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 5 of 35.
Harika ravuri said:
7 years ago
Actually he gave 50% more not 50% times.
so, the speed of train = speed of car + 50% speed of car.
= x+(50/100)*x = 3/2x.
Given that, time of car - time of train = 12.5 min,
(75/x) - (75/(3/2*x)) =125/10 min,
75/x(1 - (2/3)) =125/(10*60) hrs,
1 = 125x/600,
x = 120 kmph.
so, the speed of train = speed of car + 50% speed of car.
= x+(50/100)*x = 3/2x.
Given that, time of car - time of train = 12.5 min,
(75/x) - (75/(3/2*x)) =125/10 min,
75/x(1 - (2/3)) =125/(10*60) hrs,
1 = 125x/600,
x = 120 kmph.
Gowtham Sunny said:
9 years ago
Here, Train and car reach at the same time.
So, T1(Car) = T2(train).
Then, T1(Car) = T2(train) + 12.5
So, T1 - T2 = 12.5.
Given Distance = 75,
Let speed be x.
Speed = x -----> Car speed.
Speed = x + 50/100 of x = 3x/2 -----> Train speed.
Therefore, T1 - T2 = 12.5.
75/x - 75/(3x/2) = 12.5 min.
(75 - 150)/x = 12.5/60 hr (Since we are cal in kmph we should convert min to hrs).
x = 120 kmph.
So, T1(Car) = T2(train).
Then, T1(Car) = T2(train) + 12.5
So, T1 - T2 = 12.5.
Given Distance = 75,
Let speed be x.
Speed = x -----> Car speed.
Speed = x + 50/100 of x = 3x/2 -----> Train speed.
Therefore, T1 - T2 = 12.5.
75/x - 75/(3x/2) = 12.5 min.
(75 - 150)/x = 12.5/60 hr (Since we are cal in kmph we should convert min to hrs).
x = 120 kmph.
Jason Valentine said:
1 year ago
Speed of car = Sc ; speed of train = St.
Sc : St = 100:150 ( A train can travel 50% faster than a car)
= 2:3 -----> (1)
And therefore Tc:Tt= 3:2 -----> (2)
Delay given= 12.5 min = 12.5/60hrs.
Therefore time for car, Tc= 3 * 12.5/60 = 12.5/20 (refer (2)Tc : Tt = 3:2 )
The distance given 75km.
We know, Speed= Distance/Time.
Therefore, the speed of the car,
Sc = 75/(12.5/20),
= 120 km/hr.
Sc : St = 100:150 ( A train can travel 50% faster than a car)
= 2:3 -----> (1)
And therefore Tc:Tt= 3:2 -----> (2)
Delay given= 12.5 min = 12.5/60hrs.
Therefore time for car, Tc= 3 * 12.5/60 = 12.5/20 (refer (2)Tc : Tt = 3:2 )
The distance given 75km.
We know, Speed= Distance/Time.
Therefore, the speed of the car,
Sc = 75/(12.5/20),
= 120 km/hr.
(95)
R.Sreekanth said:
1 decade ago
@Yesoda
They given speed is 50% more than car so,
We don't know car speed,let consider car speed as X.
So, the train speed is 50% more than car(i.e, when ever we talking about anything in the form of % we must add or subtract from 100. Then convert it into [total/100], total= % + 100 ).
Therefore: the train speed is [(50+100)/100] more speed of car.
So, the train speed is = (150/100)*x.
They given speed is 50% more than car so,
We don't know car speed,let consider car speed as X.
So, the train speed is 50% more than car(i.e, when ever we talking about anything in the form of % we must add or subtract from 100. Then convert it into [total/100], total= % + 100 ).
Therefore: the train speed is [(50+100)/100] more speed of car.
So, the train speed is = (150/100)*x.
Chirag Painter said:
8 years ago
Let speed ratio=car:train=100:150=2:3.
Here distance is same so as per the rule "if distance is same then speed ratio is inversely proportional to time"
Speed ratio = 2:3
Time ratio = 3:2
Therefore car usual time = 3 * 12.5 = 37.5 min.
And train usual time = 2 * 12.5 = 25 min.
Now, speed of car = Distance/time,
= 75kmph / 37.5 min,
= 2.
Now convert into hour=2 * 60 = 120kmphr.
Here distance is same so as per the rule "if distance is same then speed ratio is inversely proportional to time"
Speed ratio = 2:3
Time ratio = 3:2
Therefore car usual time = 3 * 12.5 = 37.5 min.
