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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 dollars;
Siew Mei had .
When Siew Mei gave 1/4 of the total amount she had to Stephen, she left with .
Stephen had finally + dollars.
The final amounts are the same, so we write this equation
= + .
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 = = 210.
ANSWER. Stephen had initially 210 dollars; Siew Mei had initially 540-210 = 330 dollars.
CHECK. 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 !
Solved.