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:
345 comments Page 23 of 35.
Hong Kong said:
9 years ago
But can anyone explain to me why can't I think like this:
Let x be the time for Bus to arrive point B, then x = 12.5x2 (since train 50% faster) = 25.
So we know the train bus spend 25 mins to arrive B, and 75km/25mins = 3km/mins = 180kmph?
Let x be the time for Bus to arrive point B, then x = 12.5x2 (since train 50% faster) = 25.
So we know the train bus spend 25 mins to arrive B, and 75km/25mins = 3km/mins = 180kmph?
Vishnu b nair said:
9 years ago
The speed of both of them is 100 per cent and the train is 50% more faster than car it means the train have the 100 + 50 = 150%.
Rakshak said:
9 years ago
Can you explain, how 125/(10 * 60)?
Tesit said:
9 years ago
Can you explain, how 125/ (10 * 60)?
Rajesh.r.kzm said:
9 years ago
12.5 can be taken as 125/10. So it will be multiplied by no of minutes in an hour. So, it is 125/(10*60).
Priya said:
9 years ago
Can you exlpain how 150x/100 came?
Sagar Bhamare said:
9 years ago
The total speed of a car 100% + 50% extra,
So the speed of the train is 150%.
Which can be written as (150/100)X.
So the speed of the train is 150%.
Which can be written as (150/100)X.
Nayem said:
9 years ago
Let speed of car = x (km/h).
So, speed of train = x + 50% of x = x + x/2 = 3x/2 (km/h),
To travel 75 km, car needs time of 75/x hours,
To travel 75 km, train needs time of 75/(3x/2) hours + 12.5/60 hours,
As the two reach destination on same time on the clock, so these times should be equal. That is:
75/x = 75/(3x/2) + 12.5/60,
=> 75/x = (75*2)/3x + 12.5/60,
=> 75/x = 50/x + 12.5/60,
=> 75/x - 50/x = 12.5/60,
=> 25/x = 12.5/60,
=> 12.5x = 25 * 60,
=> x = 1500/12.5.
So, x = 120 (km/h).
So, speed of train = x + 50% of x = x + x/2 = 3x/2 (km/h),
To travel 75 km, car needs time of 75/x hours,
To travel 75 km, train needs time of 75/(3x/2) hours + 12.5/60 hours,
As the two reach destination on same time on the clock, so these times should be equal. That is:
75/x = 75/(3x/2) + 12.5/60,
=> 75/x = (75*2)/3x + 12.5/60,
=> 75/x = 50/x + 12.5/60,
=> 75/x - 50/x = 12.5/60,
=> 25/x = 12.5/60,
=> 12.5x = 25 * 60,
=> x = 1500/12.5.
So, x = 120 (km/h).
Ravi kataria said:
9 years ago
As a train is much faster, then time difference must be [distance /train speed - distance /car speed =12.5min].
Then, how it does inversely?
Then, how it does inversely?
Mithlesh said:
8 years ago
Nice explanation @Sushil.
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