Question

A man covers a distance on scooter .Had he moved 3kmph faster he would have taken 40 minutes less. If he had moved 2kmph slower he would have taken 40minutes more. The distance is?

Answer 1

Hey,

Let distance = x km and usual rate = y kmph.

X/Y-(X/Y+3)=40----I

X/(Y-2)-X/Y=40-----II

Then,

On dividing (i) by (ii), we get: x = 40.

HOPE IT HELPS YOU:-))

Answer 2

Explanation:

$\huge\bold\purple{Answer:-}$

Let distance covered by man is D , original speed is v and original time taken by him is t.

so, D = vt .......(1)

case 1 : new speed of train = (v + 4) km/h , new time taken = (t - 1/2) hr

so, D = (v + 4)(t - 1/2) ........(2)

case 2 : new speed of train = (v - 2) km/h , new time taken = (t + 1/3) hr [as 20min = 1/3 hr ]

so, D = (v - 2)(t + 1/3) ........(3)

from equations (1) and (2),

vt = vt - v/2 + 4t - 2

⇒4t - v/2 = 2 .........(4)

from equations (1) and (3),

vt = vt + v/3 - 2t - 2/3

⇒v/3 - 2t = 2/3 ........(5)

from equations (4) and (5),

2v/3 - v/2 = 2 + 4/3 = 10/3

⇒v/6 = 10/3

⇒v = 20 km/hr and t = 3 hrs

so, D = vt = 20 × 3 = 60 km