Question

An areoplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km / hr from its usual speed. Find its usual speed.

Answer 1

usual speed of plane is 300km/hr

•let the original speed of aeroplane be

x km/hr

•also time = distance /speed

• time =Ti = 1200/x hours

•now , if speed = x+100 km/hr

• time = Tj = 1200/(x+100)

• also Ti-Tj= 1

• 1200/x-1200/(x+100)=1

• (x+100-x)/x(x+100) =1/1200

• 100×1200=x(x+100)

• 120000 = x²+100x

• 0= x²+100x -120000

• 0= x²+400x-300x-120000

• 0= x(x+400)-300(x+400)

• 0= (x+400)(x-300)

• x=-400 or x=300

• speed can never be negative

Answer 2

The usual speed of Aeroplane is 300 km/h .

Step-by-step explanation:

Given as :

The distance of journey = D = 1200 km

The usual time = T hours

The usual speed of Aeroplane = S km/h

The actual time taken = (T - 1) hours

The actual speed of Aeroplane = (S + 100) km/h

According to question

∵ Distance = Speed × Time

For usual Speed and time

D = S km/h × T h

i.e  D = S T  km            .......1

Or,  ST = 1200

∴       T = $\dfrac{1200}{S}$

And

For actual speed and time

D = (S + 100) km/h × (T - 1 ) h

i.e  D = ST - S + 100 T - 100

From eq 1

ST = ST - S + 100 T - 100

Or, S - 100 T + 100 = 0

put the value of T

S - 100 × $\dfrac{1200}{S}$  + 100 = 0

i.e S² + 100 S - 120000 = 0

Solving the quadratic eq

S² + 400 S - 300 S - 120000 = 0

Or, S ( S + 400 ) - 300 ( S + 400 ) = 0

Or, ( S + 400 ) ( S - 300 ) = 0

∴   ( S + 400 ) = 0

i.e  S = -400 km/h

And  ( S - 300 ) = 0

i.e S = 300 km/h

So, The usual speed of Aeroplane = S = 300 km/h

Hence, The usual speed of Aeroplane is 300 km/h . Answer