A takes 3 hour more than B to walk 30 km. But if A doubles his pace, he is ahead of B by 3 /2 hours. Find their speed of walking

Given:
Distance = 30 km

Let speed of A be = x km/h & speed of B be = y km/h

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———–(i)

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

30/y = 30/2x + 3/2
30/y – 30/2x = 3/2
30/y – 15/x = 3/2————–(ii)

Let us assume that 1/x = p & 1/y = q

So equation (i) and equation (ii) becomes
30p – 30q = 3——–(iii)
30q – 15p = 3/2——–(iv)

On adding equation (ii) and (iv) we get,

30p – 30q = 3
+ (-15p) + 30q = 3/2
_________________
15p = 9/2
p = 9/(2×15)
p = 3/10

Substituting the value of p = 3/10 in equation(iii) we get
30p – 30q = 3
30(3/10) – 30q = 3
9 – 30q = 3
-30q = -6
q = 1/5

Now we know that , 1/y = q
1/y = 1/5
y = 5km/h

&, 1/x = p
1/x = 3/10
x = 10/3 km/h

Hence
A = 10/3 km/h
B = 5 km/h
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}