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 28 of 35.
Meetali said:
1 decade ago
If speed is x, how do you'll apply the formula in 75/x i.e distance upon speed gives time isn't it?
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
Maheshwaran said:
1 decade ago
HOW 150 comes in the initial step ?
Rahul said:
1 decade ago
It's a very easy question
suppose train running 12.5 min. & lost 12.5 mint in station
s=d/t====>
s=(75/25*60)km/s
s=1/20km/s change in m/sec
s=1000/20 m/sec =50 m/sec
change in km/hr
50*18/5=180km/h train speed.....in option of this question
only 120 car speed 50% is satisfy train speed 120+120/2=180 that's it......
suppose train running 12.5 min. & lost 12.5 mint in station
s=d/t====>
s=(75/25*60)km/s
s=1/20km/s change in m/sec
s=1000/20 m/sec =50 m/sec
change in km/hr
50*18/5=180km/h train speed.....in option of this question
only 120 car speed 50% is satisfy train speed 120+120/2=180 that's it......
Rajan said:
1 decade ago
Say, the speed of the car be 1 km/hr (0. 5 +0. 5).
And if there is 50% increase in speed means (1 +0. 5) = 1. 5 km /hr.
That's how 3/2 came.
And if there is 50% increase in speed means (1 +0. 5) = 1. 5 km /hr.
That's how 3/2 came.
Dhinesh said:
1 decade ago
Let Speed of the car=x.
Train Speed=Speed of car+50% (Speed of car).
=> x + 50/100 (x).
=> x + x/2.
=> 3x/2.
Train Speed=Speed of car+50% (Speed of car).
=> x + 50/100 (x).
=> x + x/2.
=> 3x/2.
Laxmi priya said:
1 decade ago
Let the speed of the car be 100% and the train is 50% more
100%+50%=150%
that is 150/100
100%+50%=150%
that is 150/100
Shro said:
1 decade ago
I am not understanding the main calculation part i.e. steps after finding speed pf the train.
Somebody please explain me the steps of main calculation
Somebody please explain me the steps of main calculation
Shro said:
1 decade ago
Thanks Santhiraju :).
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?
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