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.
Ajit said:
1 decade ago
It's easy...the concept goes lik this...
consider eg.. for 50km speed-takes 1hr
for 50-y->1hr+t(in problem, speed is reduced and time is increased)
therfore,
600/x-600/x+(1/2)=200(on an avg (1hr+t) increased and on an avg 200km speed less)
consider eg.. for 50km speed-takes 1hr
for 50-y->1hr+t(in problem, speed is reduced and time is increased)
therfore,
600/x-600/x+(1/2)=200(on an avg (1hr+t) increased and on an avg 200km speed less)
Nagaraju said:
1 decade ago
What a problem it is !
Firstly, I think that want to solve with the help of Relative formula is there na. 2xy/x+y but. It not suitable. So, thought another way and got the answer. Simply a good problem.
Firstly, I think that want to solve with the help of Relative formula is there na. 2xy/x+y but. It not suitable. So, thought another way and got the answer. Simply a good problem.
Hunny said:
1 decade ago
600/(x-200) - 600/x = 30/60(in hrs)
Please anyone explain.
Please anyone explain.
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
Geeth said:
1 decade ago
How? Can you explain nani?
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.
Srilu said:
1 decade ago
Where from that x/2 come ya?
Raghu said:
1 decade ago
Thanks avy.
Ram said:
1 decade ago
@avy.
Thanks I am clear now.
Thanks I am clear now.
Avy said:
1 decade ago
Guys it is very simple.
In the problem it is given that reduced speed = 200.
The initial speed is the actual speed.
And final speed is the speed after being reduced by 200km/hr.
So "initialspeed-200=finalspeed".
Speed=distance/time.
Distance=600km (both initial and final distance).
Let x be initial time taken.
Let (x+30) in min i.e. (x+1/2) in hrs be the final time.
Initial speed=initial distance/initial time=600/x.
Final speed=final distance/final time=600/ (x+1/2).
So "initialspeed-200=finalspeed".
600/x-200=600/ (x+1/2).
600/x-600/ (x+1/2) =200.
Solving this we get.
400x2+200x-600=0.
4x2 + 2x - 6= 0.
2x2 + x - 3 = 0.
2x2 - 2x + 3x - 3 = 0.
2x (x - 1) + 3 (x- 1) = 0.
(2x + 3) (x - 1) = 0.
X=3/2 or x=1.
In the problem it is given that reduced speed = 200.
The initial speed is the actual speed.
And final speed is the speed after being reduced by 200km/hr.
So "initialspeed-200=finalspeed".
Speed=distance/time.
Distance=600km (both initial and final distance).
Let x be initial time taken.
Let (x+30) in min i.e. (x+1/2) in hrs be the final time.
Initial speed=initial distance/initial time=600/x.
Final speed=final distance/final time=600/ (x+1/2).
So "initialspeed-200=finalspeed".
600/x-200=600/ (x+1/2).
600/x-600/ (x+1/2) =200.
Solving this we get.
400x2+200x-600=0.
4x2 + 2x - 6= 0.
2x2 + x - 3 = 0.
2x2 - 2x + 3x - 3 = 0.
2x (x - 1) + 3 (x- 1) = 0.
(2x + 3) (x - 1) = 0.
X=3/2 or x=1.
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