Math problem in description

Question

How much 36%

 acid solution should be added to 8%

 acid solution to make 56mL

 of a 10%

 acid solution?

Answer 1

36(X) + 8(56-X) = 10(56

36X-8X = 560-448 = 112

X = 112/28 = 28/7 = 4 mL of 36% solution

mixed with 56-X = 56-4 = 52 mL of 8% solution to get 56 mL of 10% solution

8 is much closer to 10 than 36 is, so you know you need much more 8% than 36%

8 is within 2 of 10, while 36 is within 26 of 10. 26 is 13 times as much as 2

52 is 13 times 4