A passenger train takes two hours less for a journey of 300km if its speed is increased by 5 kmph

25 kmph

let normal speed = x
Increased speed = x + 5

Initial => x = 300/t1
=> t1 = 300/x

After => (x + 5) = 300/(t1 - 2)
=> (x + 5) = 300 / [ (300/x) - 2]
=> x = 25kmph
11 years agoHelpfull: Yes(8) No(4)
300/2=2+(300/x+5)
X^2+5X-750=0
X=25 or -30(which is not possible )
10 years agoHelpfull: Yes(4) No(2)
let speed be x
distance is 300km
given (300/x)-2=300/(x+5)
solving we get 25kmph
10 years agoHelpfull: Yes(4) No(1)
Normal speed is 25 kmph

300/25=12 hrs is time taken at normal speed
300/30=10 hrs is time taken at fast speed.
300/25 = 300/30 +2
11 years agoHelpfull: Yes(3) No(0)
let the speed of train be 5 km for 2 hr.
so the speed will be 2.5 km/hr.
for 300 km it will require 300/2.5=120 hrs.
so in 25 km it will require 12 km.
and therefore if speed is increased to 25+5=03 it requires 10 hrs.
HENCE THE SPEED IS 25.
7 years agoHelpfull: Yes(2) No(0)
let speed of normal train=(x)kmph
then speed of passanger train=(x+5)kmph
2=300/x-300/x+5
x^2+5x-750
by soving this eq'n
we get
x=24
ans:25kmph
9 years agoHelpfull: Yes(0) No(0)
by verifing with options:
let normal speed=25:
then time=300/25==12hrs
inc speed=30;
now time=300/30==10hrs:::
so satisfying the 2 hrs condition ans is 25kmph
9 years agoHelpfull: Yes(0) No(0)

A passenger train takes two hours less for a journey of 300km if its speed is increased by 5 kmph from its normal speed. The normal speed is

35 kmph
50 kmph
25 kmph
30 kmph