Question

A plane left 30 minutes late than the schedule time and in order to reach its destination 1500 km away in time,it has to increase its speed by 250 km/h from its usual speed.find its usual speed.

Answer 1

Then speed of aeroplane = distance covered/time taken = 1500km/T hour ---(1)

A/C to question,
Time taken by Aeroplane in 2nd case = sedule time - 30 minutes
= (T - 1/2) hour
Now, speed of aeroplane = distance covered/time taken = 1500/(T - 1/2) ----(2)

Now, 1500/(T - 1/2) - 1500/T = 250 [according to question]
1500[ (T - T + 1/2)/(T² - T/2)] = 250
6(1/2)/(T² - T/2) = 1
3 = T² - T/2
6 = 2T² - T
2T² - T - 6 = 0
2T² - 4T + 3T - 6 = 0
(T - 2)(2T + 3) = 0
T = 2 , -3/2 but T ≠ negative so, T = 2

Hence, usual speed of aeroplane = 1500/T = 1500/2 = 750 km/h

Answer 2

Then speed of aeroplane = distance covered/time taken = 1500km/T hour ---(1)

A/C to question,

Time taken by Aeroplane in 2nd case = sedule time - 30 minutes

= (T - 1/2) hour

Now, speed of aeroplane = distance covered/time taken = 1500/(T - 1/2) ----(2)

Now, 1500/(T - 1/2) - 1500/T = 250 [according to question]

1500[ (T - T + 1/2)/(T² - T/2)] = 250

6(1/2)/(T² - T/2) = 1

3 = T² - T/2

6 = 2T² - T

2T² - T - 6 = 0

2T² - 4T + 3T - 6 = 0

(T - 2)(2T + 3) = 0

T = 2 , -3/2 but T ≠ negative so, T = 2

Hence, usual speed of aeroplane = 1500/T = 1500/2 = 750 km/h