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 4 of 21.
Sumanth said:
1 decade ago
Let me explain you ravi.
Actual speed should be(without any modifications)=dist/time=600/x
With modifications increasing 30min= 600/(x+1/2).
Difference of(actual speed - modified speed) = 200km As mentioned in question.
So (600/x)-(600/(x-1/2)) = 200.
It is as simple as it is. Why you guys are making it complicated.
Actual speed should be(without any modifications)=dist/time=600/x
With modifications increasing 30min= 600/(x+1/2).
Difference of(actual speed - modified speed) = 200km As mentioned in question.
So (600/x)-(600/(x-1/2)) = 200.
It is as simple as it is. Why you guys are making it complicated.
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)
Chetna said:
8 years ago
30 min =1/2 hr.
Now I take time T in hr.
According to the question..... Distance is same so,
Distance of Normal flight = Distance due to bad weather (reduce by 200 kmph).
We know, distance = time * speed.
T*600 = ( 600-200) (T + 1/2) ..... (600-200 bcz speed is reduce by 200 kmph).
600T = 400T +200.
T= 1 hr.
Now I take time T in hr.
According to the question..... Distance is same so,
Distance of Normal flight = Distance due to bad weather (reduce by 200 kmph).
We know, distance = time * speed.
T*600 = ( 600-200) (T + 1/2) ..... (600-200 bcz speed is reduce by 200 kmph).
600T = 400T +200.
T= 1 hr.
Ramya Chockalingam said:
7 years ago
The very simple method is everyone knows that;
Distance=speed*time.
Here the given is speed and distance.
And also time that it is increased by 30min.
Therefore t=x+30 because it gets increased to 30min.
Convert 30 min to hour because the given options are an hour.
Therefore.
X+1/2=600/200.
Therefore, X=1hour.
Distance=speed*time.
Here the given is speed and distance.
And also time that it is increased by 30min.
Therefore t=x+30 because it gets increased to 30min.
Convert 30 min to hour because the given options are an hour.
Therefore.
X+1/2=600/200.
Therefore, X=1hour.
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)
Ravi Shankar said:
1 decade ago
How can you say.
600/x-600/x+(1/2)= 200.
If '200' is the AVERAGE SPEED than the solution should be,
2*(600/x)*(600/x+(1/2))/(600/x + 600/x+(1/2)) = 200.
Average Speed = 2xy/x+y.
Where, x->first speed in km/hr.
y->second speed in km/hr.
Can anybody suggests if my answer is wrong?
600/x-600/x+(1/2)= 200.
If '200' is the AVERAGE SPEED than the solution should be,
2*(600/x)*(600/x+(1/2))/(600/x + 600/x+(1/2)) = 200.
Average Speed = 2xy/x+y.
Where, x->first speed in km/hr.
y->second speed in km/hr.
Can anybody suggests if my answer is wrong?
ARUNKUMAR S said:
7 years ago
To solve this problem within a second means. First find the equation.
Speed is reduced by 200kmph.
So,
Speed1 - speed2. = 200kmph.
D/t1 - D/(t2+ 1/2) = 200kmph.
600/x - 600(x+1/2) = 200kmph.
Substitute all the options given in the question one by one.
You will LHS =RHS.
Speed is reduced by 200kmph.
So,
Speed1 - speed2. = 200kmph.
D/t1 - D/(t2+ 1/2) = 200kmph.
600/x - 600(x+1/2) = 200kmph.
Substitute all the options given in the question one by one.
You will LHS =RHS.
(1)
Magnesh said:
8 years ago
Let the time be x.
Total speed(Dis/time) - actual speed due to weather(dis/time) = reduced speed.
600/x - 600/ x+(1/2) = 200.
Here actual speed is less because of weather condition, which means it has taken MORE TIME which is 1/2 hour or 30 min.
Solve answer you get an answer as 1 hr.
Total speed(Dis/time) - actual speed due to weather(dis/time) = reduced speed.
600/x - 600/ x+(1/2) = 200.
Here actual speed is less because of weather condition, which means it has taken MORE TIME which is 1/2 hour or 30 min.
Solve answer you get an answer as 1 hr.
Malaya Kumar said:
2 years ago
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.
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.
(18)
Ranjith said:
1 decade ago
You have done a mistake in this.
400x2+200x-600=0.
4x2 + 2x - 6= 0.
2x2 + x - 3 = 0.
2x2 - 2x + 3x - 3 = 0.
2x (x - 1) + 3 (x- 1) = 0.
(2x + 3) (x - 1) = 0.
x=3/2 or x=1.
It is (x-1)=0 and not (x+1)=0....
So that while equaling this we get as,
x=1...
Hoping you understand now.
400x2+200x-600=0.
4x2 + 2x - 6= 0.
2x2 + x - 3 = 0.
2x2 - 2x + 3x - 3 = 0.
2x (x - 1) + 3 (x- 1) = 0.
(2x + 3) (x - 1) = 0.
x=3/2 or x=1.
It is (x-1)=0 and not (x+1)=0....
So that while equaling this we get as,
x=1...
Hoping you understand now.
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