A science teacher has supply of 50% methane solution and a supply of 80% methane solution.How much of each solution should the teacher mix together to get 105 mL of 60% methane solution for an experiment.

a. 70 mL of the 50% solution and 35 mL of the 80% solution

b. 35 mL of the 50% solution and 70 mL of the 80% solution

c. 62 mL of the 50% solution and 43 mL of the 80% solution

d. 43 mL of the 50% solution and 62 mL of the 80% solution

IS IT C????

What is it

huh?? is it a??

It's A

To find the answer, we need to determine the amount of methane in each solution and then use the following formula to calculate the amount of each solution needed:

(amount of solution A)(percentage of methane in solution A) + (amount of solution B)(percentage of methane in solution B) = (total amount of mixture)(desired percentage of methane)

Let's break down the information given.

The teacher wants to create a 105 mL mixture of 60% methane solution.

Let's assume the amount of the 50% methane solution is X mL and the amount of the 80% methane solution is Y mL.

Now we can set up the equation to find the solution:

(X)(50%) + (Y)(80%) = (105 mL)(60%)

0.5X + 0.8Y = 0.6(105)

0.5X + 0.8Y = 63

To solve this equation, we can use the answer choices provided and substitute each option's values for X and Y:

a. X = 70 mL, Y = 35 mL
0.5(70) + 0.8(35) = 35 + 28 = 63

b. X = 35 mL, Y = 70 mL
0.5(35) + 0.8(70) = 17.5 + 56 = 73.5 ≠ 63

c. X = 62 mL, Y = 43 mL
0.5(62) + 0.8(43) = 31 + 34.4 = 65.4 ≠ 63

d. X = 43 mL, Y = 62 mL
0.5(43) + 0.8(62) = 21.5 + 49.6 = 71.1 ≠ 63

From this calculation, we see that only option a. satisfies the equation. Therefore, the correct answer is:

a. 70 mL of the 50% solution and 35 mL of the 80% solution