To solve this problem, we can set up a system of equations. Let x represent the amount (in mL) of the 50% sugar solution and y represent the amount (in mL) of the 80% sugar solution.
We know that the total volume of the solution is 105 mL, so we have the equation:
x + y = 105
We also know that the resulting solution should be 60% sugar, so we have the equation:
(0.50 * x + 0.80 * y) / 105 = 0.60
To simplify the second equation, we can multiply both sides by 105 to get:
0.50 * x + 0.80 * y = 0.60 * 105
Simplifying further, we get:
0.50 x + 0.80 y = 63
Now we can solve the system of equations using substitution or elimination. Let's use the elimination method. Multiply the first equation by -0.50:
-0.50 * (x + y) = -0.50 * 105
-0.50 x - 0.50 y = -52.5
Now we can add this equation to the second equation:
-0.50 x - 0.50 y + 0.50 x + 0.80 y = -52.5 + 63
0.30 y = 10.5
y = 10.5 / 0.30
y = 35
Substituting this value of y back into the first equation:
x + 35 = 105
x = 105 - 35
x = 70
Therefore, the teacher should mix 70 mL of the 50% sugar solution and 35 mL of the 80% sugar solution to get 105 mL of 60% sugar solution. Answer: 70 mL of the 50% solution and 35 mL of the 80% solution.
A science teacher has a supply of 50% sugar solution and a supply of 80% sugar solution. How much of each solution should the teacher mix together to get 105 mL of 60% sugar solution for an experiment ?
(1 point)
Responses
70 mL of the 50% solution and 35 mL of the 80% solution
70 mL of the 50% solution and 35 mL of the 80% solution
35 mL of the 50% solution and 70 mL of the 80% solution
35 mL of the 50% solution and 70 mL of the 80% solution
70 mL of the 50% solution and 70 mL of the 80% solution
70 mL of the 50% solution and 70 mL of the 80% solution
35 mL of the 50% solution and 35 mL of the 80% solution
2 answers
a,b,c, or d