And train usual time = 2 * 12.5 = 25 min.
Now, speed of car = Distance/time,
= 75kmph / 37.5 min,
= 2.
Now convert into hour=2 * 60 = 120kmphr.
Rutuja P said:
2 years ago
Speed of car=Sc ; speed of train =St
Sc:St= 100:150 ( A train can travel 50% faster than a car)
= 2:3 -----> (1)
and therefore Tc:Tt= 3:2 -----> (2)
Delay given= 12.5 min = 12.5/60hrs.
Therefore time for car, Tc= 3*12.5/60 = 12.5/20 (refer (2)Tc:Tt= 3:2 )
Distance given= 75km.
We know, Speed= Distance/Time.
Therefore, the speed of the car,
Sc= 75/(12.5/20),
= 120 km/hr.
Sc:St= 100:150 ( A train can travel 50% faster than a car)
= 2:3 -----> (1)
and therefore Tc:Tt= 3:2 -----> (2)
Delay given= 12.5 min = 12.5/60hrs.
Therefore time for car, Tc= 3*12.5/60 = 12.5/20 (refer (2)Tc:Tt= 3:2 )
Distance given= 75km.
We know, Speed= Distance/Time.
Therefore, the speed of the car,
Sc= 75/(12.5/20),
= 120 km/hr.
(109)
Shunmuga priya said:
1 decade ago
Sonam Jain said that " (125/10*60)it takes the difference b/w boths time is equal to the lost time "
But in question it is clearly given as "from point A at the same time and reach point B 75 kms away from A at the same time". The car and train starting and ending time is same so no need of subtraction but they subtracted here.
I cant get this. please explain
But in question it is clearly given as "from point A at the same time and reach point B 75 kms away from A at the same time". The car and train starting and ending time is same so no need of subtraction but they subtracted here.
I cant get this. please explain
Raihan Bangladesh said:
8 years ago
Let, car speed = X km/min.
Train speed = X + x* (50/100) Km/min.
=X+X/2 km/min.
=3X/2 km/min.
Time of car t =75/X.
Time of Train t =75/ (3X/2).
According to the question.
75/X=75/ (3X/2) +12. 5.
>> 75/X-150/3X=12. 5.
>> (225-150) /3X=12. 5.
>> 75 = 37. 5X.
>> X = 2 Km/min.
So car speed 1 minute to 2 km.
Car Speed 60 minutes to (60*2) = 120 km/h.
Train speed = X + x* (50/100) Km/min.
=X+X/2 km/min.
=3X/2 km/min.
Time of car t =75/X.
Time of Train t =75/ (3X/2).
According to the question.
75/X=75/ (3X/2) +12. 5.
>> 75/X-150/3X=12. 5.
>> (225-150) /3X=12. 5.
>> 75 = 37. 5X.
>> X = 2 Km/min.
So car speed 1 minute to 2 km.
Car Speed 60 minutes to (60*2) = 120 km/h.
(1)
Atul said:
1 decade ago
(75/x)denotes time taken by car.
this means time taken by the car to cover the distance of 75 km with speed x.
(75/(3/2)x)denotes time taken by train to cover the 75 km distance with speed 3/2x.
in the problem it is said that train rest for 12.5 min at station
and both reach 75 km at same time.
There fore difference between the time is (75/x)-(75/(3/2)x)=12.5/60 in hrs.
this means time taken by the car to cover the distance of 75 km with speed x.
(75/(3/2)x)denotes time taken by train to cover the 75 km distance with speed 3/2x.
in the problem it is said that train rest for 12.5 min at station
and both reach 75 km at same time.
There fore difference between the time is (75/x)-(75/(3/2)x)=12.5/60 in hrs.
Gurbir said:
8 years ago
Train and car covers the same distance in same time. But train 50%faster than the car.
The Train stops at stations 12.5.
It means they cover 75km in 25minutes.
So convert minutes into hr (75÷60=12/5).
So that the speed of the train = 75*12÷5=180,
50% of 120 is 60,
180 - 60 = 120.
So Speed of car is 120km/hr.
Because train is faster than car 50% is given above.
The Train stops at stations 12.5.
It means they cover 75km in 25minutes.
So convert minutes into hr (75÷60=12/5).
So that the speed of the train = 75*12÷5=180,
50% of 120 is 60,
180 - 60 = 120.
So Speed of car is 120km/hr.
Because train is faster than car 50% is given above.
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