Question

A science teacher has a supply of 20% saline solution and a supply of 50% saline solution. How much of each solution should the teacher mix together to get 60 mL of 28% saline solution for an experiment? (1 point)
A) 16 mL of the 20% solution and 44 mL of the 50% solution

B) 44 mL of the 20% solution and 16 mL of the 50% solution

C) 16 mL of the 20% solution and 16 mL of the 50% solution

D) 44 mL of the 20% solution and 44 mL of the 50% solutionhow many solutions does the system of equations have y-6x=-3 and 4y-24x=-16

Answer 1

Answer:

B)

Explanation:

$\text{Let}\\\\x-\text{a volume of}\ 20\%\ \text{saline solution}\\\\y-\text{a volume of }\ 50\%\ \text{saline solution}\\\\p\%=\dfrac{p}{100}\\\\20\%=\dfrac{20}{100}=\dfrac{2}{10}=0.2\\\\50\%=\dfrac{50}{100}=\dfrac{5}{10}=0.5\\\\0.2x-\text{a volume of salt in }\ 20\%\ \text{saline solution}\\0.5y-\text{a volume of salt in }\ 50\%\ \text{saline solution}\\\\60mL-\text{a volume of}\ 28\%\ \text{saline solution}\\\\28\%=\dfrac{28}{100}=0.28$

$0.28\cdot60mL=16.8mL-\text{a volume of salt in }\ 28\%\ \text{saline solution}\\\\\text{Equations}\\\\(1)\qquad x+y=60\\(2)\qquad0.2x+0.5y=16.8$

$\left\{\begin{array}{ccc}x+y=60\\0.2x+0.5y=168&\text{multiply both sides by 10}\end{array}\right\\\left\{\begin{array}{ccc}x+y=60&\text{multiply both sides by (-2)}\\2x+5y=168}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-2x-2y=-120\\2x+5y=168}\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad3y=48\qquad\text{divide both sides by 3}\\.\qquad y=16$

$\text{Put the value of}\ y\ \text{to the first equation:}\\\\x+16=60\qquad\text{subtract 16 from both sides}\\x=44$