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:
210 comments Page 10 of 21.
Hari said:
1 decade ago
It's so simple like.
T= D/S => 600/300=30.
As it is given + 30 mins then 30+30 is 60=1 hour.
T= D/S => 600/300=30.
As it is given + 30 mins then 30+30 is 60=1 hour.
Amala said:
1 decade ago
@Mr. saravanagopi,
we already calculate the time as x+30 in [200(x+30)=600] right. then why did you again calculating the result as 27+30.
I don't understand...
we already calculate the time as x+30 in [200(x+30)=600] right. then why did you again calculating the result as 27+30.
I don't understand...
Rahul said:
1 decade ago
very very easy question
distance 600 km.
suppose aircraft speed is 600km/hr
time =600/600=1hr
ans=1hr
check method=speed decreasing 200km/h
speed is 400km/h distance 600
600/400=1.5=======30 mint late original duration is 1 hr..ans
distance 600 km.
suppose aircraft speed is 600km/hr
time =600/600=1hr
ans=1hr
check method=speed decreasing 200km/h
speed is 400km/h distance 600
600/400=1.5=======30 mint late original duration is 1 hr..ans
Babita said:
1 decade ago
@Manik , Sunita ,
200 = 600/x+30
200(x+30)=600
200x+6000=600
200x=600-6000
200x=5400
x=5400/200
x=27
200 = 600/x+30
200(x+30)=600
200x+6000=600
200x=600-6000
200x=5400
x=5400/200
x=27
Muhammad Umair Kabir said:
1 decade ago
Let Speed is X and Time is Y,
According to formula
X = 600 / Y
From question avg Speed reduced and Time taken increased eq. becomes...
X - 200 = 600 / (Y + 1/2)
put the value of X in above eq.
600 / Y - 200 = 600 / (Y + 1/2)
Simplifying the eq. we get
(Y - 1)(2Y + 3 )...
Y = 1 Ans as time is always positive....
According to formula
X = 600 / Y
From question avg Speed reduced and Time taken increased eq. becomes...
X - 200 = 600 / (Y + 1/2)
put the value of X in above eq.
600 / Y - 200 = 600 / (Y + 1/2)
Simplifying the eq. we get
(Y - 1)(2Y + 3 )...
Y = 1 Ans as time is always positive....
Ishan said:
1 decade ago
A car covers a distance of 420 km at a certain speed. If its speed is 10 km/hr more, it will take one hour less to cover the same distance. What is its speed?
Please help me?
Please help me?
Karthilk said:
1 decade ago
A car travelling with of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car?
Racer said:
1 decade ago
@ISHAN the speed of the car is 60km/hr.
1)Let the speed be x km/hr.
2)Given distance covers =420km we know time t= distance/speed.
3)So let t= 420/x.
4)Given if speed is 10km/hr more so time (t-1)= 420/(x+10).
We have considered t=t-1 above because they have mentioned 1hr less so 420/x - 420/(x+10)=(t-t-1) i.e total distance/totalspeed = time
5)By solving we get x^2+10x-4200=0 we get x=60 &-70 ignore -70... 60 is answer.
1)Let the speed be x km/hr.
2)Given distance covers =420km we know time t= distance/speed.
3)So let t= 420/x.
4)Given if speed is 10km/hr more so time (t-1)= 420/(x+10).
We have considered t=t-1 above because they have mentioned 1hr less so 420/x - 420/(x+10)=(t-t-1) i.e total distance/totalspeed = time
5)By solving we get x^2+10x-4200=0 we get x=60 &-70 ignore -70... 60 is answer.
Alokpal said:
1 decade ago
Time = distance / speed.
600 *30 /200.
30 * 3 = 90 total time.
Exact time = total time - increase time.
= 90 - 30.
= 60.
600 *30 /200.
30 * 3 = 90 total time.
Exact time = total time - increase time.
= 90 - 30.
= 60.
Ayesha said:
1 decade ago
Still confuse about 600/x-600/(x+1/2) = 200. Please let me know something very simply.
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