Question

22. If the speed of a train is increased by 5 km/h from
its normal speed it would have taken 2 h less to
cover 300 km. What is its normal speed?
(a) 20 km/h
(b) 25 km/h
(c) 30 km/h
(d) 45 km/h

Answer 1

Answer:

Distance to be covered = 300 km,

Let normal speed of the train = n kmph. Normal duration of the journey, T = 300/n hours.

With increased speed of 5 kmhr journey time = 300/(n+5) = T-2.

T = 300/n …(1)

T-2 = 300/(n+5) …(2)

Subtract (2) from (1) to get

2 = 300/n -300/(n+5), or

2n(n+5) = 300(n+5) - 300n = 1500, or

n(n+5) = 750, or

n^2+5n-750 = 0

(n+30)(n-25) = 0

n = 25 kmph.

The normal speed of the train is 25 kmph. Answer [The negative value of n=-30 is unacceptable, here].

Check. Normal speed of 25 kmph. Duration of the journey = 300/25 = 12 hours.

With increased speed of 30 kmph, Revised duration of the journey = 300/30 = 10 hours. Correct.

Answer 2

Answer:

Explanation:

distance=300km  

let the time be t

and the speed me s

now speed=distance/time

so s=300/t or t=300/s(first)

and s+5=300/t-2

(from 1)

(s+5)(300-2s)=300s

300s-2s(square)+1500-10s=300s

-2s(square)-10s=-1500

2s(square)+10s=1500

2s(s+5)=1500

s(square)+5s=750

now

s(square)+5s-750=0

s(square)+30s-25s-750=0

s(s+30)-25(s+30)=0

(s-25)(s+30)=0

s=25and-30

speed can not be in negative so speed of train is equal to 25km perhour