Question

if raj walks at 5/4th of his usual speed, he reaches 10 min early. how long does he usually take​

Answer 1

Answer:

Step-by-step explanation:

Let RaJ's initial speed = x

then his final speed = 5x/4 ( because he is running at 5/4 of his usual speed )

initial speed : final speed = x : 5x/4 = 4 : 5 and then his time ratio = 5x/4 : x

He reached 5 min earlier so his new time minus initial time = 5x/4 - x = x/4

and x/4 = 5

so his new time will be 20 minutes and then his usual time will be 25 minutes ( because he reached 5 min earlier as given in the question so ans is 20 min + 5 min = 25 min )

Answer 2

Answer: 17 MINUTES

Step-by-step explanation:

Let the usual speed of Raj be x

Let the usual time taken by him be y

The new speed of Raj is 5x/4

and the time taken now is y-(1/6) (since 10 mins = 1/6 hour)

Now, the distance here is constant. We know distance=speed*time

:.(5x/4)(6y-1/6)=xy

Multiplying the equation by 12(since 12 is the L.C.M. of 6 and 4)

15x(12y-2)=12xy

180xy-30x=12xy

168xy=30x--eq 1

168y=30

y=30/168

y=0.17hrs

Thus , the usual time he takes is 17 mins.

Hey i hope this helps you! Feel free to correct me if I'm wrong:)

Have a great day! Please mark this as the brainliest answer:D