Answer:
John should use:
4 grams of the 30% solution and 16 grams of the 60% solution
Step-by-step explanation:
Let the number of grams of the 30% solution = x
Let the number of grams of the 60% solution = y
John needs 20 grams of 54% acid solution for his science project.
Hence,
x + y = 20 grams..... Equation 1
x = 20 - y
His school's science lab has bottles of 30% solution and bottles of 60% solution.
30% × x + 60% × y = 54% × 20
0.3x + 0.6y = 10.8......Equation 2
We substitute 20 - y for x in Equation 2
0.3(20 - y) + 0.6y = 10.8
6 - 0.3y + 0.6y = 10.8
- 0.3y + 0.6y = 10.8 - 6
0.3y = 4.8
y = 4.8/3
y = 16 grams
x = 20 - y
x = 20 - 16
x = 4 grams
Therefore, John should use:
4 grams of the 30% solution and 16 grams of the 60% solution
