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 11 of 35.
R.Sreekanth said:
1 decade ago
@Yesoda
They given speed is 50% more than car so,
We don't know car speed,let consider car speed as X.
So, the train speed is 50% more than car(i.e, when ever we talking about anything in the form of % we must add or subtract from 100. Then convert it into [total/100], total= % + 100 ).
Therefore: the train speed is [(50+100)/100] more speed of car.
So, the train speed is = (150/100)*x.
They given speed is 50% more than car so,
We don't know car speed,let consider car speed as X.
So, the train speed is 50% more than car(i.e, when ever we talking about anything in the form of % we must add or subtract from 100. Then convert it into [total/100], total= % + 100 ).
Therefore: the train speed is [(50+100)/100] more speed of car.
So, the train speed is = (150/100)*x.
Kasi Srinivas said:
1 decade ago
@ mysterious.
Thank you dude :) I understood.
Thank you dude :) I understood.
Raul said:
1 decade ago
100*(t)=150*(t+12.5)
t=37.5min
t
=37.5/60hr
speed=(75*60)/37.5
=120
t=37.5min
t
=37.5/60hr
speed=(75*60)/37.5
=120
Bhupat said:
1 decade ago
All righ but my question is. No body has taken 12.5 minus loss in train speed calculation. Why?
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
Sravya said:
1 decade ago
Thank you so much @kanchana.
Sravya said:
1 decade ago
@kanchana
How did x+50/100 became 100x+50x as it will become (100x+50)/100x ?
How did x+50/100 became 100x+50x as it will become (100x+50)/100x ?
Amit said:
1 decade ago
Sarvya Kanchna is right:-)
dont consider x with 100 consider with 50 like this.
=x+50x/100
=(100x+50x)/100
=150x/100
If you consider x with 100 then when we take LCM then eq.
=(100x^2+50)/100x
:-)
dont consider x with 100 consider with 50 like this.
=x+50x/100
=(100x+50x)/100
=150x/100
If you consider x with 100 then when we take LCM then eq.
=(100x^2+50)/100x
:-)
Dhruva & hitesh said:
1 decade ago
First we assume that the speed of car is 100% so as per Q.speed of train is 150%.
Ajit said:
1 decade ago
In given problem ,u ve to equate total or exact time of both...
Let t be time tkn by car to cover 75kms... n t' be time takn by train...
i.e. train's total time=stoppage time+time in which it covers 75kms...nw t=t'....
Let t be time tkn by car to cover 75kms... n t' be time takn by train...
i.e. train's total time=stoppage time+time in which it covers 75kms...nw t=t'....
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