times should be zero. How come it is 125/10*60?
Sreenadh reddy said:
6 years ago
The train speed is 50% faster than the car. If the car is moving in 100 then the train moves 50% fast i.e. (100car speed+ 50faster speed) so they taken 150/100.
Hannu said:
6 years ago
If the speed of car is 100 then 50% of 100 is 50.
And the speed of the train is 50% more than of car speed.
so 150*x/100.
And the speed of the train is 50% more than of car speed.
so 150*x/100.
KArthik said:
6 years ago
How did 75/x -75/(3/2x) = 125/(10*60) came? Please explain this.
Sandhu said:
6 years ago
x+50% of x's speed.
x+50x/100 = 150x/100.
x+50x/100 = 150x/100.
Sameer said:
5 years ago
Let the speed of the car=x kmph.
Then, speed of the train is 50% more than car = 50x/100 + x = 3x/2 kmph. (50x/100+x/1=50x+100x/100 =150x/100= 3x/2).
Time is taken by the car to travel from A to B= 75x hours.
Time is taken by the train to travel from A to B=75/ (3x/2) +12.5/60 hours. 1 min = 1/60 hrs. So 12.5/60 hrs.
Since both start from A at the same time and reach point B at the same time,
75/x =75/ (3x/2) +12.5/60.
75/x = 150/3x + 12.5/60.
75/x= 50/x + 12.5/60.
75/x-50/x = 12.5/60.
25/x =12.5/60.
x=25*60/12.5...... 1/x=12. 5/60*25 so, x = 60*25/12.5.
X=2*60.
x = 120.
Then, speed of the train is 50% more than car = 50x/100 + x = 3x/2 kmph. (50x/100+x/1=50x+100x/100 =150x/100= 3x/2).
Time is taken by the car to travel from A to B= 75x hours.
Time is taken by the train to travel from A to B=75/ (3x/2) +12.5/60 hours. 1 min = 1/60 hrs. So 12.5/60 hrs.
Since both start from A at the same time and reach point B at the same time,
75/x =75/ (3x/2) +12.5/60.
75/x = 150/3x + 12.5/60.
75/x= 50/x + 12.5/60.
75/x-50/x = 12.5/60.
25/x =12.5/60.
x=25*60/12.5...... 1/x=12. 5/60*25 so, x = 60*25/12.5.
X=2*60.
x = 120.
Jarvis said:
5 years ago
If it's 50x.
It says that train travels in 50% less speed than the car.
It says that train travels in 50% less speed than the car.
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