Jerry and Tom had $720 altogether. Jerry gave 20% of his money to Tom. Tom then gave 25% of his money back to Jerry. In the end both boys had an equal amount of money each. How much money did each boy had at first?
Let's say Jerry had x dollars at first.
After giving 20% of his money to Tom, Jerry remained with 0.8*x dollars.
Tom received 0.2*x dollars.
Together, Jerry and Tom had (0.8*x + 0.2*x) = 0.8*x + 0.2*x = 1*x = x + 0.2*x + 0.8*x = 1*x = 720 dollars
Tom therefore had 0.2*x = 0.2*720 = 144 dollars
Since both boys had the same amount of money in the end, Jerry had 0.8*x = 0.8*720 = 576 dollars at first. Answer: \boxed{576}.
Let's solve this step by step.
Let's assume that Jerry had x amount of money initially.
According to the given information, Jerry gave 20% of his money to Tom. This means Jerry had 80% of his original money (0.8x) left.
Tom received the money from Jerry and had (720 - x) + 0.2x = (720 + 0.2x - x) = (720 - 0.8x) at this point.
Next, Tom gave 25% of his money back to Jerry. This means Tom had 75% of his remaining money (0.75 * (720 - 0.8x)) left.
According to the problem, both boys had an equal amount of money each at this point.
So, we can set up the equation as follows:
0.8x = 0.75 * (720 - 0.8x)
Let's solve this equation to find the value of x.
0.8x = 0.75 * 720 - 0.6x
0.8x + 0.6x = 0.75 * 720
1.4x = 540
x = 540 / 1.4
x ≈ 385.71
So, Jerry initially had about $385.71.
To find Tom's initial amount, we can subtract Jerry's initial amount from the total initial amount:
Tom's initial amount = 720 - 385.71 = $334.29
Therefore, Jerry initially had about $385.71, and Tom initially had about $334.29.