Respuesta :
Let x be the ounces of 30 cents worth iodine.
We know that we have 50 ounces of 18 cents worth leonine, so we have (50)(18) = 900 cents of iodine.
Now, we can set up our equation and solve for x to get the ounces that should be mixed:
[tex]30x+900=20(x+50)[/tex]
[tex]30x+900=20x+1000[/tex]
[tex]10x=100[/tex]
[tex]x= \frac{100}{10} [/tex]
[tex]x=10[/tex]
We can conclude that we should mix 10 ounces of 30 cents worth iodine, so we can sell the mixture for 20 cents an ounce.
We know that we have 50 ounces of 18 cents worth leonine, so we have (50)(18) = 900 cents of iodine.
Now, we can set up our equation and solve for x to get the ounces that should be mixed:
[tex]30x+900=20(x+50)[/tex]
[tex]30x+900=20x+1000[/tex]
[tex]10x=100[/tex]
[tex]x= \frac{100}{10} [/tex]
[tex]x=10[/tex]
We can conclude that we should mix 10 ounces of 30 cents worth iodine, so we can sell the mixture for 20 cents an ounce.
Let x denote number of ounces of 30-cent iodine.
Mixing of 50 ounces of 18-cent iodine and x ounces of 30-cent iodine will result in (50+x) mixture of iodine. We want the mixture to be worth 20 cent per ounce. The value of 50 ounces of 18-cent iodine is of course
50â‹…18=900.
Similarly, x ounces of 30-cent iodine is worth 30â‹…x.
Finally, 50+x ounces of 20-cent iodine will cost (50+x)â‹…20.
30x+50â‹…18=(50+x)â‹…20
30x+900=1000+20x
10x=100
x=10
Mixing of 50 ounces of 18-cent iodine and x ounces of 30-cent iodine will result in (50+x) mixture of iodine. We want the mixture to be worth 20 cent per ounce. The value of 50 ounces of 18-cent iodine is of course
50â‹…18=900.
Similarly, x ounces of 30-cent iodine is worth 30â‹…x.
Finally, 50+x ounces of 20-cent iodine will cost (50+x)â‹…20.
30x+50â‹…18=(50+x)â‹…20
30x+900=1000+20x
10x=100
x=10