Brandon and Albert had $240 altogether. Albert gave 2/5 of his money money to Brandon. After that, Brandon gave 3/5 of his money to Albert. In the end, Brandon had 1/4 of the total sum of money. How much did Brandon have at first?
Let B be the amount Brandon had at first.
Then, Albert had 240 - B at first and 2/5 * (240 - B) after giving some money to Brandon.
After Albert gave money to Brandon, he had (240 - B) - 2/5 * (240 - B) = 3/5 * (240 - B) left.
After Brandon gave some money to Albert, he had (1 - 3/5) * (240 - B) = 1/4 * (240 - B).
Thus, (1/4) * (240 - B) = B.
240 - B = 4B.
5B = 240.
B = <<48=48>>48. Answer: \boxed{48}.
Let's break down the information given:
Let's assume that Brandon had x dollars at first.
According to the given information, Albert gave 2/5 of his money to Brandon. This means that Albert gave (2/5) * x = 2x/5 dollars to Brandon.
After Albert gave money to Brandon, Brandon had x + (2x/5) = (5x + 2x) / 5 = 7x/5 dollars.
Next, we know that Brandon gave 3/5 of his money to Albert. This means that Brandon gave (3/5) * (7x/5) = 21x/25 dollars to Albert.
After Brandon gave money to Albert, Albert had (2x/5) + (21x/25) = (10x + 21x) / 25 = 31x/25 dollars.
Finally, we know that Brandon had 1/4 of the total sum of money. This means that (7x/5) = (1/4) * (240) = 60.
To find the value of x, we can solve the equation (7x/5) = 60 for x:
7x/5 = 60
7x = 60 * 5
7x = 300
x = 300/7
So, Brandon had $300/7 at first.