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:
354 comments Page 36 of 36.
SutharsanBabu said:
3 months ago
Most Members doubt is how come this speed equation =(3/2)x.
speed =x.
Eg 1:
Car speed =50.
Train speed =>50 + (50/2)=> 50 + 25 = 75
Speed equation => train speed/car speed => 75/50 =3/2.
Ex 2:
Car speed =100.
Train speed =>100+(100/2) => 100 + 50 = 150.
Speed equation =>train speed/car speed => 150/100 = 3/2.
Time equation for train:
However, the train lost about 12.5 minutes while stopping at the stations.
Train's Total Time = Run time + Stopping time.
Convert 12.5 minutes in Hours as follows by 12.5 => 12.5/60 => 125/600 => 25/120 = 5/24.
Want to find the real speed of both the car and the train:
Formula : time = distance/speed.
speed = x.
car = train.
75/x = 75/(3/2)x + 5/24 simplify ->(75/1)*(2/3)->150/3=50/1.
75/x = 50/1x + 5/24 -> 1x and x both are same.
(75/x)-(50/x) = 5/24.
75-50/x = 5/24.
25x = 5/24.
x = 5/24/25, simplify -> (24*25)/5 -> 600/5 -> 120/1.
x= 120/1, don't consider 1.
Answer = 120.
speed =x.
Eg 1:
Car speed =50.
Train speed =>50 + (50/2)=> 50 + 25 = 75
Speed equation => train speed/car speed => 75/50 =3/2.
Ex 2:
Car speed =100.
Train speed =>100+(100/2) => 100 + 50 = 150.
Speed equation =>train speed/car speed => 150/100 = 3/2.
Time equation for train:
However, the train lost about 12.5 minutes while stopping at the stations.
Train's Total Time = Run time + Stopping time.
Convert 12.5 minutes in Hours as follows by 12.5 => 12.5/60 => 125/600 => 25/120 = 5/24.
Want to find the real speed of both the car and the train:
Formula : time = distance/speed.
speed = x.
car = train.
75/x = 75/(3/2)x + 5/24 simplify ->(75/1)*(2/3)->150/3=50/1.
75/x = 50/1x + 5/24 -> 1x and x both are same.
(75/x)-(50/x) = 5/24.
75-50/x = 5/24.
25x = 5/24.
x = 5/24/25, simplify -> (24*25)/5 -> 600/5 -> 120/1.
x= 120/1, don't consider 1.
Answer = 120.
(5)
Udit Sharma said:
3 months ago
Speed of train: speed of car = 150: 100 = 3: 2.
(As speed is inversely proportional to time)
Train time: car time = 2 : 3.
1 part = 12.5min, 2part = 25min, 3part = 37.5min.
Convert it into hr 375/600 = 5/8 hr.
Speed = distance/time.
Speed of car = 75/(5/8) = 75 * 8/5 = 120km/hr.
(As speed is inversely proportional to time)
Train time: car time = 2 : 3.
1 part = 12.5min, 2part = 25min, 3part = 37.5min.
Convert it into hr 375/600 = 5/8 hr.
Speed = distance/time.
Speed of car = 75/(5/8) = 75 * 8/5 = 120km/hr.
(30)
Devaraj R A said:
1 month ago
For every 10km he walked, he walked an extra 4km.
So the question is, if he walked an extra 20 km, what would be the actual distance he covers?
Then 10 × 20/4 = 50.
So the question is, if he walked an extra 20 km, what would be the actual distance he covers?
Then 10 × 20/4 = 50.
(2)
Siddardha said:
3 weeks ago
@ALL.
Time, distance, and speed problems always assume one variable as constant and calculate the others.
See in this question, started at A and reached at B (75 km).
Both the car and the train reached point B at the same time, even though their speeds are not the same, because the train had waited 12.5 mins in between. So the train has travelled 12.5 minutes less than the total time taken by car.
So, the difference in their times is just 12.5 mins.
t1 - t2 = 12.5 mins (t=d/s)
d1/s1 - d2/s2 = 12.5 mins. ( let x is speed of the car, and 50% more will be 1.5x )
75/x - 75/(1.5x) = 12.5/60 hr.
Solve this as it has only 1 unknown, we will get the speed of the car.
Time, distance, and speed problems always assume one variable as constant and calculate the others.
See in this question, started at A and reached at B (75 km).
Both the car and the train reached point B at the same time, even though their speeds are not the same, because the train had waited 12.5 mins in between. So the train has travelled 12.5 minutes less than the total time taken by car.
So, the difference in their times is just 12.5 mins.
t1 - t2 = 12.5 mins (t=d/s)
d1/s1 - d2/s2 = 12.5 mins. ( let x is speed of the car, and 50% more will be 1.5x )
75/x - 75/(1.5x) = 12.5/60 hr.
Solve this as it has only 1 unknown, we will get the speed of the car.
(5)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers