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 | ![]() |
100 | 2 |
![]() |
75 | - | 75 | = | 125 |
x | (3/2)x | 10 x 60 |
![]() |
75 | - | 50 | = | 5 |
x | x | 24 |
![]() |
![]() |
25 x24 | ![]() |
= 120 kmph. |
5 |
Discussion:
342 comments Page 4 of 35.
Jasmine said:
6 years ago
Distance between A and B = 75kms.
Let 'x' be the speed of the car.
Given that the train can travel 50% faster than the car i.e.,
Car speed = x.
Train speed = x+x*50/100 = 3/2*x.
Distance = speed*time.
Tc(Time of the car) = 75/x.
Tt(Time of the train) = 75/(3/2*x)+12.5 min(while stopping at the station).
=>75/(3/2*x)+12.5/60(into hours).
=>75/x=75/(3/2*x)+12.5/60.
By solving x = 120kmph.
Let 'x' be the speed of the car.
Given that the train can travel 50% faster than the car i.e.,
Car speed = x.
Train speed = x+x*50/100 = 3/2*x.
Distance = speed*time.
Tc(Time of the car) = 75/x.
Tt(Time of the train) = 75/(3/2*x)+12.5 min(while stopping at the station).
=>75/(3/2*x)+12.5/60(into hours).
=>75/x=75/(3/2*x)+12.5/60.
By solving x = 120kmph.
Vidhya said:
1 decade ago
We need to convert 12.5 min to hours because the options given are in kmph (kilometre per hour).
To convert 12.5 min to hours you divide it by 60.
To make calculation easier 12.5 is converted into 125 by multiplying it by 10. If you multiply numerator with a number, then you also have to multiply denominator by that number.
Hence in this case it will be 12.5*10/60*10 = 125/600 which is actually the same as 12.5/60.
125/600 = 5/24.
12.5/60 = 5/24.
To convert 12.5 min to hours you divide it by 60.
To make calculation easier 12.5 is converted into 125 by multiplying it by 10. If you multiply numerator with a number, then you also have to multiply denominator by that number.
Hence in this case it will be 12.5*10/60*10 = 125/600 which is actually the same as 12.5/60.
125/600 = 5/24.
12.5/60 = 5/24.
T S SOUNDHAR said:
4 months ago
If the car speed is 100kmph then the train speed is 150 kmph.
Because train 50% extra speed than a car.
Train in station stopping time is 12.5min.
But a car and train cover the same distance.
In case the train running non-stoping train cover more distance.
car: train 100%:150%.
car : train 100% ====(100% + 12.5min)
That 50% is 12.5min.
100% = 2 * 12.5min.
100% = 25min..
37.5min =75kmph
1min = 75/37.5
1min = 2 kmph
1hour = 60min = 2 * 60 = 120kmph.
Because train 50% extra speed than a car.
Train in station stopping time is 12.5min.
But a car and train cover the same distance.
In case the train running non-stoping train cover more distance.
car: train 100%:150%.
car : train 100% ====(100% + 12.5min)
That 50% is 12.5min.
100% = 2 * 12.5min.
100% = 25min..
37.5min =75kmph
1min = 75/37.5
1min = 2 kmph
1hour = 60min = 2 * 60 = 120kmph.
(24)
Mysterious said:
1 decade ago
Friends......
let speed of car be x
thus, speed of train (150/100)x (stated that train is 50% faster than car)
now since the train lost 12.5 mins. i.e. the difference between time taken by car and time taken by train is 12.5 mins,
and since TIME=DISTANCE/SPEED
time taken by car= 75/x
time taken by train= 75/(150/100)x or 75/(3/2)x
and the difference between time taken by car and time taken by train is 75/x - 75/(3/2)x = 125/10*60.
x=120 kmph
let speed of car be x
thus, speed of train (150/100)x (stated that train is 50% faster than car)
now since the train lost 12.5 mins. i.e. the difference between time taken by car and time taken by train is 12.5 mins,
and since TIME=DISTANCE/SPEED
time taken by car= 75/x
time taken by train= 75/(150/100)x or 75/(3/2)x
and the difference between time taken by car and time taken by train is 75/x - 75/(3/2)x = 125/10*60.
x=120 kmph
Sushil pal said:
1 decade ago
Let Car's speed be x Kmph
Train speed = (Carspeed + (50*Carspeed)/100)
" = (x+(50x)/100)
" = 150x/100
" = 3/2xKmph
Now the train lost about 12.5min, so time subtraction betn Car and train will give us distance covered by car in that much period of time (lost period)and to know the speed we have to equate it with lost period
Distance covered = lostPeriod
75/x-75/(3/2)x = 125/600
75/x-50/x = 5/24
x = (25*24/5)=120 Kmph.
Train speed = (Carspeed + (50*Carspeed)/100)
" = (x+(50x)/100)
" = 150x/100
" = 3/2xKmph
Now the train lost about 12.5min, so time subtraction betn Car and train will give us distance covered by car in that much period of time (lost period)and to know the speed we have to equate it with lost period
Distance covered = lostPeriod
75/x-75/(3/2)x = 125/600
75/x-50/x = 5/24
x = (25*24/5)=120 Kmph.
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.
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.
Rakkesh said:
5 years ago
The car and train reached 75km at the same time.the train was late for 12.5 mins.
The train could reach 12.5 mins faster than the car.
* 75/x is the timing of car to reach 75kms.
*75/(3x/2) +12.5/60(time for stops) is the time taken by train to reach 75kms.
*Both the timings are equal since the time taken by both car and train are equal so then you can get the equation.
75/x = 75/(3x/2) + 12.5/60.
75/x - 75/(3x/2) = 12.5/60.
The train could reach 12.5 mins faster than the car.
* 75/x is the timing of car to reach 75kms.
*75/(3x/2) +12.5/60(time for stops) is the time taken by train to reach 75kms.
*Both the timings are equal since the time taken by both car and train are equal so then you can get the equation.
75/x = 75/(3x/2) + 12.5/60.
75/x - 75/(3x/2) = 12.5/60.
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.
Arpit Mahor said:
3 years ago
Let the speed of the car = x.
So, the speed of the train will be 50% faster than the car i.e;
x + 50% of x = 1.5 x.
The Time taken by the car and train is same and the train lost 12.5 min or 12.5/60 hours on stops.
So,
Time taken by train = Time taken by car.
Time taken by train to cover 75km = (75/1.5x) - (12.5/60)Time lost.
Time taken by car to cover 75km = (75/x).
equate these equations and find x.
So, the speed of the train will be 50% faster than the car i.e;
x + 50% of x = 1.5 x.
The Time taken by the car and train is same and the train lost 12.5 min or 12.5/60 hours on stops.
So,
Time taken by train = Time taken by car.
Time taken by train to cover 75km = (75/1.5x) - (12.5/60)Time lost.
Time taken by car to cover 75km = (75/x).
equate these equations and find x.
(43)
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