Question

plz solve thus in proper steps ( iv)

Answer 1

Hope it helps you :)

Answer 2

Let the speed of train be a km/h

Distance = 480 km

Time taken in first time = (480 / a) hours

T1 = (480/a)

Time taken in second time when speed is 8 km/h less = (480/ a-8) hours

T2 = (480/a-8)


According to question

T2 - T1 = 3

$\frac{480}{a - 8} - \frac{480}{a} = 3 \\ 480( \frac{1}{a - 8} - \frac{1}{a} ) = 3 \\ \frac{a - a+ 8}{a(a - 8)} = \frac{3}{480} \\ \frac{8}{ {a }^{2} - 8a} = \frac{1}{160} \\ {a}^{2} - 8a = 1280 \\ {a}^{2} - 8a - 1280 = 0$

a^2 - 8a - 1280 = 0

=> a^2 - 40a +32a - 1280 = 0

=> a(a-40)+32(a-40) = 0

=> (a-40)(a+32) = 0

a = 40 and - 32

But speed can't be negative.

So, a = 40

Speed of train = 40 km/h