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:
1 hour
2 hours
3 hours
4 hours
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 20 of 21.

Vishnu said:   1 decade ago
Thank you kiran.

Priya said:   1 decade ago
Yes, please tell me if any short method is available.
(1)

Rafi said:   1 decade ago
Your right shanti but is there any easy and time saving solution for it?

Jaanaki said:   1 decade ago
Thank you shanthi.

Shanthi said:   1 decade ago
Its simple only...take LCM of (600/x)-(1200/(2x+1))=200
you get (2x+1)(600)-1200x=200(2x^2 +x)
2x^2 +x=3
which is nothing but x(2x+1)=3

Sivakumaran said:   1 decade ago
X (2x + 1) = 3.

How this step comes along.

Neelima said:   1 decade ago
60 mins = 1 hr

Then, 30 mins = 1/2 hr.

Sivaram said:   1 decade ago
Hi Bhavana, I can't get the answer why you got 1/2. How you get it?

Bhavana said:   2 decades ago
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.

Hope you got d answer Sundar.

Sundar raj said:   2 decades ago
@Kiran

your answer is right only. But can you briefly explain how to find the value X from 2x2 + x - 3 = 0.
(1)


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