Aptitude - Time and Distance - Discussion

D = 600 Km,
S = 200 Km/hr
Let the initial speed be x.
So, new speed = x - 200Km/hr,
S = D/T, T = D/S.
So, T= 600/ x-200.
x - 200 = 600,
x = 800,
T = 600/800 - 200 = 600/600 = 1 hour.

Answer is 1.5hours.

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.

D=600 Km,
S=200 Km/hr
Let initial speed be x.

So, new speed = x-200Km/hr
S = D/T, T=D/S.
So, T= 600/ x-200.
x-200 = 600
x = 800
T = 600/800-200 = 600/600 = 1 hour.

Let, The speed of flight is:xkm/hr.

The total distance travelled by flight is = 600km
The speed is reduced to = x - 200km/hr,
Time was taken with speed x = 600/x.
Similarly,

Time taken by flight with reduced speed = 600/(x-200)
By Given Condition,
(600/x) + 1/2 = 600/(x-200)
1200 + x/2x = 600/(x-200)
(x-200) * (1200+x) = 1200x
So, x^2 - 200x - 240000 = 0,
(x+400) * (x - 600) = 0,
So x = 600.
Time taken is = 600/600 = 1hr.

D = 600km.
Average speed is 200km/hr,
s = 600/3 = 200km/hr.
200km time taken 30min.
The remaining time is 400km = 30 + 30 = 60min = time taken = 1hr.

D = S * T.
600 = 200 * (T+30),
600/200 = T+30.
T = 27.
TOTAL DURATION = T+30= 27+30 = 57 ~1HOUR.

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.

New speed (old speed - change in speed) = ΔS/ΔT.
200/.5h = 400km/h.
new time(old time + extra time) = 600/400 = 1.5h,
Initial time is taken 1.5 - 0.5 = 1h.

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.