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:
205 comments Page 1 of 21.
Moustafa Ehosiny said:
1 year 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.
(323)
Mary grace said:
3 years ago
D = 600km.
Speed = 200km/hr,
Time = ?( 30 mins increases).
30T = d/s.
30t = 600/200.
30t = 3,
T = 1.
Speed = 200km/hr,
Time = ?( 30 mins increases).
30T = d/s.
30t = 600/200.
30t = 3,
T = 1.
(207)
Shyam kumar said:
1 year 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.
(100)
Sivanityam said:
1 year 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.
(80)
Vpk said:
2 years ago
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 =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.
(76)
Kaviprema said:
2 years ago
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 - 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.
(72)
Legend said:
2 years ago
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.
Therefore,
x+1/2 = 600/200
x = 3-1/2
x = 5/2.
Therefore total time = x+1/2
So, total time = 3 hr.
(49)
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.
(43)
Ravi said:
4 years ago
S=D/T(30 min increased).
200 = 600/(x+30).
x+30 = 600/200,
x+30 = 3,
x = 27.
Time is always positive.
So, time increses by 30=x+30.
= 27 + 30 = 57 = 1hr.
200 = 600/(x+30).
x+30 = 600/200,
x+30 = 3,
x = 27.
Time is always positive.
So, time increses by 30=x+30.
= 27 + 30 = 57 = 1hr.
(24)
Parvezkhan Pathan said:
1 year 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.
(21)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers