Question 1189297
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Stephen and Siew Mei had $540 altogether. 
Stephen gave 1/7 of his money Siew Mei. 
In return Siew Mei gave 1/4 of the total amount she had to Stephen . 
Then they both had the same amount of money. How much did Stephen had at first ?
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Let x be the amount which Stephan had at first.

Then Siew Mei had (540-x) dollars at first.


When Stephen gave 1/7 of his money to Siew Mei, Stephen left with {{{(6/7)x}}} dollars;

Siew Mei had {{{(540-x) + (1/7)x}}}.


When Siew Mei gave 1/4 of the total amount she had to Stephen, she left with {{{(3/4)*((540-x) + (1/7)x)}}}.

Stephen had finally {{{(6/7)x}}} + {{{(1/4)*((540-x)+(1/7)x)}}} dollars.


The final amounts are the same, so we write this equation

    {{{(3/4)*((540-x)+(1/7)x)}}} = {{{(6/7)x}}} + {{{(1/4)*((540-x)+(1/7)x)}}}.


Simplify this equation and find x. First multiply both sides by 4*7 = 28
    
    3*7*((540-x) + (1/7)*x) = 6*4x + 7*((540-x)+(1/7)x)

    21*(540-x) + 3x = 24x + 7*(540-x) + x

    14*(540-x) = 22x

    14*540 = 22x + 14x

    14*540 = 36x

    x = {{{(14*540)/36}}} = 210.


<U>ANSWER</U>.  Stephen had initially 210 dollars;  Siew Mei had initially 540-210 = 330 dollars.


<U>CHECK</U>.  After first giving, Stephen had 210-30 = 180 dollars; Siew Mei had 330+30 = 360 dollars.

        After second giving, Stephen had 180 + 90 = 270 dollars; Siew Mei had 360-90 = 270 dollars.   ! Equal amounts, correct !
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Solved.