Ascientist has two solutions, which she has labeled solution a and solution b. each contains salt. she knows that solution a is  20% salt and solution b is  70% salt. she wants to obtain  110 ounces of a mixture that is  30% salt. how many ounces of each solution should she use?

ANSWER:  

88 ounces of 20% salt solution A must be mixed with 22 ounces of 70% salt solution B to obtain 110 ounces of 30% solution.

SOLUTION:  

First, set up a table.  fill in the unknowns with variables x and y.  The table is attached below.

From the table, we can easily set up the two equations.  

Sum of values of two salts = Value of mixture  

0.2x+0.7y=33  

For convenience, we'll multiply the entire equation by 10,

2 x + 7 y = 330       ------ eqn (1)  

Now, Sum of amounts of each salt = Amount of mixture  

x + y = 110         --------- eqn (2)

multiply eqn (2) with “2” for easy calculation and derive the equation into one variable.

2x + 2y = 220  --- eqn 3

Subtracting equation (3) from (1), we get  

0 + 5y = 110  

Thus, 5y = 110  

Substituting y = 22 in (2),

2x + 2(22) = 220  

2x + 44 = 220  

2 x = 220 - 44 = 176  

So, we have x = 88 and y = 22

We can conclude that 88 ounces of 20% salt solution A must be mixed with 22 ounces of 70% salt solution B to obtain 110 ounces of 30% solution.

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2