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.

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

4

step-by-step explanation: