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 9 of 35.
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)
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
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.
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
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) .
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