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 28 of 35.
Ashish patil said:
7 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.
Bhavi said:
7 years ago
Speed of train =( 1 + 50% )of speed of car.
75/t-12.5 = (1+1/2)75/t,
t=37.5 minutes,
Car speed = (2/3) * train speed,
Car speed = (2/3)*75/(37.5-12.5),
Car speed = 2 kmpm,
2 * 60 = 120 kmph.
75/t-12.5 = (1+1/2)75/t,
t=37.5 minutes,
Car speed = (2/3) * train speed,
Car speed = (2/3)*75/(37.5-12.5),
Car speed = 2 kmpm,
2 * 60 = 120 kmph.
Sameer azmi said:
7 years ago
The train speed of the train is 50% more the speed of the car = x + 50%x/100.
Then, (100x+50x) /100.
Then, (100x+50x) /100.
Gete Basu Chakma said:
7 years ago
The speed of car = x.
speed of train = 1.5x.
The lost time=12.5/60 = 5/24 hr.
The time taken by car = lost time +running time of train.
75/x = (5/24) +75/1.5x.
Then, x =120km/hr.
speed of train = 1.5x.
The lost time=12.5/60 = 5/24 hr.
The time taken by car = lost time +running time of train.
75/x = (5/24) +75/1.5x.
Then, x =120km/hr.
Zama said:
7 years ago
@All.
Explanation of this part, 12.5/10 * 60.
12.5 is 125/10.
=> 125/10*60,
=> 5*25/10*60, #since 125 can be written as 5*25.
=> 25/10*12, #after cancelling denominator 60 by numerator 5 as 5*12=60.
=>25/120, #here cancellation goes as 5*5=25 & 5*24=120.
=>5/24.
Hence, we got 5/24.
Explanation of this part, 12.5/10 * 60.
12.5 is 125/10.
=> 125/10*60,
=> 5*25/10*60, #since 125 can be written as 5*25.
=> 25/10*12, #after cancelling denominator 60 by numerator 5 as 5*12=60.
=>25/120, #here cancellation goes as 5*5=25 & 5*24=120.
=>5/24.
Hence, we got 5/24.
RAMJAN RAEEN said:
7 years ago
Can anyone explain to me how 75/X - 75/ (3x/2) = 12.5?
How the time is the same in both case car and train if a halt is not considering?
How the time is the same in both case car and train if a halt is not considering?
Ansh gupta said:
7 years ago
Can you explain how 150/100x came?
Ivdurp said:
7 years ago
We always consider percentage to 100 right.
They have given that train speed 50 % more than a car. here they take car speed as x.
We already we knew the percentage is (100+50)*x/100.
They have given that train speed 50 % more than a car. here they take car speed as x.
We already we knew the percentage is (100+50)*x/100.
Ameer said:
6 years ago
The speed of the car - speed of the train = loss time.
Abraham said:
6 years ago
Speed of car be x.
Speed of train= x+x/2 = 3x/2 ie; 1.5x.
Train didnt run for 12.5min ie; 12.5/60 hrs.
Difference between time ran by both is 12.5/60hr.
ie: (75/x) - (75/1.5x) = 12.5/60.
Upon solving x= 120kmph.
Speed of train= x+x/2 = 3x/2 ie; 1.5x.
Train didnt run for 12.5min ie; 12.5/60 hrs.
Difference between time ran by both is 12.5/60hr.
ie: (75/x) - (75/1.5x) = 12.5/60.
Upon solving x= 120kmph.
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