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 32 of 35.
Divs said:
1 decade ago
Raj would you please tell me how come the speed be 25?
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.
Jinto said:
1 decade ago
Can anybody explain if both the vehicle reaches at the same time then where is the delay arises i am talking about if two vehicles reaches at the same time then how this equation came
(75/x)-(75/(3/2)x)=12.5/60, 12.5 minutes delay will not be ther if both the vehicles reaches at the same time pls help
(75/x)-(75/(3/2)x)=12.5/60, 12.5 minutes delay will not be ther if both the vehicles reaches at the same time pls help
Sushil pal said:
1 decade ago
Let Car's speed be x Kmph
Train speed = (Carspeed + (50*Carspeed)/100)
" = (x+(50x)/100)
" = 150x/100
" = 3/2xKmph
Now the train lost about 12.5min, so time subtraction betn Car and train will give us distance covered by car in that much period of time (lost period)and to know the speed we have to equate it with lost period
Distance covered = lostPeriod
75/x-75/(3/2)x = 125/600
75/x-50/x = 5/24
x = (25*24/5)=120 Kmph.
Train speed = (Carspeed + (50*Carspeed)/100)
" = (x+(50x)/100)
" = 150x/100
" = 3/2xKmph
Now the train lost about 12.5min, so time subtraction betn Car and train will give us distance covered by car in that much period of time (lost period)and to know the speed we have to equate it with lost period
Distance covered = lostPeriod
75/x-75/(3/2)x = 125/600
75/x-50/x = 5/24
x = (25*24/5)=120 Kmph.
Shunmuga priya said:
1 decade ago
Sonam Jain said that " (125/10*60)it takes the difference b/w boths time is equal to the lost time "
But in question it is clearly given as "from point A at the same time and reach point B 75 kms away from A at the same time". The car and train starting and ending time is same so no need of subtraction but they subtracted here.
I cant get this. please explain
But in question it is clearly given as "from point A at the same time and reach point B 75 kms away from A at the same time". The car and train starting and ending time is same so no need of subtraction but they subtracted here.
I cant get this. please explain
Suja said:
1 decade ago
@ jamal:
We know that time=(distance/speed).
Here distance=75kmph and speed=3x/2.
Therefore time=75/(3x/2)
We know that time=(distance/speed).
Here distance=75kmph and speed=3x/2.
Therefore time=75/(3x/2)
Jamal akhter said:
1 decade ago
Could you explain how time for train you had taken 75/(3/2.x) ?
Prasad said:
1 decade ago
Train speed is 50% more than car, so it is 1/2 % more speed than car .
So it is 1 and 1/2 == 3/2 ( as car 100% it will be 1).
So it is 1 and 1/2 == 3/2 ( as car 100% it will be 1).
Senhil said:
1 decade ago
Given train is 50% greater than car. Let car speed will be 50km/hr then train speed ll be 100km/hr. So 100+50=150 km/hr.
Rohita said:
1 decade ago
Explain the simple way, I can't understand train speed.
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