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:
345 comments Page 14 of 35.
Rohit said:
1 decade ago
Let speed of car be = x km/hr.
Then, speed of train will be = 50% more than car.
50% means half of speed of car.
Half of car speed can be written as "x/2.
Then, speed of train will be = x+x/2 = 3x/2.
Total distance = 75km.
Then, according to question Time = Distance/Speed,
75/3x/2-75/x = 12.5.
300/3x-75/x = 125/10/60.
300-225/3x = 125/600.
75/3x = 5/24.
3x*5 = 75*24.
3x = 75*24/5.
Therefore, x = 75*24/15.
x = 5*24 = 120.
Then, speed of train will be = 50% more than car.
50% means half of speed of car.
Half of car speed can be written as "x/2.
Then, speed of train will be = x+x/2 = 3x/2.
Total distance = 75km.
Then, according to question Time = Distance/Speed,
75/3x/2-75/x = 12.5.
300/3x-75/x = 125/10/60.
300-225/3x = 125/600.
75/3x = 5/24.
3x*5 = 75*24.
3x = 75*24/5.
Therefore, x = 75*24/15.
x = 5*24 = 120.
Fazil said:
1 decade ago
Can't we say speed of train is 50% faster than car.
If speed of car is x, then train must move with 2x speed. Is it?
If speed of car is x, then train must move with 2x speed. Is it?
Mukund said:
1 decade ago
I think the question is technically wrong. It is given that both the car & train reach point B at the same time.
Then what is the significance of lagging of train by 12.5 min. And if it is so (if they reach in same time) then the equation this should be valid.
75/x = 75/x+ (x/2).
Where x = Speed of car.
x+(x/2) = Speed of train.
Then what is the significance of lagging of train by 12.5 min. And if it is so (if they reach in same time) then the equation this should be valid.
75/x = 75/x+ (x/2).
Where x = Speed of car.
x+(x/2) = Speed of train.
Aliya said:
1 decade ago
Its not in same time it is at same time if we consider car starts at 10 to reach at 12 then train also have at same time but it is only possible with different speed if have time lag.
NITISH GULERIA said:
1 decade ago
Hello friends don't confuse 50% more means x+50% = x+50/100, = x+1/2.
= LCM are 2 and whole equation are 3x/2.
= LCM are 2 and whole equation are 3x/2.
Shyam said:
1 decade ago
Can anyone explain in detail?
Kishan B said:
10 years ago
Here is simple way of solving this problem.
We have Velocity = Distance/Time.
For car let x be the speed, then speed of the train is 3/2x as per the problem.
Now x = 75/t for car.
1.5x = 75/(t-(5/24)) as train stops for a duration of 5/24 hours.
5/24 obtained by converting 12.5 minute to hour.
Now divide two equations resulting in:
1/1.5 = (24t-5)/24t which upon solving gives t = 0.625.
Put t value in v = d/t where d = 75 km and t = 0.625. Thus you get v = 120 kmph.
Hope this was useful.
We have Velocity = Distance/Time.
For car let x be the speed, then speed of the train is 3/2x as per the problem.
Now x = 75/t for car.
1.5x = 75/(t-(5/24)) as train stops for a duration of 5/24 hours.
5/24 obtained by converting 12.5 minute to hour.
Now divide two equations resulting in:
1/1.5 = (24t-5)/24t which upon solving gives t = 0.625.
Put t value in v = d/t where d = 75 km and t = 0.625. Thus you get v = 120 kmph.
Hope this was useful.
Abhishek Banik said:
10 years ago
A to B=75 km.
Distance T = Time.
S = Speed, D = Distance.
Let S of car = x kmph.
S of train = 50% of car speed = x+{(50/100)*x}.
= x+{(1/2)*x}.
= x+x/2 = 3x/2.
T = D/S.
T of car - T of train = 12.5 mins as car speed is low that of train speed.
= 75/x-75/(3x/2) = 12.5/60 (converting 12.5 mins to 12.5/60 hr).
= 75/x-75*2/3x = 12.5/60.
= 75/x-50/x = 12.5/60.
= 25/x = 12.5/60.
x = 25*60/12.5 = 25*60*10/125 = 120.
x = 120 kmph.
Distance T = Time.
S = Speed, D = Distance.
Let S of car = x kmph.
S of train = 50% of car speed = x+{(50/100)*x}.
= x+{(1/2)*x}.
= x+x/2 = 3x/2.
T = D/S.
T of car - T of train = 12.5 mins as car speed is low that of train speed.
= 75/x-75/(3x/2) = 12.5/60 (converting 12.5 mins to 12.5/60 hr).
= 75/x-75*2/3x = 12.5/60.
= 75/x-50/x = 12.5/60.
= 25/x = 12.5/60.
x = 25*60/12.5 = 25*60*10/125 = 120.
x = 120 kmph.
Ravi said:
10 years ago
Can you explain please?
Pallavi said:
10 years ago
Can you explain from second step?
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