Please help

Dave had some 50-cent coins. Wayne had some $1 coins. Dave gave 50% of his coins to Wayne and Wayne gave 50% of his coins to Dave. Wayne used some of the 50-cent coins to buy a book that cost $5. Dave used some of his $1 coins to buy a bag that cost $16. After that the ratio of the number of 50-cent coins to $1 coin Dave had was 1:2 and the number of 50-cent coins to the number of $1 coins Wayne had was 1:5. How much did Dave have at first?

 

slight simplification:   

Dave had some 50-cent coins.     I let that be  2x  (50c) coins   

Wayne had some $1 coins.          I let that be  2y  ($1) coins   

Dave gave 50% of his coins to Wayne and Wayne gave 50% of his coins to Dave.

Now they each have    x (50c)  +  y ($1)

 

Wayne used some of the 50-cent coins to buy a book that cost $5.  So wayne spends  10 of his 50c peices

Dave used some of his $1 coins to buy a bag that cost $16.             So Dave spends  16 of his $1 coins

 

So now 

Wayne has   (x - 10)  (50c coins)   +  y  ($1coins)                 which means that Wayne has   $  \((\frac{x-10}{2}+y)\)

 

Dave has          x       (50c coins)   +  (y-16)    ($1coins)        which means that Dave has      \($(\frac{x}{2}\;+\;\;y-16)\)

 

 

After that the ratio of the number of 50-cent coins to $1 coin Dave had was 1:2

 

\(\frac{x}{y-16}=\frac{1}{2}\\ 2x=y-16\\ y=2x+16\\\)

 

and the number of 50-cent coins to the number of $1 coins Wayne had was 1:5. 

 

\(\frac{x-10}{y}=\frac{1}{5}\\ y=5x-50\)

 

Solve simultaneously

\(2x+16=5x-50\\ 66=3x\\ x=22\\ y=44+16=60\)

 

 

How much did Dave have at first?  

He had 2x   50c coins so initially he had    44  fifty cent peices    =    $22

 

 

Check:

Started with Dave having 44  fifties and Wayne having 120 dollars

 

Dave

44 fifties

22 fifties and 60 ones

22 fifties and 44 ones         Ratio 1 to 2 good

 

Wayne

120 dollars

22 fifties and 60 ones

12 fifties and 60 ones        Ratio  1 to 5  good