At the start of the game, Peter and Jack had 48 marbles altogether. In the first round, Peter lost 2/5 of his marbles to Jack. In the second round, Jack lost 1/9 of his marbles to Peter. At the end of the second round, they had the same number of marbles. How many marbles did Jack have at first?
Let no. of marbles Jack had = x
no. of marbles Peter had = y
x + y = 48. — (1) y = 48 - x or x = 48 - y
After In 1 round, Peter had y - 2/5y = y1
Jack had, x1 = x + 2/5y
After 2nd round, Jack had x2 = x, - 1/9x,
x2 = x1 - 1/9x,
x2 = x + 2/5y - 1/9(x + 2/5)y
= x + 2/5y - 1/9x - 2/45y
x2 = 8x/9 + 16/45y
After second round, Peter had y2 = y1 + 1/9x,
y2 = y1 + 1/9x,
y2 = y - 2/5y + 1/9 (x + 2/5y)
= y - 2y/5 + 1/9x + 2y/45
= x/9 + (45 - 18 + 2)/45 y
y2 = x/9 + 29/45y
Given x2 = y2
=> 8x/9 + 16/45y = x/9 + 29/45y
=> 8x/9 - x/9 = 29/45y - 16/45y
=> 7x/9 = 13/45y
Sub y * 48 - x in above equation
=> 7x/9 = 13/45 (48 - x) =>
=> 7x/9 + 13/45x = (13 * 48)/45 => 48/45x = (13 * 48)/45
x = 13
Jack had 13 marbles at first