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.

Distance: 600km
Let the original speed: s.

(avg speed reduced) so: s - 200.
let original time: T.
(time increased by 30 min => 30/60 => 1/2) so: T+1/2
s = d/t;
s = 600/t ----> (1);

New speed is
s - 200 = 600/(T +1/2) -----> (2)
Sub (1) in (2);
600/T - 200 = 600 / (T+1/2);
(600 - 200T)/ T = (600 *2)/(2T+1);

Cross multiply:
(2T+1)(600-200T) = (1200)*T;
1200T - 400T^2 + 600 -200T -1200T = 0;
400T^2 + 200T - 600 = 0;
2T^2 + T - 3 = 0;
(2T + 3 ) (T-1) 0
T-1 = 0 => T = 1.
2T+3 =0 => T=-3/2(neglect the negative value)
Hence time is 1 HOUR.

The speed of the second train =400/4 = 100km/h.
Distance is same in both cases:

Then,
The ratio of speed = reverse ratio of time
7/8 = 8/7.

Speed of first train = 100/8/7= 87.5km/h.

Simple,
Here is a trick, for such question, always break distance in its multiple,

For e,g.
600km = 300kmph x 2 hour.
Or
600km = 600kmph x1 hour

So we can see for 300kmph it takes 2 hours and for 600kmph it takes 1 hour.
And that's the answer.

Let's denote the original speed of the aircraft as "S" km/hr and the original duration of the flight as "T" hours. We're given that the flight was slowed down by 200 km/hr, so its reduced speed is (S - 200) km/hr.

We know that:

Distance = Speed × Time

So, the original duration of the flight can be expressed as:

T = 600 km / S

And the duration of the flight with the reduced speed is:

T + 0.5 hours (since it increased by 30 minutes, which is 0.5 hours).

Now, we can set up an equation for the changed situation:
600 km = (S - 200) km/hr × (T + 0.5 hours).

Now, substitute the value of T from the original duration equation:
600 km = (S - 200) km/hr × (600 km/S + 0.5 hours).

To solve for S, let's get rid of the fractions by multiplying both sides by S:
600S = (S - 200) × (600 + 0.5S).

Now, distribute on the right side of the equation:
600S = 600S - 200S + 0.5S^2 - 100S.

Combine like terms:
600S = 500S + 0.5S^2 - 100S,

Now, simplify further:
600S = 400S + 0.5S^2.

Rearrange the equation:
0.5S^2 = 200S.

Now, divide both sides by 0.5 to isolate S:
S^2 = 400S.
Divide both sides by S:
S = 400 km/hr.

Now that we have the original speed, we can find the original duration of the flight:
T = 600 km / 400 km/hr = 1.5 hours.
So, the original duration of the flight was 1.5 hours.

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

Lcm is
You get (2x+1)(600) - 1200x = 200(2x^2 +x).
1200x+600-1200x = 200(2x^2 +x),
600/200(2x^2+1),
3/(2x^2+x),
2x^2+x = 3.
X = 1.

@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.

Simple way to solve this problem:

Distance = speed * time.

Let the distance be d and the time taken by the plane at 600 km/hr be a
Then:

d=600*a (1) at regular speed
and, d=(600-200) * (a+.5 hr) (2) at reduced speed
Equating the two we get;

600*a = 400(a+.5)
600a = 400a + 200
200a=200
a=1.

The time taken is 1hr.

Speed = distance/time.
Time = distance/speed.
Speed=600-200 = 400 km/hr.
Distance to cover = 600 km.

Duration required(time) = 600/400 = 3/2 hr.
3/2*60 = 60 min = 1hr.