Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 6)
6.
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
Answer: Option
Explanation:
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
2x2 + x - 3 = 0
(2x + 3)(x - 1) = 0
x = 1 hr. [neglecting the -ve value of x]
Discussion:
208 comments Page 1 of 21.
Shiv said:
2 months ago
Average speed = total distance/total time
Average speed = 200km/hr.
Total distance = 600km.
otal time = x+(x+30),
= 2x+30 min,
= 2x+30/60,
= 2x+1/2 hr.
And put these values into the formula.
200 = 600/2x + 1/2.
x = 1 hour.
Average speed = 200km/hr.
Total distance = 600km.
otal time = x+(x+30),
= 2x+30 min,
= 2x+30/60,
= 2x+1/2 hr.
And put these values into the formula.
200 = 600/2x + 1/2.
x = 1 hour.
(3)
Mahadeb Das said:
4 months ago
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.
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.
(22)
Mohamed Niyas said:
6 months ago
Answer is 1.5hours.
(4)
Circle said:
1 year ago
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.
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.
(16)
Moustafa Ehosiny said:
2 years ago
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.
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.
(372)
Parvezkhan Pathan said:
2 years ago
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.
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.
(23)
Sivanityam said:
2 years ago
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.
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.
(88)
Shyam kumar said:
2 years ago
D = S * T.
600 = 200 * (T+30),
600/200 = T+30.
T = 27.
TOTAL DURATION = T+30= 27+30 = 57 ~1HOUR.
600 = 200 * (T+30),
600/200 = T+30.
T = 27.
TOTAL DURATION = T+30= 27+30 = 57 ~1HOUR.
(115)
Anisha said:
2 years ago
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.
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.
(16)
Nitin Kataria said:
2 years ago
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.
200/.5h = 400km/h.
new time(old time + extra time) = 600/400 = 1.5h,
Initial time is taken 1.5 - 0.5 = 1h.
(45)
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