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
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?
1 answer