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 7 of 21.
Simi said:
1 decade ago
@ISHAN. The speed of the car is 70 km/hr.
Solve the following eqquation:
t-1 = 420/x+10 km/hr.
= 420/x - 420/ x + 10 = t-t-1.
= 420 x + 4200- 420 x = t-t-1.
= 4200 = -1.
= 4200 / 60 (mins)= 70 kms/hr.
Solve the following eqquation:
t-1 = 420/x+10 km/hr.
= 420/x - 420/ x + 10 = t-t-1.
= 420 x + 4200- 420 x = t-t-1.
= 4200 = -1.
= 4200 / 60 (mins)= 70 kms/hr.
Ravi Shankar said:
1 decade ago
How can you say.
600/x-600/x+(1/2)= 200.
If '200' is the AVERAGE SPEED than the solution should be,
2*(600/x)*(600/x+(1/2))/(600/x + 600/x+(1/2)) = 200.
Average Speed = 2xy/x+y.
Where, x->first speed in km/hr.
y->second speed in km/hr.
Can anybody suggests if my answer is wrong?
600/x-600/x+(1/2)= 200.
If '200' is the AVERAGE SPEED than the solution should be,
2*(600/x)*(600/x+(1/2))/(600/x + 600/x+(1/2)) = 200.
Average Speed = 2xy/x+y.
Where, x->first speed in km/hr.
y->second speed in km/hr.
Can anybody suggests if my answer is wrong?
Sumanth said:
1 decade ago
Let me explain you ravi.
Actual speed should be(without any modifications)=dist/time=600/x
With modifications increasing 30min= 600/(x+1/2).
Difference of(actual speed - modified speed) = 200km As mentioned in question.
So (600/x)-(600/(x-1/2)) = 200.
It is as simple as it is. Why you guys are making it complicated.
Actual speed should be(without any modifications)=dist/time=600/x
With modifications increasing 30min= 600/(x+1/2).
Difference of(actual speed - modified speed) = 200km As mentioned in question.
So (600/x)-(600/(x-1/2)) = 200.
It is as simple as it is. Why you guys are making it complicated.
Arun said:
1 decade ago
@Ravi, 200 is not the avg speed, it says average speed is reduced by 200, not to 200.
Jay said:
1 decade ago
Here A Shortcut method.
Initial speed = 600km/hr = s1.
Final Speed = 600-200 = 400km/hr = s2.
So, (Taking ratio of speed),
s1 : s2 = 600 : 400= 3 : 2.
So, time taken,
t1 : t2 = 2 : 3 (time and speed inversely proportional).
1 part time= 30 min.
So t1= 2*30 = 60 min = 1hr.
Initial speed = 600km/hr = s1.
Final Speed = 600-200 = 400km/hr = s2.
So, (Taking ratio of speed),
s1 : s2 = 600 : 400= 3 : 2.
So, time taken,
t1 : t2 = 2 : 3 (time and speed inversely proportional).
1 part time= 30 min.
So t1= 2*30 = 60 min = 1hr.
Rahim Shaikh said:
1 decade ago
Another method:
let x = original speed.
And y= original time taken.
So 600/x=y--------(1).
Now we know if speed reduces 30 minutes more will be taken.
i.e 30/60 hours.
So,
600/x-200= y+30/60 here 30/60 means 30 minutes.
Now put the value of y as shown in (1),
So equation will be,
600/x-200= 600/x - 30/60.
You will get,
x2-200x-240000=0 here x2 means x square.
(x-600)(x+200) = 0.
X = 600.
That's why
Time =600/600.
=1 hours.
Hope you guys understand
let x = original speed.
And y= original time taken.
So 600/x=y--------(1).
Now we know if speed reduces 30 minutes more will be taken.
i.e 30/60 hours.
So,
600/x-200= y+30/60 here 30/60 means 30 minutes.
Now put the value of y as shown in (1),
So equation will be,
600/x-200= 600/x - 30/60.
You will get,
x2-200x-240000=0 here x2 means x square.
(x-600)(x+200) = 0.
X = 600.
That's why
Time =600/600.
=1 hours.
Hope you guys understand
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...;).
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 "-" ?
Dinesh said:
1 decade ago
2x^2+x-3=0 how to solve this factorisation? So please help me.
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?
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