Answer:
$48
Explanation:
The given relations can be used to write and solve a pair of equations for the amounts each has. Knowing those amounts, the total can be found.
Setup
Let b and s represent the amounts of money initially held by Ben and Shane, respectively. Then we have ...
b = (1/3)s . . . . . . . . . . Ben has 1/3 Shane's amount
(s -8) -(b +8) = 8 . . . . after transferring $8, Shane has $8 more than Ben
Solution
Substituting for b in the second equation gives ...
s -8 -(1/3)s -8 = 8
2/3s = 24 . . . . . . . . . collect terms, add 16
s = 36 . . . . . . . . . . . multiply by 3/2
b = (1/3)(36) = 12
b +s = 12 +36 = 48 . . . . . . the amount they have in all
Shane and Ben have $48 in all.
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Check
Ben starts with $12 and ends with $20. Shane starts with $36 and ends with $28, which is $8 more than Ben ends with—as required.
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Additional comment
Sometimes the problem can be solved mentally by considering "ratio units." The ratio of original amounts is b : s = 1 : 3. That is, Shane has 3 -1 = 2 more ratio units of money than Ben has.
If the difference in their initial amounts were $16, the $8 transfer from Shane to Ben would make the difference zero. Here, the remaining difference is $8, so Shane must start with $24 more than Ben.
We know that difference corresponds to 2 ratio units, and the total is 1+3 = 4 ratio units (twice the difference). So, the total of their money must be 2×24 = $48.