Hi there! On wednesday I'm having a really important test at University and in the practice-book I found a puzzle I just can't solve, it goes like this:
In a coffee storea new coffee is being prepared. It contains: 10kg of a $20/kg coffee and some of $12/kg. How many Kilos of $12/kg coffeemust the mix contain so that charging it $18/kg they obtain 10% profit over the cost?
Sorry if it's grammatically wrong.
Thanx !
Quite unclear; "charging it" means charging the mixture or only the $12 coffee?Originally Posted by nicoT30
Is the $20:$12 including the 10%, or will it be sold at $22:$13.20 ?
Anyhow, leaving the "profit out":
x = kg of $20 coffee
[20x + 12(10 - x)] / 10 = 18
Solve to get x = 7.5 kg
Do you know the answer? We can probably figure out the meaning from that.
Hello, nicoT30!
The information is incomplete . . .
In a coffee store a new coffee is being prepared.
It contains: 10kg of a $20/kg coffee and some of $12/kg.
How many kilos of $12/kg coffee must the mix contain
so that charging $18/kg they obtain 10% profit over the cost?
Ignoring profit, we can reason as follow . . .
We have 10 kg of Brand A at $20/kg.
. . Its value is:.dollars.
We addkg of Brand B at $12/kg.
. . Its value is:.dollars.
The total value of the mixture is:.dollars. .[1]
The mixture will bekg at $18/kg.
. . Its value is:.dollars. .[2]
We just described the value of the mixture in two ways.
There is our equation! ,
Solve for
. . Hence: .
To determine 10% profit, we must know
. . the unit cost of Brand A and Brand B.
And this was not provided.
Really interestingI've burnt my mind and couldn't do it. About the 10% I got to this..
Real price of Mix = X
So.. $18= X + 10/100 X
$18= 110/100 X
$18 / (110/100) = X
$16,36 = X
Sorry about the way I wrote the equation, I did not learn the other way yet, I have to read the tutorial. I hope you have at least enjoyed a bit this problem
Thanx for helping!
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