Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 6)
6.
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
Answer: Option
Explanation:
Let the duration of the flight be x hours.
| Then, | 600 | - | 600 | = 200 |
| x | x + (1/2) |
 |
600 | - | 1200 | = 200 |
| x | 2x + 1 |
x(2x + 1) = 3
2x2 + x - 3 = 0
(2x + 3)(x - 1) = 0
x = 1 hr. [neglecting the -ve value of x]
Discussion:
206 comments Page 18 of 21.
Parimalanathan said:
6 years ago
I can't understand this.
(1)
Neeru sharma said:
5 years ago
How come 1/2?
(1)
Prabi said:
5 years ago
@Neeru.
1/2 is because here told that 30 minutes. If it is 60 minutes then it will be 1.
1/2 is because here told that 30 minutes. If it is 60 minutes then it will be 1.
Diya said:
5 years ago
Given distance =600.
Let initial speed and time be s & t.
So,speed s=600/t.
Now given speed is reduced by 200,so s becomes s-200 and time is increased by 30 mins, So t becomes t+(30/60) in hrs.
Distance= speed*time.
600=(s-200)(t+(30/60)).
Sub for s ,600=(600/t -200)(t+(30/60)).
600=200[(3/t)-1](3t/2).
1=(3/2)-(t/2).
t = 1.
Let initial speed and time be s & t.
So,speed s=600/t.
Now given speed is reduced by 200,so s becomes s-200 and time is increased by 30 mins, So t becomes t+(30/60) in hrs.
Distance= speed*time.
600=(s-200)(t+(30/60)).
Sub for s ,600=(600/t -200)(t+(30/60)).
600=200[(3/t)-1](3t/2).
1=(3/2)-(t/2).
t = 1.
(2)
Aadarsh said:
5 years ago
@Diya.
Good explanation. Thanks.
Good explanation. Thanks.
Manish said:
5 years ago
Simple way to solve this problem:
Distance = speed * time.
Let the distance be d and the time taken by the plane at 600 km/hr be a
Then:
d=600*a (1) at regular speed
and, d=(600-200) * (a+.5 hr) (2) at reduced speed
Equating the two we get;
600*a = 400(a+.5)
600a = 400a + 200
200a=200
a=1.
The time taken is 1hr.
Distance = speed * time.
Let the distance be d and the time taken by the plane at 600 km/hr be a
Then:
d=600*a (1) at regular speed
and, d=(600-200) * (a+.5 hr) (2) at reduced speed
Equating the two we get;
600*a = 400(a+.5)
600a = 400a + 200
200a=200
a=1.
The time taken is 1hr.
(5)
Patel said:
5 years ago
Here we get x=1, but the question is to find time duration of a plane in bad weather, which is x+0.5 hrs. So the answer must be 1.5hrs. Right?
(1)
Priyanka said:
5 years ago
Can anyone explain in a simple way? please.
(1)
Gayathri said:
5 years ago
Can anyone explain in a simple method?
(1)
Harika said:
4 years ago
It's simple way using average model.
2 * X * Y/(X+Y) = 2 * 200 * 600/200 * 600.
=>2.
In 30minutes we have to find?
2 * 30minutes = 60minutes
60minutes is gone to 1hour.
Ans: 1hour.
2 * X * Y/(X+Y) = 2 * 200 * 600/200 * 600.
=>2.
In 30minutes we have to find?
2 * 30minutes = 60minutes
60minutes is gone to 1hour.
Ans: 1hour.
(3)
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