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:
344 comments Page 20 of 35.
Siva said:
9 years ago
12.5/10 * 60,
= time/speed.
= 12.5/10 in minutes.
= 12.5/10 * 60 in seconds.
= time/speed.
= 12.5/10 in minutes.
= 12.5/10 * 60 in seconds.
Yeshi said:
9 years ago
Can anyone suggest me the logic? Please.
Gayuverma said:
9 years ago
Can this question be solved using relative speed concept? If no, why? If yes, please show it.
Bdboy Hasan said:
9 years ago
Let,
Time needed for car = 3x minute.
Time needed for train = 2x minute.
3x - 2x = 12.5 minute.
So, x = 12.5 minute.
Time needed for car = 3*12.5.
= 37.5 minute.
So, speed of car = 75km/37.5 minute.
= 2 km per minute.
= 60 * 2kmph.
= 120 kmph.
Time needed for car = 3x minute.
Time needed for train = 2x minute.
3x - 2x = 12.5 minute.
So, x = 12.5 minute.
Time needed for car = 3*12.5.
= 37.5 minute.
So, speed of car = 75km/37.5 minute.
= 2 km per minute.
= 60 * 2kmph.
= 120 kmph.
Rahul said:
9 years ago
Since the ratio of speed => train : car.
150 : 100 -----> A.
Then the ratio of time will be 100 : 150 ----> B.
(speed = distance/time).
Now given 12.5 min is the difference of time as the train stopped for the same then, from equation B.
=> 50 = 12.5 (ratio difference).
then 1=12.5/5.
So time taken by train = (12.5/50)x100.
=25 min (25/60 hr) ----> C.
Now the speed of train = distance covered by train/time taken.
= (75km) x 25/60 hr (from C).
= 180km/hr.
Hence, the answer will be 180km/hr.
150 : 100 -----> A.
Then the ratio of time will be 100 : 150 ----> B.
(speed = distance/time).
Now given 12.5 min is the difference of time as the train stopped for the same then, from equation B.
=> 50 = 12.5 (ratio difference).
then 1=12.5/5.
So time taken by train = (12.5/50)x100.
=25 min (25/60 hr) ----> C.
Now the speed of train = distance covered by train/time taken.
= (75km) x 25/60 hr (from C).
= 180km/hr.
Hence, the answer will be 180km/hr.
Koushik roy said:
9 years ago
Can anybody explain me how to came 3/2 ?
Nisha said:
9 years ago
Can anybody tell me why we multiply the speed with 25?
Snehesh said:
9 years ago
An alternative Method:
Speed of Train = S Train; Speed of Car = S Car; Time of Train = T Train; Time of Car = T Car;
S Train = 3/2 S Car.
Therefore, T Train = 2/3 T Car (Speed inversely Proportional to Time since Distance is constant).
T Car - T Train = 12.5 mins.
Substituting, T Car - 2/3.
T Car = 12.5.
T Car = 37.5 minutes.
S Car = Distance/ T Car.
S Car = 75/(37.5/60). Converting minutes to hours.
S Car = 120 kmph.
Speed of Train = S Train; Speed of Car = S Car; Time of Train = T Train; Time of Car = T Car;
S Train = 3/2 S Car.
Therefore, T Train = 2/3 T Car (Speed inversely Proportional to Time since Distance is constant).
T Car - T Train = 12.5 mins.
Substituting, T Car - 2/3.
T Car = 12.5.
T Car = 37.5 minutes.
S Car = Distance/ T Car.
S Car = 75/(37.5/60). Converting minutes to hours.
S Car = 120 kmph.
Snehesh said:
9 years ago
Car speed = x.
Train speed is 50% more.
Train speed = x + (50/100)x = x + 1/2x = 3/2x.
Train speed is 50% more.
Train speed = x + (50/100)x = x + 1/2x = 3/2x.
Bhumi said:
9 years ago
Car speed = x.
Train speed is 50% more.
Train speed = x+ (50/100) x = 100x + 50x/100.
= 150x/100.
= 3/2x Kmph.
Train speed is 50% more.
Train speed = x+ (50/100) x = 100x + 50x/100.
= 150x/100.
= 3/2x Kmph.
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