Question 122250: The average of Alex's, Bonnie's, and Carl's ages is 9. Four years ago Carl was the same age as Alex is now. In three years, Bonnie's age will be two-thirds Alex's age. How old is each now?
Answer by MaxLiebling(13)   (Show Source):
You can put this solution on YOUR website! The average of Alex's (A), Bonnie's (B), and Carl's (C) ages is 9:
(A+B+C)/3 = 9
-> A+B+C =27; (1)
Four years ago Carl was (C-4) the same age as Alex is now:
C-4 = A
-> C = A+4; (2)
In three years, Bonnie's age will be two-thirds Alex's age:
In three years B+3, A+3
B+3 = 2/3 * (A+3)
-> B = -3 + (2/3)*(A+3)
-> B = -3 +(2/3)*A + (2/3)*3
-> B = -3 +(2/3)*A + 2
-> B = -1 +(2/3)*A; (3)
Put results (2) and (3) to (1):
27 = A+B+C = A + [-1+(2/3)*A] + A+4 = A - 1 +(2/3)*A + A + 4 = 2A+(2/3)*A - 1 + 4 = (8/3)*A + 3 = 27
-> (8/3)*A = 24
-> 8*A = 72
-> A = 9
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Use (2):
C = A+4 = 9+4 = 13
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Use (3):
B = -1 +(2/3)*A = -1 +(2/3)*9 = -1 + 6 = 5
... or ... use (1)
A+B+C =27 ->
B = 27 - A - C = 27 - 9 - 13 = 18-13 = 5
B = 5
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Proof:
A+B+C = 9+5+13 = 14+13 = 27 ... okey.
C-4 = 13-4 = 9 = A ... okey.
B+3 = 5+3 = 8
2/3 * (A+3) = (2/3) * (9+3) = (2/3) * 12 = 24/3 = 8 = B+3 ... okey.
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