a plane left 30 minutes later than the scheduled time in order to reach its destination 1500 km away in time it has to increase the speed by 250km/hr .find its usual speed
Answers
Answered by
2
Let the speed be x km/h
Time=Distance/speed
Therefore,
1500/x+1500/(x+250)=1/2
then simplify by cross multiplication
write it down only then you will understand!!
Time=Distance/speed
Therefore,
1500/x+1500/(x+250)=1/2
then simplify by cross multiplication
write it down only then you will understand!!
rajpreet2:
plese solve it full
it is simple maths so i thought you could do it. Anyway,
there is a mistake
its 1500/x-1500/(x+250)=1/2 after cross multiplication, 1500(x+250)-1500(x)/(x*(x+250))=1/2. Now at least you can do it
Answered by
10
Solution :-
Let the original speed of train be x km/hr
New speed = (x + 250) km/hr
We know that,
Time = Distance / Speed
Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.
According to the question,
=> 1500/x - 1500/(x + 250) = 1/2
=> (1500x + 37500 - 1500x)/x(x + 250) = 1/2
=> 2(37500) = x(x + 250)
=> 75000 = x² + 250x
=> x² + 250x - 75000 = 0
=> x² + 1000x - 750x - 75000 = 0
=> x(x + 1000) - 750(x + 1000) = 0
=> (x - 750) (x + 1000) = 0
=> x = 750 or x = - 1000
∴ x ≠ - 1000 (Because speed can't be negative)
Hence,
Its usual speed = 750 km/hr
Let the original speed of train be x km/hr
New speed = (x + 250) km/hr
We know that,
Time = Distance / Speed
Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.
According to the question,
=> 1500/x - 1500/(x + 250) = 1/2
=> (1500x + 37500 - 1500x)/x(x + 250) = 1/2
=> 2(37500) = x(x + 250)
=> 75000 = x² + 250x
=> x² + 250x - 75000 = 0
=> x² + 1000x - 750x - 75000 = 0
=> x(x + 1000) - 750(x + 1000) = 0
=> (x - 750) (x + 1000) = 0
=> x = 750 or x = - 1000
∴ x ≠ - 1000 (Because speed can't be negative)
Hence,
Its usual speed = 750 km/hr
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