Two solutions of different concentrations of acid are mixed creating 40 mL of a solution that is 32% acid. One-quarter of the
solution is made up of a 20% acid solution. The remaining three-quarters is made up of a solution of unknown
concentration, c.
Which equation can be used to determine c, the unknown concentration?
6 30c + 10(0.2) = 40(0.32)
02()+ . (0.2) = 40(0.32)
O
(C)
(0.2)) = 40(0.32)
O 30(c)(10(0.2)) = 40(0.32)

Let

x = amount of 20% solution used (mL)

y = amount of solution used with unknown concentration (also mL)

c = the unknown concentration (%)

There are 40 mL of the new solution, so

x + y = 40

We use 10 mL of the 20% solution, so x = 10.

Each mL of either solution contributes c/100 mL of acid. So if 1/4 of the new solution (which is 10 mL) is made from the 20% solution, then this solution contributes

(20/100) * (10 mL) = 2 mL

of acid to the new one.

The new solution has a concentration of 32% acid, meaning it contains

(32/100) * (40 mL) = 12.8 mL

of acid. This means (12.8 - 2) mL = 10.8 mL of acid are provided by the second solution of unknown concentration.

We used 30 mL of the unknown solution, so the concentration is c such that

(c/100) * (30 mL) = 10.8 mL  ==>  c = 36

Putting everything together, we found c by setting up the equation

0.2 * 10 + c/100 * 30= 0.32 * 40

From your given options, pick whatever is equivalent. It's hard to tell what each option is, so I'll leave it to you.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3