Bill wants to make 10Lcof a 30% saline solution by mixing together a 60% saline solution and a 10% saline solution. How much of each solution must he use?

10 L of 30 % saline solution can be formed by mixing 4 L of 60 % saline solution and 6 L of  10 % saline solution.

Step-by-step explanation:

Let x be the number of liters of 60% saline solution

Now we require 10 L of 30% saline solution.

Liter soln %  liters  saline %  

30 %                 10        0.3  

60 %                   x        0.6

10 %                  10-x      0.1

Now forming the algebraic equation,

0.6x + 0.1 (10-x) = 10 (0.3)

0.6x + 1 - 0.1 x  = 3

0.5 x = 2

x = 4 ( 4 l of 60 % solution is required. So 10 % saline solution required is 10 - 4 = 6 L).

Hence, 10 L of 30 % saline solution can be formed by mixing 4 L of 60 % saline solution and 6 L of  10 % saline solution.

Each solution required for the Bill is: 5L and 70L of saline solution.

Step-by-step explanation:

30% of 10L = × 10 = 3 L

Now,

60% of saline solution = 3L

So, saline solution = 3 × = 5L

Similarly,

10% of other saline solution = 7L

so, other saline solution =  7 × = 70L

Hence required solution is:

5 L of 60% saline solution

70 L of 10% other saline solution

i not know

step-by-step explanation:

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]