How many ml of a 35% acid mixture and a 95% acid mixture should be mixed to get 120 ml of a 40% acid mixture?

ANSWER:

110ML of 35% acid solution must be mixed with 10ML of 95% solution to obtain 120ML of 40% solution.

SOLUTION:  

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

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

Sum of values of two acids = Value of mixture  

0.35x + 0.95y = 48  

For convenience, we will multiply the entire equation by 100,

35 x + 95y = 4800     ------ (1)  

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

x + y = 120    --------- (2)

Multiply eqn 2  with 35 for easy calculation and derive the equation into one variable.

35 = 35x + 35y = 4200  

Subtracting equation (2) from (1),  we get

0 + 60y = 600  

Thus, 60y = 600  

Substituting y = 10 in (2),

35x + 35(10) = 4200  

35x + 350 = 4200  

35x = 4200 - 350  

35x = 3850  

So, we have x = 110 and y = 10

We can conclude that 110ML of 35% acid solution must be mixed with 10ML of 95% solution to obtain 120 ML of 40% solution.

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.