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 2 of 21.
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)
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)
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)
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)
Shraddha Patel said:
3 years ago
Thanks, @Ravi.
(3)
Varun said:
3 years ago
@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.
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.
(6)
Yash Soni said:
3 years ago
Can we take the difference of time instead of speed? Please anyone explain me.
ANKIT said:
4 years ago
Thanks @Kiran.
(3)
Nitu said:
4 years ago
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.
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.
(8)
Nitu said:
4 years ago
Thank you everyone for explaining.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers