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 4 of 21.
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.
Geeth said:
1 decade ago
How? Can you explain nani?
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
Hunny said:
1 decade ago
600/(x-200) - 600/x = 30/60(in hrs)
Please anyone explain.
Please anyone explain.
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.
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)
Sunita said:
1 decade ago
Hey Saravanakgopi. How did you get 5400? please explain.
Amit said:
1 decade ago
We know that 1/x inc in one quantity leds to 1/(x+1) dec in other quantity
1/(x+1)= 1/2
=> x=1
1/(x+1)= 1/2
=> x=1
Nivas said:
1 decade ago
@saravanakgopi.
What he said is wrong. Because he didn't convert hr into min. So that his method is wrong.
What he said is wrong. Because he didn't convert hr into min. So that his method is wrong.
Mukesh Singh said:
1 decade ago
Sorry Dear...
But I found sentence mistake is this Qs.
If in Qs it is asked that speed is reduced to 200km/hr then given solution is right.
But in qs it asked that speed is reduced by, so soln should be
600/(t+1/2) = 200
t = 5/2 hr
But I found sentence mistake is this Qs.
If in Qs it is asked that speed is reduced to 200km/hr then given solution is right.
But in qs it asked that speed is reduced by, so soln should be
600/(t+1/2) = 200
t = 5/2 hr
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