At first,A had 2/3 p then B.Then A buy another 8 p,B lost 5 p.A has 4/5 p then B.How many p A had at first?

Answer 1

Let's say A had x amount of p at first
and B had y amount of p at first

Then x = 2/3 y
and x+8 = 4/5 (y-5)

x+8 = 4/5 y - 4
x = 4/5 y - 4 - 8
= 4/5 y - 12

x = 2/3 y = 4/5 y - 12

Therefore 2/3 y = 4/5 y - 12
Make them the same denominators:
10/15 y = 12/15 y - 12
2y = 12
y = 6
x = 2/3 (6)
=4

Thus, A had 4p at first

Answer 2

Let at first A have x p and B y p

So, x-y = 2/3

x+8 -(y-5) = 4/5

or x-y = 4/5-13

Answer 3

2/3?