Question 143324: Lucinda bought two record albums and a compat disc for a total of $28.00. The price of each record album was 66 2/3% (sixty-six and two-thirds) of the price od the compact disc. How much did Lucinda pay for a record album? (Note: Both record albums cost the same)
Answer by jojo14344(1513)   (Show Source):
You can put this solution on YOUR website! First let us assign variables for the following items:
"x" = price of 1st record album
"y" = price of 2nd record album
"z" = price of compact disc
So, "x+y+z=$28" right? ------------------ eqn 1
But there are conditions, price of "x" & "y" is 66-2/3% of "z". To show,
"x" = 66-2/3%(z) ------------ condition 1
"y" = 66-2/3%(z) ------------ condition 2
Substitute these conditions in eqn 1,
66-2/3%(z) + 66-2/3%(z) + z =$28
2/3(z) + 2/3(z) + z =$28
Note: how we get 2/3? divide 66-2/3% by 100 to remove the %
continuing,
2-1/3(z) = $28
z= $12, price for the compact disc
For the recorded albums, $28-$12 = $16, price for the 2 recorded albums.
Since they have same price, $16/2 = $8, price for each album
In doubt? Go back condition 1 or 2,
x= 66-2/3%(z) = 2/3($12)
x= $8, also price of "y"
Also thru eqn 1,
$8+$8=$12=$28
$28=$28, cool!
Thank you,
Jojo
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