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:
211 comments Page 10 of 22.
Arun said:
8 years ago
If the average speed is increased by 200km/hr and time of flight is decreased by 30mins. What is the equation?
(600/x)-(600/x+1/2))=-200 is this correct?
(600/x)-(600/x+1/2))=-200 is this correct?
Shakila said:
1 decade ago
I agree to @Deepak Jain's word and yes bad weather has it be considered and as tony mentioned 400 is the speed.
Its just simple as much as 1.5 hours.
Its just simple as much as 1.5 hours.
Sridhar reddy said:
2 decades ago
One more similar method.
Let x be the speed
600/(x-200) - 600/x = 30/60(in hrs)
solving we get x=600
then Time taken= distance/speed = 600/600 =1hr
Let x be the speed
600/(x-200) - 600/x = 30/60(in hrs)
solving we get x=600
then Time taken= distance/speed = 600/600 =1hr
Sivanityam said:
2 years ago
D = 600km.
Average speed is 200km/hr,
s = 600/3 = 200km/hr.
200km time taken 30min.
The remaining time is 400km = 30 + 30 = 60min = time taken = 1hr.
Average speed is 200km/hr,
s = 600/3 = 200km/hr.
200km time taken 30min.
The remaining time is 400km = 30 + 30 = 60min = time taken = 1hr.
(97)
Varsh said:
5 years ago
Its simple only, take LCM of (600/x)-(1200/(2x+1)) = 200,
you get (2x+1)(600) - 1200x = 200(2x^2 +x),
2x^2 +x = 3.
which is nothing but x(2x+1) = 3.
you get (2x+1)(600) - 1200x = 200(2x^2 +x),
2x^2 +x = 3.
which is nothing but x(2x+1) = 3.
(3)
Sahin said:
1 decade ago
@Rahim Shaikh.
There is a small mistake in your answer.
Tt will be:
(x - 600) (x + 400) = 0.
x2 - 200x - 240000 = 0.
Not x+200. Check this once.
There is a small mistake in your answer.
Tt will be:
(x - 600) (x + 400) = 0.
x2 - 200x - 240000 = 0.
Not x+200. Check this once.
Patel said:
6 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)
Nani said:
2 decades ago
Time is inversely proportional to speed so:
t1/t2=s2/s1
t2=t1+0.5 and s2=s1-200
and now solve general eqn d=t*s using above two to get t1.
t1/t2=s2/s1
t2=t1+0.5 and s2=s1-200
and now solve general eqn d=t*s using above two to get t1.
Shanthi said:
2 decades ago
Its simple only...take LCM of (600/x)-(1200/(2x+1))=200
you get (2x+1)(600)-1200x=200(2x^2 +x)
2x^2 +x=3
which is nothing but x(2x+1)=3
you get (2x+1)(600)-1200x=200(2x^2 +x)
2x^2 +x=3
which is nothing but x(2x+1)=3
Rohit said:
10 years ago
Let, x is original speed.
(600/x - 200) - (600/x) = 30/60,
Solving this equation, we get x = 600.
Duration of flight = 600/600 = 1 hr.
(600/x - 200) - (600/x) = 30/60,
Solving this equation, we get x = 600.
Duration of flight = 600/600 = 1 hr.
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