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.
Firoz Rahman said:
1 decade ago
The Train lost 12.5 min to reach the destination.By this we understood that train took more time than car to reach the destination. So the car time should be subtracted from train's time and the result we get is -120. As speed will not be negative, the answer is 120. Is this right one?
Anonymous said:
5 years ago
@VIcky045.
Actually,
To take delay time of train we need to have express time in hr(given is in min).
To convert Minute to Hour (x minutes/60).
To convert Hour to Minute (x hours*60).
So,
Converting 12.5 mins to 12.5/60 hour.
125/10*60 also correct since it only removes decimals.
Actually,
To take delay time of train we need to have express time in hr(given is in min).
To convert Minute to Hour (x minutes/60).
To convert Hour to Minute (x hours*60).
So,
Converting 12.5 mins to 12.5/60 hour.
125/10*60 also correct since it only removes decimals.
Sarath said:
1 decade ago
In one way it can be understood as difference of times is 12.5 min.
i.e 75/x-50/x = 12.5 min.
But, if the time taken by train is excess 12.5 mins then car reaches before 12.5 mins.
We write this as car time + 12.5 mins excess = Train time i.e 75/x+ 12.5 min = 50/x.
i.e 75/x-50/x = 12.5 min.
But, if the time taken by train is excess 12.5 mins then car reaches before 12.5 mins.
We write this as car time + 12.5 mins excess = Train time i.e 75/x+ 12.5 min = 50/x.
Shubham Boni said:
5 years ago
The Answer method, I correct with 1 simple mistake in calculations.
75/x - 75/(3x/2) = 12.5 minutes.
Now, the equation is correct but Now since time is in Minutes we need to convert it into seconds.
We Know, 1min = 60seconds this means, 12.5min = (12.5X 60)seconds.
75/x - 75/(3x/2) = 12.5 minutes.
Now, the equation is correct but Now since time is in Minutes we need to convert it into seconds.
We Know, 1min = 60seconds this means, 12.5min = (12.5X 60)seconds.
Khanchana said:
1 decade ago
Suppose speed of the car = xkm/hr(1)
speed of the train = 50% more than that of train
= x + 50/100x
= (100x + 50x)/100
= 150x/100 = 3/2x ------------>(2)
delay of train = 12.5 mins= 12.5 * 10/60*10 hrs
speed of the train = 50% more than that of train
= x + 50/100x
= (100x + 50x)/100
= 150x/100 = 3/2x ------------>(2)
delay of train = 12.5 mins= 12.5 * 10/60*10 hrs
Tarun said:
1 decade ago
Its easy guys OK I explain. Let speed of car is X.
Then speed of train is 150/100 multiply by X. Then after solve this 3/2X kmph.
Taking difference between car and train speed minutes.
= 75/x-75/3/2x = 12.5/60. Then speed of car means X is equal to 120 kmph.
Then speed of train is 150/100 multiply by X. Then after solve this 3/2X kmph.
Taking difference between car and train speed minutes.
= 75/x-75/3/2x = 12.5/60. Then speed of car means X is equal to 120 kmph.
Ankit said:
9 years ago
Let me explain : assume car travel at speed x so speed of train 50% more than car 50/100 * x or 0.5 x.
So total speed of train faster than car is x + 0.5x which is 1.5x. Meaning of 150/100 * x = 1.5x.
So total speed of train faster than car is x + 0.5x which is 1.5x. Meaning of 150/100 * x = 1.5x.
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?
Saranya said:
1 decade ago
@maheshwaran
take x as a spd of car
T=75/x -->*
spd of train is x+x/2(train 50% faster than car)
and time reduce by 12.5 min( by qus)
12.5min=12.5/60hr= 2.5/12hr
T-2.5/12=75/(x+x/2)
T=(75/x+0.5x)+ 2.5/12
-->#
by solving * and# we will get ans as 120
take x as a spd of car
T=75/x -->*
spd of train is x+x/2(train 50% faster than car)
and time reduce by 12.5 min( by qus)
12.5min=12.5/60hr= 2.5/12hr
T-2.5/12=75/(x+x/2)
T=(75/x+0.5x)+ 2.5/12
-->#
by solving * and# we will get ans as 120
Bdboy Hasan said:
9 years ago
Let,
Time needed for car = 3x minute.
Time needed for train = 2x minute.
3x - 2x = 12.5 minute.
So, x = 12.5 minute.
Time needed for car = 3*12.5.
= 37.5 minute.
So, speed of car = 75km/37.5 minute.
= 2 km per minute.
= 60 * 2kmph.
= 120 kmph.
Time needed for car = 3x minute.
Time needed for train = 2x minute.
3x - 2x = 12.5 minute.
So, x = 12.5 minute.
Time needed for car = 3*12.5.
= 37.5 minute.
So, speed of car = 75km/37.5 minute.
= 2 km per minute.
= 60 * 2kmph.
= 120 kmph.
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