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 6 of 35.
Ayush said:
3 years ago
Please anyone can explain since the speed of the train is 1.5 * speed of the car, so how in the solution we are taking that difference of time is 12.5 min. How we come to know that if 12.5min added to the time taken by car will equal to the time taken by train?
Since the speed of both is different so how can they have time taken by car = T and by train =T+12.5 * 60.
Since the speed of both is different so how can they have time taken by car = T and by train =T+12.5 * 60.
Vivek said:
7 years ago
The speed of car : speed of train :100:150=2:3.
The time needed for the car: time needed for train =3:2.
i.e., the train only takes 23 of the time taken by car.
Since both the car and train start and reach at the same time,
13 of the time needed by car is 12.5 minutes.
Time needed by the car =3*12.5 min.
Therefore, speed of the car =75(3 * 12.560)=120 km/hr.
The time needed for the car: time needed for train =3:2.
i.e., the train only takes 23 of the time taken by car.
Since both the car and train start and reach at the same time,
13 of the time needed by car is 12.5 minutes.
Time needed by the car =3*12.5 min.
Therefore, speed of the car =75(3 * 12.560)=120 km/hr.
Ashish patil said:
7 years ago
Dist = speed * time.
Let, the speed of car=100x.
Speed of train=150x (as 50% faster).
So, Time taken by car= dist/speed=75/100x.
And by train=75/150x.
And train lost 12.5 min=12.5/60 hr =125/600 hr.
This time is equal to the time taken b/w car and train i.e;
75/100x - 75/150x =125/600.
Solving this eq we get x=6/5.
So, the speed of car is=100*(6/5)= 120km/hr.
Let, the speed of car=100x.
Speed of train=150x (as 50% faster).
So, Time taken by car= dist/speed=75/100x.
And by train=75/150x.
And train lost 12.5 min=12.5/60 hr =125/600 hr.
This time is equal to the time taken b/w car and train i.e;
75/100x - 75/150x =125/600.
Solving this eq we get x=6/5.
So, the speed of car is=100*(6/5)= 120km/hr.
Rakesh Selvam A said:
3 years ago
By ratio of Speed.
T :C = 150: 100,
= 3:2.
Since, Distance is constant.
Time ratio = 2:3.
Diff between time ratios 2 and 3 is 1, which means 1 = 12.5 mins.
1 -----> 12.5 min,
3------> 37.5min,
2 ------> 25 min,
,
Speed of Car = Distance/Time.
= (75 /37.5) * 60( For min to hr),
= 120km/hr.
T :C = 150: 100,
= 3:2.
Since, Distance is constant.
Time ratio = 2:3.
Diff between time ratios 2 and 3 is 1, which means 1 = 12.5 mins.
1 -----> 12.5 min,
3------> 37.5min,
2 ------> 25 min,
,
Speed of Car = Distance/Time.
= (75 /37.5) * 60( For min to hr),
= 120km/hr.
(94)
Pavan said:
2 years ago
Let time is t.
Suppose Car is going at speed x,
Train speed will be x+(x/2) = 3x/2 -------> 1
Speed of Train(dist/time) = 75/(t-12.5) ------> 2
1 =2.
3x/2 = 75/t ----->3
Speed of car = 75/t.
x = 75/t ----> 4.
substitute 4 in 3.
By solving you get;
time of car(t) = 37.5(in min)
By converting into hr.
t = 0.625 hr.
Speed of car = 75/0.625.
Ans : 120kmph
Suppose Car is going at speed x,
Train speed will be x+(x/2) = 3x/2 -------> 1
Speed of Train(dist/time) = 75/(t-12.5) ------> 2
1 =2.
3x/2 = 75/t ----->3
Speed of car = 75/t.
x = 75/t ----> 4.
substitute 4 in 3.
By solving you get;
time of car(t) = 37.5(in min)
By converting into hr.
t = 0.625 hr.
Speed of car = 75/0.625.
Ans : 120kmph
(62)
Trishita said:
7 years ago
Dist=speed*time.
Let, speed of car=100x.
Speed of train=150x (as 50% faster).
So..Time taken by car= dist/speed=75/100x.
And by train=75/150x.
And train lost 12.5 min=12.5/60 hr =125/600 hr.
This time is equal to the time taken b/w car and train i.e;
75/100x - 75/150x =125/600.
Solving this eq we get x=6/5.
So, the speed of car is=100*(6/5)= 120km/hr.
Let, speed of car=100x.
Speed of train=150x (as 50% faster).
So..Time taken by car= dist/speed=75/100x.
And by train=75/150x.
And train lost 12.5 min=12.5/60 hr =125/600 hr.
This time is equal to the time taken b/w car and train i.e;
75/100x - 75/150x =125/600.
Solving this eq we get x=6/5.
So, the speed of car is=100*(6/5)= 120km/hr.
Rameez said:
1 decade ago
Let t' is the exact time of the train without stopping...
Let t is the exact time of the car
12.5 minutes is the wastage of time because of train stopping
Since in given problem it is said that both car & train arrives at same time....
Therefore t'+(125/600)=t
75/(3x/2)+125/600=75/x
75/x-75/(3x/2)=125/600
By solving it we get x=120Kmph
Let t is the exact time of the car
12.5 minutes is the wastage of time because of train stopping
Since in given problem it is said that both car & train arrives at same time....
Therefore t'+(125/600)=t
75/(3x/2)+125/600=75/x
75/x-75/(3x/2)=125/600
By solving it we get x=120Kmph
Astik kumar said:
7 years ago
Let the speed of the car be 100 then speed of the train be 150.
The ratio of the speed of car and train be = 2/3.
When the distance is constant the time is inversly proprotional,
So ratio of time = 3:2,
Diffrence in time =12.5min=5/24hours,
Total time is taken by car=3*5/24=15/24,
Speed = d/t => dis=75km,
Speed of car=75/(15/24)=120km/h.
The ratio of the speed of car and train be = 2/3.
When the distance is constant the time is inversly proprotional,
So ratio of time = 3:2,
Diffrence in time =12.5min=5/24hours,
Total time is taken by car=3*5/24=15/24,
Speed = d/t => dis=75km,
Speed of car=75/(15/24)=120km/h.
Sonam jain said:
1 decade 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:
10 years 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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers