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

Question

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

Answer 1

To solve this problem, we can set up a system of equations.

Let x represent the amount of the 50% sugar solution and y represent the amount of the 80% sugar solution.

We are given two pieces of information:

1) The total volume of the mixture is 105 mL: x + y = 105.

2) The resulting mixture is 60% sugar: 0.5x + 0.8y = 0.6(105).

Now we can solve this system of equations.

From the first equation, we can solve for x: x = 105 - y.

Substituting this expression for x into the second equation, we get:

0.5(105 - y) + 0.8y = 63
52.5 - 0.5y + 0.8y = 63
0.3y = 10.5
y = 35

Substituting this value for y back into the first equation, we can solve for x:

x + 35 = 105
x = 70

So, the teacher should mix 70 mL of the 50% sugar solution and 35 mL of the 80% sugar solution.

Answer 2

To solve this problem, you can use the concept of weighted averages.

Let x represent the amount of the 50% sugar solution to be mixed, and y represent the amount of the 80% sugar solution to be mixed.

We can set up a system of equations based on the information given:

Equation 1: x + y = 105 (Total volume of the solution should be 105 mL)
Equation 2: 0.50x + 0.80y = 0.60(105) (The total sugar content of the mixture should be 60% of 105 mL)

Now we can solve this system of equations.

From Equation 1, we can express x in terms of y: x = 105 - y.

Substituting this into Equation 2, we get:

0.50(105 - y) + 0.80y = 0.60(105)
52.5 - 0.50y + 0.80y = 63
0.30y = 63 - 52.5
0.30y = 10.5
y = 10.5 / 0.30
y = 35

Now, substitute the value of y back into Equation 1 to find x:

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 obtain 105 mL of a 60% sugar solution.

Answer 3

To solve this problem, we can apply the principle of "amount of solute = concentration x volume."

Let's denote the volume of the 50% sugar solution as x mL and the volume of the 80% sugar solution as (105 - x) mL.

Since the goal is to obtain a 60% sugar solution, we can set up the equation:
0.5x + 0.8(105 - x) = 0.6(105)

Now, let's solve for x:

0.5x + 0.8(105 - x) = 0.6(105)
0.5x + 84 - 0.8x = 63
-0.3x = -21
x = -21 / -0.3
x = 70

Therefore, the teacher should mix 70 mL of the 50% sugar solution and (105 - 70) mL, which is 35 mL, of the 80% sugar solution to obtain a 105 mL 60% sugar solution.

So, the answer is: 70 mL of the 50% solution and 35 mL of the 80% solution.