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:
205 comments Page 3 of 21.
Bala said:
9 years ago
Simple friends, we solve by using S = D/T.
S - speed.
D - distance.
T - Time.
S = 420/t --> (1).
S + 10 = 420/(t-1) --> (2).
From(2) St- S+ 10t-10 = 420 --> (3)
Let equate (1) and (3), from this we can find speed.
St = ST - S + 10t - 10.
(OR) T = S+10/10.
Now S = D/T.
Substitute the values S = 420/S + 10/10.
S = (420*10)/(S+10).
S = -70 and 60. Neglect the negative S=60.
S - speed.
D - distance.
T - Time.
S = 420/t --> (1).
S + 10 = 420/(t-1) --> (2).
From(2) St- S+ 10t-10 = 420 --> (3)
Let equate (1) and (3), from this we can find speed.
St = ST - S + 10t - 10.
(OR) T = S+10/10.
Now S = D/T.
Substitute the values S = 420/S + 10/10.
S = (420*10)/(S+10).
S = -70 and 60. Neglect the negative S=60.
Babar said:
9 years ago
One hour answer is logically wrong. If we assume that original speed was 800 km/hr, then, 600/800 = 45mints will be the time taken at original speed.
And after reduction of 200 km/hr new speed will be 600 km/hr, then time would reduce by 15 mints, not 30 mints.
30 mints can only be possible if original speed was 600km/hr which is reduced to 400km/hr. then 600/400 = 1.50, which is 1 hour and 30 mins.
And after reduction of 200 km/hr new speed will be 600 km/hr, then time would reduce by 15 mints, not 30 mints.
30 mints can only be possible if original speed was 600km/hr which is reduced to 400km/hr. then 600/400 = 1.50, which is 1 hour and 30 mins.
Khrish said:
1 decade ago
Suppose duration of flight is = x.
Speed = distance/time.
200=(600/x)-(600/(x+30mins)).
200=(600/x)-[600/(x+(30/60)hr)].
200=(600/x)-[600/(x+(1/2))].
200=(600/x)-[600/((2x+1)/2)] by LCM,
200=(600/x)-[1200/2x+1].
200=600[(1/x)-(2/2x+1)].
200/600=(1/x)-(2/2x+1).
1/3=(2x+1-2x)/(2x^2+x).
1/3=1/(2x^2+x).
2x^2+x-3=0.
(2x+3)(x-1)=0.
X=-3/2, x=1.
I hope you got it...;).
Speed = distance/time.
200=(600/x)-(600/(x+30mins)).
200=(600/x)-[600/(x+(30/60)hr)].
200=(600/x)-[600/(x+(1/2))].
200=(600/x)-[600/((2x+1)/2)] by LCM,
200=(600/x)-[1200/2x+1].
200=600[(1/x)-(2/2x+1)].
200/600=(1/x)-(2/2x+1).
1/3=(2x+1-2x)/(2x^2+x).
1/3=1/(2x^2+x).
2x^2+x-3=0.
(2x+3)(x-1)=0.
X=-3/2, x=1.
I hope you got it...;).
Shalz said:
1 decade ago
Hello friends.
Here D = 600 km.
Average speed = 200.
Here in the problem he did not mentioned about time so lets take it as x.
But he given increased time as 30 min.
So according to the problem actual time = x+30 min.
So, he is asking duration T=D/S.
x+30 = 600/200.
x+30 = 3.
x = 30-3 = 27 min.
But actual time = x+30 = 27+30 = 57 min approx 1 hr.
Here D = 600 km.
Average speed = 200.
Here in the problem he did not mentioned about time so lets take it as x.
But he given increased time as 30 min.
So according to the problem actual time = x+30 min.
So, he is asking duration T=D/S.
x+30 = 600/200.
x+30 = 3.
x = 30-3 = 27 min.
But actual time = x+30 = 27+30 = 57 min approx 1 hr.
Venkat Bitra said:
9 years ago
Assume the normal time of flight was X.
Due to bad weather it was X+30 convert it into hours it would be X+ 1/2.
Normal speed of flight be S.
Due to bad weather, it was S-200.
First equation would be S = 600/X,
Second one would be S-200= 600/X+ 1/2.
Solve them equate Find the value of X by Equating in terms of X to each other
600/X=(1200/ 2X+1 )+ 200.
Due to bad weather it was X+30 convert it into hours it would be X+ 1/2.
Normal speed of flight be S.
Due to bad weather, it was S-200.
First equation would be S = 600/X,
Second one would be S-200= 600/X+ 1/2.
Solve them equate Find the value of X by Equating in terms of X to each other
600/X=(1200/ 2X+1 )+ 200.
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)
Vijaya lakshmi said:
6 years ago
We know that speed = distance/time;
Let us assume time as x,then actual speed=600/x;
Now the time is incremented to 30 min right,
So, now the speed=600/(x+1/2)
Here we take the time in hrs so (x+1/2).
In the problem, the speed is reduced by 200km/hr
So the diff between speed is equal to 200km/hr.
i.e. 600/x-600/x+(1/2) = 200.
Let us assume time as x,then actual speed=600/x;
Now the time is incremented to 30 min right,
So, now the speed=600/(x+1/2)
Here we take the time in hrs so (x+1/2).
In the problem, the speed is reduced by 200km/hr
So the diff between speed is equal to 200km/hr.
i.e. 600/x-600/x+(1/2) = 200.
(3)
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)
Kiran.p said:
2 decades ago
We know that speed=distance/time;
Let us assume time as x,then actual speed=600/x;
Now the time is incremented to 30 min right,
So, now the speed=600/(x+1/2)
Here we take the time in hrs so (x+1/2).
In the problem the speed is reduced by 200km/hr
So the diff btwn speed is equal to 200km/hr.
i.e. 600/x-600/x+(1/2)=200
Let us assume time as x,then actual speed=600/x;
Now the time is incremented to 30 min right,
So, now the speed=600/(x+1/2)
Here we take the time in hrs so (x+1/2).
In the problem the speed is reduced by 200km/hr
So the diff btwn speed is equal to 200km/hr.
i.e. 600/x-600/x+(1/2)=200
(3)
Muhammad Umair Kabir said:
1 decade ago
Let Speed is X and Time is Y,
According to formula
X = 600 / Y
From question avg Speed reduced and Time taken increased eq. becomes...
X - 200 = 600 / (Y + 1/2)
put the value of X in above eq.
600 / Y - 200 = 600 / (Y + 1/2)
Simplifying the eq. we get
(Y - 1)(2Y + 3 )...
Y = 1 Ans as time is always positive....
According to formula
X = 600 / Y
From question avg Speed reduced and Time taken increased eq. becomes...
X - 200 = 600 / (Y + 1/2)
put the value of X in above eq.
600 / Y - 200 = 600 / (Y + 1/2)
Simplifying the eq. we get
(Y - 1)(2Y + 3 )...
Y = 1 Ans as time is always positive....
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers