Question

answer the question please

Answer 1

Step-by-step explanation:

s =360 km

let v be the speed & t be time taken by train

v= s/t.

t = 360/v. (1)

v+5 = 360/(t-1) ( 2)

substitute value of t from (1) into (2)

v+5 = 360 /{(360/v) -1}

v+5 = 360v/360-v

360v+1800-v^2-5v -360v = 0

v^2+5v-1800=0

solving above quadratic equation, we get

v=40 km/h & v = -45 km/h

v=-45 km/h is not possible, since speed is never negative

hence v= 40 km/h is the answer

Answer 2

Answer:

40km/hr

Step-by-step explanation:

let the initial speed of train be x km/hr.

final speed of train=(x+5)km/hr

distance=d=360km

According to the condition,

360/(x+5) - 360/x= 1 since,time=distance/speed

360(1/(x+5)- 1/x)=1

1/x+5 - 1/x = 1 /360

x-x+5/x(x+5) =1/360 the

5/x^2+5x =1/360

5×360 =x^2 +5x

1800 =x^2+5x

x^2 +5x-1800 =0

x^2 -40x+45x-1800=0

x(x-40)+45(x-40) =0

(x-40)(x+45) =0

so, x-40=0 or x+45=0

so, x =40 or x =-45

but ,speed cannot be negative

therefore,x=40 is valid

hence,speed= 40km/hr

So, the speed of train is 40km/hr.

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