Aptitude - Time and Distance - Discussion

See we know D = ST.
Now,
d = 600km.
Let,
Old speed = x km/hr <- taken time by T hr.
Original speed = (x-200)Km/hr <- taken time 3T/2.
(3T/2=> T + extra 30mins )=>(1+30/60)T).
So,
600 = xT & 600 = (x-200)3T/2 => (3xT/2) - 300T
Now 300T = 900 - 600.
Finally, T = 1hr.

Let's assume the original average speed of the aircraft for the 600 km trip was "x" km/hr.

When the aircraft was slowed down due to bad weather, its average speed was reduced by 200 km/hr.
So, the new average speed for the trip would be "(x - 200)" km/hr.

The time of flight increased by 30 minutes, which can be converted to hours by dividing by 60. So, the additional time is 30/60 = 0.5 hours.
We can use the formula: Time = Distance/Speed to calculate the duration of the flight.
For the original speed, the time taken would be 600 km/x km/hr = 600/x hours.
For the reduced speed, the time taken would be 600 km/(x - 200) km/hr = 600/(x - 200) hours.

Since the time taken increased by 0.5 hours, we can set up the equation:
600/x + 0.5 = 600/(x - 200).

To solve this equation, we can cross-multiply and simplify:
600(x - 200) + 0.5x(x - 200) = 600x,
600x - 120000 + 0.5x^2 - 100x = 600x,
0.5x^2 - 100x - 120000 = 0.

Dividing the equation by 0.5 to simplify further:
x^2 - 200x - 240000 = 0.

Now we can solve this quadratic equation. Using the quadratic formula:
x = (-b ± √(b^2 - 4ac))/2a
Where a = 1, b = -200, and c = -240000.

Plugging in the values and solving for x:

x = (-(-200) ± √((-200)^2 - 4(1)(-240000)))/(2(1)).
x = (200 ± √(40000 + 960000))/2,
x = (200 ± √1000000)/2,
x = (200 ± 1000)/2.

Now we have two possible values for x:
x1 = (200 + 1000)/2 = 600 km/hr.
x2 = (200 - 1000)/2 = -400 km/hr (rejecting this negative value).

Therefore, the original average speed of the aircraft was 600 km/hr.
To find the duration of the flight, we can substitute this value back into the equation:
Time = Distance / Speed = 600 km / 600 km/hr = 1 hour,
So, the duration of the flight is 1 hour.

Let X be time for duration of the flight;
Time =x + (30/60) --) x + 1/2;
Reduced speed = 200km
Speed directly proportional to distance.
Usual speed = x;
Distance = usual speed ~ reduced speed;
600= x ~ 200;
x = 600 - 200.
So, usual speed= 400km/hr;


T = D/S;
x + 1/2 = 600/400;
x+1/2 = 3/2;
x = 3/2 - 1/2;
x=2/2;
×=1.

Time = Distance/Speed.
Therefore,
x+1/2 = 600/200
x = 3-1/2
x = 5/2.

Therefore total time = x+1/2
So, total time = 3 hr.

Soln,
D - 600 km.
S - 200 km/ hr.
T - ?(but increased by 30 minutes)

Let us assume the time is x,
T - x.
T - > 1 hr - 60 min.
1/2 hr - 30 min.
T - (x + 1/2).

Formula:
T = d/s.
(X + 1/2) = 600/200.
2x + 1 = 3.
2x= 3 - 1,
2x = 2,
X = 1.

So the ans is T = x.
T = 1 hr.

D = 600km.
Speed = 200km/hr,
Time = ?( 30 mins increases).
30T = d/s.
30t = 600/200.
30t = 3,
T = 1.

Thanks, @Ravi.

@Saravanakgopi.

It's wrong.

When you got 200 (x+30) =600. Then the next step would be 200x=600-6000. The result you get is negative.

So the final answer you get if you follow that method would be x= -27. So when you add it with increased time then -27+30 Which is 3.

Can we take the difference of time instead of speed? Please anyone explain me.

Thanks @Kiran.