SOLUTION: Hi bob 3/7 of his money.john spent 1/5 of his and alan spent 1/3 of his.all of them had the same amount of money left. If the 3 of them had spent $144 , how much did each have at

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Hi bob 3/7 of his money.john spent 1/5 of his and alan spent 1/3 of his.all of them had the same amount of money left. If the 3 of them had spent $144 , how much did each have at       Log On

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Question 1105659: Hi
bob 3/7 of his money.john spent 1/5 of his and alan spent 1/3 of his.all of them had the same amount of money left. If the 3 of them had spent $144 , how much did each have at first.
thanks

Answer by math_helper(2461) About Me  (Show Source):
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Bob starts with B dollars: (4/7)B = k (1) (Bob spent 3/7 so he has 4/7 left…)
John starts with J dollars: (4/5)J = k (2)
Alan starts with A dollars: (2/3)A = k (3)
(3/7)B + (1/5)J + (1/3)A = 144 (4)
From (1) and (2): (4/7)B = k = (4/5)J —> J = (5/7)B
From (1) and (3): (4/7)B = k = (2/3)A —> A = (6/7)B
Now substitute these last two equations for J and A in (4) to get one equation in one unknown:
(3/7)B + (1/5)(5/7)B + (1/3)(6/7)B = 144
(3/7)B + (1/7)B + (2/7)B = 144
(6/7)B = 144
B = 144*7 / 6 = 168 —> A = 144, J = 120
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Ans: Bob started with $168, Alan with $144, and John with $120
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Check:
(3/7)(168) + (1/5)(120) + (1/3)(144)
= 72 + 24 + 48
= 144 (ok)