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.
Reeta said:
1 decade ago
600/x-1200/(2x+1) = 200
divide eq by 100,we get the eq.
6/x-12/(2x+1) = 2
Now its simple to solve
divide eq by 100,we get the eq.
6/x-12/(2x+1) = 2
Now its simple to solve
Saravanakgopi said:
1 decade ago
Another method: We know
Speed = Distance covered/ time taken
speed = 200km/hr
Distance=600km
time=x+30 (30 min. increase)
Apply the formule:
200 =600/x+30
200(x+30)=600
200x =5400
x = 27 min.
Now Time = x+30 so 27+30= 57 Min. Approx 1 Hr.
Speed = Distance covered/ time taken
speed = 200km/hr
Distance=600km
time=x+30 (30 min. increase)
Apply the formule:
200 =600/x+30
200(x+30)=600
200x =5400
x = 27 min.
Now Time = x+30 so 27+30= 57 Min. Approx 1 Hr.
Pavan said:
1 decade ago
Helloo kiran I have 1 doubt you say that x is in hr so 3 you taken as x/2 but we dnt't know what is x value it may be 2hr 3hr. Then how you will take 3min=x/2.
Madhu said:
1 decade ago
Excellent saravanakgopi.
Anu said:
1 decade ago
@Saravanakgopi tell me how you get 5400 in your result.
Manasi said:
1 decade ago
Awesome @Saravanakgopi.
Thanks dude.
Thanks dude.
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.
Ram said:
1 decade ago
@avy.
Thanks I am clear now.
Thanks I am clear now.
Raghu said:
1 decade ago
Thanks avy.
Srilu said:
1 decade ago
Where from that x/2 come ya?
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