Question

Ngoc needs to mix a 10% fungicide solution with a 50% fungicide solution to create 200 millileters of a 26% solution. How many millileters of each solution must Ngoc use?

Answer 1

Answer:

• 80milliliters of the 10% fungicide solution

• 170milliliters of the 50% fungicide solution

Step-by-step explanation:

let A represent the amount of the 10% fungicide solution

let B represent the amount of the 50% fungicide solution

•  

B milliliters = 50% of B = (50/100)A = 0.5B

Total milliliters = 26% of 200 milliliters = 0.26 * 200 = 52

A milliliters + B milliliters = 200 milliliters

0.1A + 0.5B = 52

 

• A + B = 200
• 0.1A + 0.5B = 52

using substitution method to solve A and B

• from equation 1    A = 200-B

• insert  A = 200 - B in equation 2

0.1(200-B) + 0.5B = 52

20 - 0.1B + 0.5B = 52

20 + 0.4B = 52

0.4B = 52 -20 = 32

0.4B = 32

B = 32/0.4 = 80

since B = 80

A = 200 -B = 200 - 80 = 120

• 80milliliters of the 10% fungicide solution

• 170milliliters of the 50% fungicide solution

Answer 2

Answer: he must use 120 milliliters of the 10% solution and 80 milliliters of the 50% solution.

Step-by-step explanation:

Let x represent the amount of 10% fungicide solution that Ngoc must use.

Let y represent the amount of 50% fungicide solution that Ngoc must use.

The total volume of the fungicide solution that he wants to create is 200 milliliters. It means that

x + y = 200

Ngoc needs to mix a 10% fungicide solution with a 50% fungicide solution to create 200 millileters of a 26% solution. This means that

(10/100 × x) + (50/100 × y) = (26/100 × 200)

0.1x + 0.5y = 52 - - - - - - - - - -1

Substituting x = 200 - y into equation 1, it becomes

0.1(200 - y) + 0.5y = 52

20 - 0.1y + 0.5y = 52

- 0.1y + 0.5y = 52 - 20

0.4y = 32

y = 32/0.4 = 80 milliliters

x = 200 - y = 200 - 80

x = 120 milliliters