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:
206 comments Page 15 of 21.
Naziya said:
10 years ago
Any one solve this simple and shortly? How this x+1/2 came? I'm not getting.
Manig said:
10 years ago
Its simple yeah.
Why because just try think about question.
Any question read at least 2 times.
Ans:
Speed is reduced by 200km per hour.(not reduced to).
So delay extra 30 minutes i.e. in hours 30/60 = 1/2 = 0.5;.
As we know s=d/t; d=600km; without delay time req to reach destination t=T.
So finally,
(600/t) = speed without any delay.
(600/T+0.5) = speed with delay.
From question,
200 = (600/t) - 600/T+0.5).
Then you will get the answer.
Why because just try think about question.
Any question read at least 2 times.
Ans:
Speed is reduced by 200km per hour.(not reduced to).
So delay extra 30 minutes i.e. in hours 30/60 = 1/2 = 0.5;.
As we know s=d/t; d=600km; without delay time req to reach destination t=T.
So finally,
(600/t) = speed without any delay.
(600/T+0.5) = speed with delay.
From question,
200 = (600/t) - 600/T+0.5).
Then you will get the answer.
AnilKumarReddy said:
10 years ago
When solving the equation 2x^2+x-3 = 0.
We get two factors for x = 1 and x = -3/2.
Here time is always positive. So we are taking x = 1 there is no logic.
We get two factors for x = 1 and x = -3/2.
Here time is always positive. So we are taking x = 1 there is no logic.
NILESH DADARAO PAHURKAR said:
10 years ago
Flight 1st speed = 600 and due to bad weather average speed = 200 increased by = 30 minutes duration.
200 = 600/(x+30).
200(x+30) = 600.
200x + 6000 = 600.
200x = 600-6000 = 5400
x = 5400/200
x = 27.
x+30 = 27+30 = 57.
Answer: 57 min.
200 = 600/(x+30).
200(x+30) = 600.
200x + 6000 = 600.
200x = 600-6000 = 5400
x = 5400/200
x = 27.
x+30 = 27+30 = 57.
Answer: 57 min.
Suresh said:
10 years ago
Please tell me the formula of these type of problems.
Azly said:
10 years ago
Friends, we have to use trigonometry to solve this problems. And why they have mentioned average speed is reduced by 200 km/h since it the reduced "speed".
Shubham said:
10 years ago
Let the speed be 'S' kmph and time 't' hr.
So S = 600/t --- (1).
And since speed got reduced by 200 and time increased but 30 min.
So (s - 200) = 600/ (t + 30/60) --- (2).
Solving 1 and 2 equation we will get:
2t^2 + t-3 = 0 --- (3).
By equation 3 we will get:
t = -3/2 or t = 1/2.
Since time can't be negative. So t = 1/2.
Putting this value in equation 1.
We will get t = 1/2.
But we have to find how much time it will take after it got slowed.
So total time = 1/2 + 30/60 = 1/2 * 2 = 1 hours;
So S = 600/t --- (1).
And since speed got reduced by 200 and time increased but 30 min.
So (s - 200) = 600/ (t + 30/60) --- (2).
Solving 1 and 2 equation we will get:
2t^2 + t-3 = 0 --- (3).
By equation 3 we will get:
t = -3/2 or t = 1/2.
Since time can't be negative. So t = 1/2.
Putting this value in equation 1.
We will get t = 1/2.
But we have to find how much time it will take after it got slowed.
So total time = 1/2 + 30/60 = 1/2 * 2 = 1 hours;
Diya said:
10 years ago
How can we add. Minutes to the hours like 200 hours and 30 minutes.
Bala said:
9 years ago
Simple friends, we solve by using S = D/T.
S - speed.
D - distance.
T - Time.
S = 420/t --> (1).
S + 10 = 420/(t-1) --> (2).
From(2) St- S+ 10t-10 = 420 --> (3)
Let equate (1) and (3), from this we can find speed.
St = ST - S + 10t - 10.
(OR) T = S+10/10.
Now S = D/T.
Substitute the values S = 420/S + 10/10.
S = (420*10)/(S+10).
S = -70 and 60. Neglect the negative S=60.
S - speed.
D - distance.
T - Time.
S = 420/t --> (1).
S + 10 = 420/(t-1) --> (2).
From(2) St- S+ 10t-10 = 420 --> (3)
Let equate (1) and (3), from this we can find speed.
St = ST - S + 10t - 10.
(OR) T = S+10/10.
Now S = D/T.
Substitute the values S = 420/S + 10/10.
S = (420*10)/(S+10).
S = -70 and 60. Neglect the negative S=60.
Sahin said:
9 years ago
@Rahim Shaikh.
There is a small mistake in your answer.
Tt will be:
(x - 600) (x + 400) = 0.
x2 - 200x - 240000 = 0.
Not x+200. Check this once.
There is a small mistake in your answer.
Tt will be:
(x - 600) (x + 400) = 0.
x2 - 200x - 240000 = 0.
Not x+200. Check this once.
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