If the speed of an aeroplane is reduced by 40km/hr, it takes 20 minutes more to cover 1200m. Find the time taken by aeroplane to cover 1200m initially.
We know from the distance formula (d) using rate (r) and time (t) that:
d = rt
Regular speed:
1200 = rt
Divide each side by t, we get:
r = 1200/t
Reduced speed. 20 minutes = 60/20 = 1/3 of an hour. So we multiply 1,200 by 3
3600 = (r - 40)(t + 1/3)
If we multiply 3 by (t + 1/3), we get:
3t + 1
So we have:
3600 = (r - 40)(3t + 1)
Substitute r = 1200/t into the reduced speed equation:
3600 = (1200/t - 40)(3t + 1)
Multiply through and we get:
3600 = 3600 - 120t + 1200/t - 40
Subtract 3,600 from each side
3600 - 3600 = 3600 - 3600 - 120t + 1200/t - 40
The 3600's cancel, so we get:
- 120t + 1200/t - 40 = 0
Multiply each side by t:
-120t^2 - 40t + 1200 = 0
We've got a quadratic equation. To solve for t, we type this in our search engine and we get:
t = -10/3 or t = 3. Since time cannot be negative, our final answer is:
t = 3
We know from the distance formula (d) using rate (r) and time (t) that:
d = rt
Regular speed:
1200 = rt
Divide each side by t, we get:
r = 1200/t
Reduced speed. 20 minutes = 60/20 = 1/3 of an hour. So we multiply 1,200 by 3
3600 = (r - 40)(t + 1/3)
If we multiply 3 by (t + 1/3), we get:
3t + 1
So we have:
3600 = (r - 40)(3t + 1)
Substitute r = 1200/t into the reduced speed equation:
3600 = (1200/t - 40)(3t + 1)
Multiply through and we get:
3600 = 3600 - 120t + 1200/t - 40
Subtract 3,600 from each side
3600 - 3600 = 3600 - 3600 - 120t + 1200/t - 40
The 3600's cancel, so we get:
- 120t + 1200/t - 40 = 0
Multiply each side by t:
-120t^2 - 40t + 1200 = 0
We've got a quadratic equation. To solve for t, we type this in our search engine and we get:
t = -10/3 or t = 3. Since time cannot be negative, our final answer is:
t = 3