SOLUTION: The ratio of Jack to Chris's money was 12 : 5 at first. After Jack saved another $340 and Chris saved another $174, Jack had twice of Chris. How much did Jack have in the end ?

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The ratio of Jack to Chris's money was 12 : 5 at first. After Jack saved another $340 and Chris saved another $174, Jack had twice of Chris. How much did Jack have in the end ?      Log On


   

Question 983040: The ratio of Jack to Chris's money was 12 : 5 at first. After Jack saved another $340 and Chris saved another $174, Jack had twice of Chris. How much did Jack have in the end ?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The ratio of Jack to Chris's money was 12 : 5 at first.
After Jack saved another $340 and Chris saved another $174, Jack had twice of Chris.
How much did Jack have in the end ?
:
Let x = the multiplier
then
12x = Jacks original amt
5x = Chris's amt
:
%28%2812x%2B340%29%29%2F%28%285x%2B174%29%29 = 2 (Jack has twice as much as Chris)
12x + 340 = 2(5x + 174)
12x + 340 = 10x + 348
12x - 10x = 348 - 340
2x = 8
x = 4 is the multiplier
:
then
12*4 = $48 Jack's amt originally
48 + 340 = $388 Jack amt in the end