Aptitude - Time and Distance - Discussion

When solving the equation 2x^2+x-3 = 0.

We get two factors for x = 1 and x = -3/2.

Here time is always positive. So we are taking x = 1 there is no logic.

Flight 1st speed = 600 and due to bad weather average speed = 200 increased by = 30 minutes duration.

200 = 600/(x+30).

200(x+30) = 600.

200x + 6000 = 600.

200x = 600-6000 = 5400

x = 5400/200

x = 27.

x+30 = 27+30 = 57.

Answer: 57 min.

Please tell me the formula of these type of problems.

Friends, we have to use trigonometry to solve this problems. And why they have mentioned average speed is reduced by 200 km/h since it the reduced "speed".

Let the speed be 'S' kmph and time 't' hr.

So S = 600/t --- (1).

And since speed got reduced by 200 and time increased but 30 min.

So (s - 200) = 600/ (t + 30/60) --- (2).

Solving 1 and 2 equation we will get:

2t^2 + t-3 = 0 --- (3).

By equation 3 we will get:

t = -3/2 or t = 1/2.

Since time can't be negative. So t = 1/2.

Putting this value in equation 1.

We will get t = 1/2.

But we have to find how much time it will take after it got slowed.

So total time = 1/2 + 30/60 = 1/2 * 2 = 1 hours;

How can we add. Minutes to the hours like 200 hours and 30 minutes.

Simple friends, we solve by using S = D/T.
S - speed.
D - distance.
T - Time.
S = 420/t --> (1).
S + 10 = 420/(t-1) --> (2).
From(2) St- S+ 10t-10 = 420 --> (3)

Let equate (1) and (3), from this we can find speed.
St = ST - S + 10t - 10.
(OR) T = S+10/10.

Now S = D/T.
Substitute the values S = 420/S + 10/10.
S = (420*10)/(S+10).
S = -70 and 60. Neglect the negative S=60.

@Rahim Shaikh.

There is a small mistake in your answer.

Tt will be:
(x - 600) (x + 400) = 0.
x2 - 200x - 240000 = 0.
Not x+200. Check this once.

How you got 1200/2x+1? Please explain it.

Dear all.

I think ,duration of the flight is 1.5 hour.

With the calculations, normal speed = 600 km/hr.
So reduced speed = 600 - 200 = 400 km/hr.
To cover the journey of 600 km @ 400 km/hr time taken = 600/400 = 1.5 hour.