Aptitude - Time and Distance - Discussion

One hour answer is logically wrong. If we assume that original speed was 800 km/hr, then, 600/800 = 45mints will be the time taken at original speed.

And after reduction of 200 km/hr new speed will be 600 km/hr, then time would reduce by 15 mints, not 30 mints.

30 mints can only be possible if original speed was 600km/hr which is reduced to 400km/hr. then 600/400 = 1.50, which is 1 hour and 30 mins.

Let, x is original speed.

(600/x - 200) - (600/x) = 30/60,
Solving this equation, we get x = 600.
Duration of flight = 600/600 = 1 hr.

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.
2x^2 + x - 3 = 0.
(2x + 3) (x - 1) = 0.

Then, x = 1 hr.

600/x - 600/x + 1/2 = 200,

First solve the denominaor of 600/x + 1/2,

600/x - 600 * 2/2(x) + 1 = 200,

600/x - 1200/2x + 1 = 200,

Take LCM, 600(2x + 1) - 1200(x)/x (2x + 1) = 200,

1200x + 600 - 1200x/x (2x + 1) = 200,

As we can see 1200x substract from 1200x,

So the new equation is 600/x (2x + 1) = 200,

Or we can say 600/200 = x(2x + 1),

Now 3 = x(2x + 1),

2x^2 + x - 3 = 0,

2x^2 - 2x + 3x - 3 = 0.

2x (x - 1) + 3 (x- 1) = 0.

(2x + 3) (x - 1) = 0.

x = 3/2 or x = 1.

Thanks for your solution @Kiran.

Let: Time = T, Speed = x

600/x = t ----> (1)
600/x - 200 = T + 1/2,
600/x - 200 = 3/2 T,
600 * 2/(x - 200) * 3 = T.
1200/3x - 600 = T---> (2)

BySolving..

1200/3x - 600 = 600/x.
3x - 600 = 2x.
x = 600.

Hence Speed is 600.

Since D/S = T.
D = 600.
S = 600.
T = 1.

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.

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

@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.

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.