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:
210 comments Page 10 of 21.
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:
1 decade 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:
1 decade 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.
NEHAL ANSARI said:
9 years ago
I have one doubt, due to bad weather the time increased by 30 min then we should add 30 min in x ie (x+30/60) or not?
Please help me.
Please help me.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers