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 10 of 35.
Hong Kong said:
9 years ago
But can anyone explain to me why can't I think like this:
Let x be the time for Bus to arrive point B, then x = 12.5x2 (since train 50% faster) = 25.
So we know the train bus spend 25 mins to arrive B, and 75km/25mins = 3km/mins = 180kmph?
Let x be the time for Bus to arrive point B, then x = 12.5x2 (since train 50% faster) = 25.
So we know the train bus spend 25 mins to arrive B, and 75km/25mins = 3km/mins = 180kmph?
Suvasish Singha said:
8 years ago
Let, Speed of the train be X kmph.
Than, Speed of the train= x + (x * 50/100).
= x + x/2,
= (2x+x) / 2,
= 3x/2.
Than, Speed of the train= x + (x * 50/100).
= x + x/2,
= (2x+x) / 2,
= 3x/2.
Shavra yaqub shah said:
5 years ago
T1 - T2 = 12.5 minutes
speed1/distance1 - speed2/distance2 = 12.5 minutes .
@All.
This is how you form the equation and solve the problem related to the time difference (early/late than usual),or speed difference or distance difference.
speed1/distance1 - speed2/distance2 = 12.5 minutes .
@All.
This is how you form the equation and solve the problem related to the time difference (early/late than usual),or speed difference or distance difference.
Jyothi said:
7 years ago
Here we take 150/100 because it is given 50% more and always we have to take 100 as a base value here it is 50%more so 100+50/100 = 150/100 = 3/2.
(If it is given less than 50% then 100-50/100 = 50/100 = 1/2.
I hope you all understand.
(If it is given less than 50% then 100-50/100 = 50/100 = 1/2.
I hope you all understand.
PALAK said:
4 years ago
Let the car speed be x.
Speed of train = 150/100x = 3/2xkmph(explain 25*3/25*2).
75/x-75/3/2x = 12.5/60.
75/x-75/3/2x = 125/600.
75/x-75*2/3x = 5/24.
75/x-150/3x = 5/24.
75/x-50/x = 5/24.
25/x = 5/24.
x = 25*24/5.
x = 5*24.
x = 120.
Speed of train = 150/100x = 3/2xkmph(explain 25*3/25*2).
75/x-75/3/2x = 12.5/60.
75/x-75/3/2x = 125/600.
75/x-75*2/3x = 5/24.
75/x-150/3x = 5/24.
75/x-50/x = 5/24.
25/x = 5/24.
x = 25*24/5.
x = 5*24.
x = 120.
(1)
Ashish Katoch said:
1 decade ago
150/100 come using this strategy.
<-------------simple logic:------------->
Suppose car speed =x km/hr.
Then train speed= x + 0.5x.
1.5x or 150/100 or 3/2.
{0.5x comes by breaking the 1x into half (50%) i.e 1/2 i.e 0.5}.
<-------------simple logic:------------->
Suppose car speed =x km/hr.
Then train speed= x + 0.5x.
1.5x or 150/100 or 3/2.
{0.5x comes by breaking the 1x into half (50%) i.e 1/2 i.e 0.5}.
MikesMatE said:
8 years ago
@ALL.
150x/100 is to say that if the car travelled at x speed which is 100%x=x, then the train was 50% faster than that, 2x is an addition of 100% to x not 50%.
Therefore to add 50% to x would be to add 1/2x which makes it 3/2x.
150x/100 is to say that if the car travelled at x speed which is 100%x=x, then the train was 50% faster than that, 2x is an addition of 100% to x not 50%.
Therefore to add 50% to x would be to add 1/2x which makes it 3/2x.
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) .
Prawinn said:
1 decade ago
Simple train and car reach the place in a same time if car takes 8 hours to reach the place train also take 8 hours that 8 hour be t+somewaste of time the diff b/w both train and car is 0.
So, t1-(t+x) = 0
=> t1-t = x .
So, t1-(t+x) = 0
=> t1-t = x .
ABHISHEK said:
2 months ago
The speed of car be x km/h.
Then speed of the train will be = 3x/2km/h,
and the time taken to cover up 75 km is equal, but the train lost about 12.5 minutes while stopping at the stations.
So 75/x - 50/x = 5/24,
x= 120km/h.
Then speed of the train will be = 3x/2km/h,
and the time taken to cover up 75 km is equal, but the train lost about 12.5 minutes while stopping at the stations.
So 75/x - 50/x = 5/24,
x= 120km/h.
(2)
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