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 14 of 21.
Udiet said:
1 decade ago
How come 600/x-1200/2x+1=200?
Deepak Jain said:
1 decade ago
It is not mentioned in the problem that at what speed duration of flight to be find out because you are giving the answer for the speed before the bad weather.
According to me speed after bad weather should be taken and answer will be 1.5 hours.
According to me speed after bad weather should be taken and answer will be 1.5 hours.
Tony said:
1 decade ago
Hello friends !
Let be simple ...200 reduce speed and we got 30 mnt more time.
Means 400 km speed need how much time ?
200 ---- 30 mnt need.
400 ---- ?
(400 * 30)/200 = 60 mnt.
Let be simple ...200 reduce speed and we got 30 mnt more time.
Means 400 km speed need how much time ?
200 ---- 30 mnt need.
400 ---- ?
(400 * 30)/200 = 60 mnt.
Ranjith said:
1 decade ago
You have done a mistake in this.
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.
It is (x-1)=0 and not (x+1)=0....
So that while equaling this we get as,
x=1...
Hoping you understand now.
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.
It is (x-1)=0 and not (x+1)=0....
So that while equaling this we get as,
x=1...
Hoping you understand now.
Khushbu said:
1 decade ago
@Avy.
In equation (2x-3)(x+1) =0.
Then x=3/2 and x = -1.
So how x=1?
In equation (2x-3)(x+1) =0.
Then x=3/2 and x = -1.
So how x=1?
Ramesh said:
1 decade ago
Why the difference in speed matched with 200km/hr?
Nandasena said:
1 decade ago
It is said that average speed is reduced by 200. Not just speed is reduced by 200. Then we know there is another equation to calculate average speed. But in here it has not used. Can someone please explain?
Dinesh said:
1 decade ago
2x^2+x-3=0 how to solve this factorisation? So please help me.
Kush said:
1 decade ago
Why there is always subtraction in the formula?
i.e. 600/x - 600/(x+.5) = 200.
Why "-" ?
i.e. 600/x - 600/(x+.5) = 200.
Why "-" ?
Khrish said:
1 decade ago
Suppose duration of flight is = x.
Speed = distance/time.
200=(600/x)-(600/(x+30mins)).
200=(600/x)-[600/(x+(30/60)hr)].
200=(600/x)-[600/(x+(1/2))].
200=(600/x)-[600/((2x+1)/2)] by LCM,
200=(600/x)-[1200/2x+1].
200=600[(1/x)-(2/2x+1)].
200/600=(1/x)-(2/2x+1).
1/3=(2x+1-2x)/(2x^2+x).
1/3=1/(2x^2+x).
2x^2+x-3=0.
(2x+3)(x-1)=0.
X=-3/2, x=1.
I hope you got it...;).
Speed = distance/time.
200=(600/x)-(600/(x+30mins)).
200=(600/x)-[600/(x+(30/60)hr)].
200=(600/x)-[600/(x+(1/2))].
200=(600/x)-[600/((2x+1)/2)] by LCM,
200=(600/x)-[1200/2x+1].
200=600[(1/x)-(2/2x+1)].
200/600=(1/x)-(2/2x+1).
1/3=(2x+1-2x)/(2x^2+x).
1/3=1/(2x^2+x).
2x^2+x-3=0.
(2x+3)(x-1)=0.
X=-3/2, x=1.
I hope you got it...;).
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