Aptitude - Time and Distance - Discussion

@Ben You are right.

I am just copying a comment listed above.

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 why 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.

In the above equation we solved x to be 600 where x is the average speed when there isn't any bad weather. But there was bad weather and the flight slowed down to 400 kmph (600kmph-200kmph).

So the time taken for the actual flight due to slow down is 600km/(600kmph-200kmph).

i.e 1 and 1/2 hours.

Any one solve this simple and shortly? How this x+1/2 came? I'm not getting.

Its simple yeah.

Why because just try think about question.
Any question read at least 2 times.

Ans:

Speed is reduced by 200km per hour.(not reduced to).
So delay extra 30 minutes i.e. in hours 30/60 = 1/2 = 0.5;.
As we know s=d/t; d=600km; without delay time req to reach destination t=T.

So finally,

(600/t) = speed without any delay.
(600/T+0.5) = speed with delay.

From question,
200 = (600/t) - 600/T+0.5).

Then you will get the answer.

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.