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:
342 comments Page 8 of 35.
Aishwarya said:
6 years ago
Let the Speed of car be=x km/hr.
Speed of train = 1.5x km/hr.
Time taken by car to travel 75km = 75/x h.
Time taken by train to travel 75 km = 75/1.5 x-12.5/60.
Both are equal.
75/x = 75/1.5x-12.5/60.
1/x(75"75/1.5) = 12.5/60.
25/x = 12.5/60
1500 = 12.5x.
So, x = 1500/12.5 = 120 km/hr is the Answer.
Speed of train = 1.5x km/hr.
Time taken by car to travel 75km = 75/x h.
Time taken by train to travel 75 km = 75/1.5 x-12.5/60.
Both are equal.
75/x = 75/1.5x-12.5/60.
1/x(75"75/1.5) = 12.5/60.
25/x = 12.5/60
1500 = 12.5x.
So, x = 1500/12.5 = 120 km/hr is the Answer.
Jinto said:
1 decade ago
Can anybody explain if both the vehicle reaches at the same time then where is the delay arises i am talking about if two vehicles reaches at the same time then how this equation came
(75/x)-(75/(3/2)x)=12.5/60, 12.5 minutes delay will not be ther if both the vehicles reaches at the same time pls help
(75/x)-(75/(3/2)x)=12.5/60, 12.5 minutes delay will not be ther if both the vehicles reaches at the same time pls help
Justin said:
1 decade ago
Hi @Mancy,
OK here the answer for your question,
Ex:1% of 2 = 1/100*2 = 1/50.
10% of 100 = 10/100*100 = 10.
So now come to our problem ok.
Take speed of the car is X ok. (Consider 100).
Train is 50% faster ok so 100+50 = 150;.
So we need find that 150%X which gives 150/100*X.
All the best.
OK here the answer for your question,
Ex:1% of 2 = 1/100*2 = 1/50.
10% of 100 = 10/100*100 = 10.
So now come to our problem ok.
Take speed of the car is X ok. (Consider 100).
Train is 50% faster ok so 100+50 = 150;.
So we need find that 150%X which gives 150/100*X.
All the best.
Sai Durga said:
2 years ago
Let the speed of the train be x.
So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr,
Time delayed = 12.5 mins = 12.5/60 sec.
Now;
D/s = T; (here speed = speed of the car -speed of the train),
So,
75/x -75/3x/2 = 12.5/60.
75/x - 50/x = 5/24.
x = 25 x 24/5.
x = 120 km/hr.
So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr,
Time delayed = 12.5 mins = 12.5/60 sec.
Now;
D/s = T; (here speed = speed of the car -speed of the train),
So,
75/x -75/3x/2 = 12.5/60.
75/x - 50/x = 5/24.
x = 25 x 24/5.
x = 120 km/hr.
(13)
Neranjan said:
6 years ago
A to B have 75km.
We think, Train not stopped,
It can go 112.5km.
So, train speed =112.5km-75km/12.5m.
12.5m covert to hour 12.5/60.
= 37.5km * 60/12.5,
= 180kmph.
So car speed =180kmph * 100/150.
=120kmph.
We think, Train not stopped,
It can go 112.5km.
So, train speed =112.5km-75km/12.5m.
12.5m covert to hour 12.5/60.
= 37.5km * 60/12.5,
= 180kmph.
So car speed =180kmph * 100/150.
=120kmph.
Sai Durga said:
2 years ago
Let the speed of the train be x.
So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr,
time delayed = 12.5 mins==12.5/60 sec.
Now;
D/s = T; (here speed = speed of the car -speed of the train),
So,
75/x -75/3x/2 = 12.5/60.
75/x - 50/x = 5/24.
x = 25 x 24/5
x = 120 km/hr.
So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr,
time delayed = 12.5 mins==12.5/60 sec.
Now;
D/s = T; (here speed = speed of the car -speed of the train),
So,
75/x -75/3x/2 = 12.5/60.
75/x - 50/x = 5/24.
x = 25 x 24/5
x = 120 km/hr.
(29)
Sreekanth said:
1 decade ago
Think... Y 3/2x came with an example...
if u think car has x speed, then train ll have 2x.
then substitute x with simple value..
x:2x
if x=10 then
10:20
it mean train has double value of car speed.
but x:3/2x
then 10:15x
In which train has 50% faster than car.
50% of car is 5. so 10+5=15.
if u think car has x speed, then train ll have 2x.
then substitute x with simple value..
x:2x
if x=10 then
10:20
it mean train has double value of car speed.
but x:3/2x
then 10:15x
In which train has 50% faster than car.
50% of car is 5. so 10+5=15.
Aadhithyan A said:
3 years ago
Let the speed of the train be x.
So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr, time delayed = 12.5 mins==12.5/60 sec. Now;
D/s = T; (here speed = speed of the car -speed of the train),
So,
75/x -75/3x/2 = 12.5/60.
75/x - 50/x = 5/24.
x = 25 x 24/5
x = 120 km/hr.
So, train speed 50% faster than car means x + x/2 = 3x/2, speed = 75km/hr, time delayed = 12.5 mins==12.5/60 sec. Now;
D/s = T; (here speed = speed of the car -speed of the train),
So,
75/x -75/3x/2 = 12.5/60.
75/x - 50/x = 5/24.
x = 25 x 24/5
x = 120 km/hr.
(50)
Dan said:
8 years ago
Train spent extra 12.5 minutes then a car. So we have to subtract bigger time from lesser time to get 12.5. It means that time spent by car must be subtracted from the time taken by car to get the excess time of 12.5 minutes. Thus, it should be like this:
75/(1.5*x)-75/x=0.21 (in hours).
75/(1.5*x)-75/x=0.21 (in hours).
Shubham said:
4 years ago
It's given that both Car and the train covers 75kms at the same time (t1=t2)
t=d/s
Let's assume car speed as 'x' km/hr.
Train Speed = x + x/2.
=3/2x.
t1=75/x.
t2=75/(3/2)*x.
t1 = t2+12.5mins extra time.
Then solve by converting 12.5 mins to hr by dividing by 60.
t=d/s
Let's assume car speed as 'x' km/hr.
Train Speed = x + x/2.
=3/2x.
t1=75/x.
t2=75/(3/2)*x.
t1 = t2+12.5mins extra time.
Then solve by converting 12.5 mins to hr by dividing by 60.
(1)
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