solve it fast plzz
it's my paper tomorrow
Attachments:

Answers
Answered by
7
Important Formulas:
(i) Speed = Distance/Time
(ii) Time = Distance/Speed.
According to the Question,
Let the usual speed be 'x'.
Given that it has increased its speed by 100 km/hr.
So, the speed of the plane = (x + 100) km/hr.
Given that it left 30 minutes later = (30/60) = 1/2.
Given Distance = 1500 km.
Now, We have Distance and Speed. Apply the 2nd formula:
⇒ (1500/x) - (1500/x + 100) = 1/2
⇒ (1500x + 150000 - 1500x)/x(x + 100) = 1/2
⇒ 150000 * 2 = x(x + 100)
⇒ 300000 = x^2 + 100x
⇒ x^2 + 100x - 300000 = 0
⇒ x^2 + 600x - 500x - 300000 = 0
⇒ x(x + 600) - 500(x + 600) = 0
⇒ (x - 500)(x + 600) = 0
⇒ x = 500,-600 {Speed cannot be negative}
⇒ x = 500.
Therefore, the usual speed of the plane = 500 km/hr.
Hope this helps!
siddhartharao77:
:-)
Answered by
3
Let to be usual speed of plane X km/hr
Increased speed of the plane = (x + 100) km/hr
Time taken to reach the destination at usual speed,
T1=1500/x hr
Time taken to reach the destination at increased speed,
T2=1500/100+x
Given,
t 1 – t 2 = 30 minutes


Note:- Speed Never be Negative so I taken positive value.
Thanks!!!
Increased speed of the plane = (x + 100) km/hr
Time taken to reach the destination at usual speed,
T1=1500/x hr
Time taken to reach the destination at increased speed,
T2=1500/100+x
Given,
t 1 – t 2 = 30 minutes
Note:- Speed Never be Negative so I taken positive value.
Thanks!!!
Similar questions