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:
355 comments Page 6 of 36.
Rutuja P said:
3 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.
(110)
Shunmuga priya said:
2 decades 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:
9 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:
2 decades 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:
9 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.
Ayush said:
4 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:
8 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:
8 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:
4 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:
3 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)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers