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:
350 comments Page 7 of 35.
Sonam jain said:
2 decades ago
In 125/10*60
Actually what happen in the question they said train lost 12.5 min. so first of all we change it in hour thats y divided by 60. now bcz it lost thats y it takes the difference b/w boths time is equal to the lost time. 12.5 is equal to the 125/10
So final eq is diff b/w time of train and car is = to lost time i.e. 125/10*60
Actually what happen in the question they said train lost 12.5 min. so first of all we change it in hour thats y divided by 60. now bcz it lost thats y it takes the difference b/w boths time is equal to the lost time. 12.5 is equal to the 125/10
So final eq is diff b/w time of train and car is = to lost time i.e. 125/10*60
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.
Kishorebabu said:
1 year ago
Convert km/h to m/s.
S(c) = 75000/T -->(1)
S(tr) = 75000/(T-12.5x60)-->(2)
S(tr) = 3/2S(c) -->(3)
Sub (3) in (2).
3/2s(c) = 75000/(T-12.5x60).
sub (1) in above;
(3/2).(75000/T) = 75000/(T-12.5x60).
T = 3x12.5x60 ---->(4)
Sub (4 in (1) [to get the speed of the car].
S(c) = (75000/3x12.5x60 ).(18/5).
s(c) = 120km/h.
S(c) = 75000/T -->(1)
S(tr) = 75000/(T-12.5x60)-->(2)
S(tr) = 3/2S(c) -->(3)
Sub (3) in (2).
3/2s(c) = 75000/(T-12.5x60).
sub (1) in above;
(3/2).(75000/T) = 75000/(T-12.5x60).
T = 3x12.5x60 ---->(4)
Sub (4 in (1) [to get the speed of the car].
S(c) = (75000/3x12.5x60 ).(18/5).
s(c) = 120km/h.
(15)
Smith matsiko said:
1 decade ago
Since the train is 50% faster than than the car and also delayed for 12.5 minutes,
It implies it would have used the time delayed to cover half of the distance=75/2 km.
Therefore its speed =180 km/hr (75/2*60/12.5).
Since trains' speed assuming x to be the speed of the car,
Then x + 0.5x = 180 km,
Thus x = 120 km/hr.
It implies it would have used the time delayed to cover half of the distance=75/2 km.
Therefore its speed =180 km/hr (75/2*60/12.5).
Since trains' speed assuming x to be the speed of the car,
Then x + 0.5x = 180 km,
Thus x = 120 km/hr.
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......
GIS said:
1 decade ago
Let Speed of Car = x.
Train is 50% faster than car = x + x/2 = 3x/2.
Time difference = 12. 5 mins = 12. 5/60 hours = 125/600 = 5/24.
Distance = 75 km.
(Distance/Car speed) - (Distance/Train speed) = 12. 5 min = 5/24 hr.
(75/x) - (75/ (3/2x) ) = 5/24.
(75/x) - (150/3x) = 5/24.
25/x = 5/24.
X = (25*24) /5 = 120kmph.
Train is 50% faster than car = x + x/2 = 3x/2.
Time difference = 12. 5 mins = 12. 5/60 hours = 125/600 = 5/24.
Distance = 75 km.
(Distance/Car speed) - (Distance/Train speed) = 12. 5 min = 5/24 hr.
(75/x) - (75/ (3/2x) ) = 5/24.
(75/x) - (150/3x) = 5/24.
25/x = 5/24.
X = (25*24) /5 = 120kmph.
Piyush said:
4 years ago
@All.
So many of us are confused in where the hell this (150x/100) came from right.
So, my simple explanation is that see the train is 50% faster than car.
Now, consider car speed is X so the train speed will be (X+ 50% of X) Now calculate the train speed.
And you will get (X + 50X/100) = 150X/100.
Hope this helps.
So many of us are confused in where the hell this (150x/100) came from right.
So, my simple explanation is that see the train is 50% faster than car.
Now, consider car speed is X so the train speed will be (X+ 50% of X) Now calculate the train speed.
And you will get (X + 50X/100) = 150X/100.
Hope this helps.
(9)
Singh said:
2 decades ago
Here percentage is given means we have to take any unknown quantity as 100 thats why x is taken as 100. here train speed is 50% more than car means it travels with 100+100/2 speed.total speed is 150.now covert 150 into percentages as 150/100.now 150/100=3/2.if car speed is let xkm/h.then train speed is 3/2x km/h.
DevB007 said:
8 years ago
Train speed = 50% faster than car speed.
Train speed = car speed + extra 50 % of car speed.
(car speed = X),
(50% extra of X = X/2),
Train speed = X + X/2 = (X/1 + X/2) = (2X+X)/2;
Train speed = 3X/2.
Train speed = car speed + extra 50 % of car speed.
(car speed = X),
(50% extra of X = X/2),
Train speed = X + X/2 = (X/1 + X/2) = (2X+X)/2;
Train speed = 3X/2.
Deepak porwal said:
1 decade ago
As the train speed is 50% more than car
So train speed is (car speed + 50% of car speed)
i.e. (x+(50/100)*x)= 3x/2;
now time diffrence between them is 12.5 minute
converting it in to hour 12.5/60;
now (distance/speed of train) - (distance /speed of car)= 12.5/60;
as time diffrence between them is 12.5 min;
So train speed is (car speed + 50% of car speed)
i.e. (x+(50/100)*x)= 3x/2;
now time diffrence between them is 12.5 minute
converting it in to hour 12.5/60;
now (distance/speed of train) - (distance /speed of car)= 12.5/60;
as time diffrence between them is 12.5 min;
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