Question

A takes 3 hrs more than B to walk a distance of 30 KM but A doubles its speed{pace].he is ahead by 1 and ahalf hrs.find the speed of walking?

Answer 1

Let's assume that A's pace be 'a' km/hr
and B's speed be 'b' km/hr

A/Q , 30/a = 30/b + 3      .⇒10/a = 10/b + 1 ......(1)

and when A double his pace a→2a ,

30/2a + 1 1/2 = 30b ⇒ 30/2a + 3/2 = 30/b
⇒  30/a + 3 ⇒ 60/b ⇒ 10/a + 1 = 20 ....(2)

Put  1/a = x and 1/b = y

we get,   10x = 10y +1         and        10x + 1 = 20y

Solving both these fetches x = 3/10         and y = 1/5

Therefore a = 10/3 km/hr & b = 1/5 km/hr

Hope this gonna assist you.

Answer 2

Answer:

Step-by-step explanation:

Time taken by A to cover 30  km = 30/x

[ Time = Distance /speed]

Time taken by B to cover 30 km = 30/y

30/x = 30/y + 3

30/x - 30/y = 3…………(1)

[Given : A takes 3 hours more than B to walk ]

When A doubles his speed then, time taken by A = 30/2x

[Given: time taken by B =  Time taken by A + 3/2 h]

30/y = 30/2x + 3/2

30/y - 30/2x  = 3/2

30/y - 15/x  = 3/2 ……..(2)

Let  1/x= p  and 1/y = q

so eqs 1 & 2 become ,then On adding eq 3 & 4

30p -  30q = 3……….(3)

-15p +  30q = 3/2………(4)

---------------------------

15 p = 3 + 3/2

15 p =( 6 + 3)/2 = 9/2

15p = 9/2

p = 9 /(2×15) = 3/10

p = 3/10

On Putting the value of p = 3/10 in eq 3

30p -  30q = 3

30 (3/10) - 30q = 3

9 - 30q = 3

9 - 3 = 30q

6 = 30q

q = 6/30= 1/5

1/y = q               [ Let 1/y = q]

1/y = ⅕

y = 5 km/h

1/x = p             [Let 1/x= p ]

1/x = 3/10

x = 10/3 km/h

so x = 10/3 km/h. &  y = 5 km/h

Hence, the speed of A is 10/3 km/h & B is 5 km/h.

HOPE THIS WILL HELP YOU….