A solution needs to contain between 46% glucose and 50% glucose. - MATH INEQUALITIES

Question

A solution needs to contain between 46% glucose and 50% glucose. Find the least and greatest amount of a 60% glucose solution that should be mixed with a 40% glucose solution in order to end up with 30 kilograms of a solution containing the allowable percentage of glucose.

Answer 1

15 of each, 15 kg of 60% and 15 kg of 40% will yield 30 kg of 50%

You can see that immediately as 50 is the average of 40 and 60

so, 15 kg is the maximum amount of 60% you need

for the minimum allowed, to keep it at least 46%, it will be a little less than 15 kg

60 is within 14 of 46, 40 is within 6 of 46. 14/6 is the ratio of 40% needed with 60% 14/6 = 7/3 set that equal to x/(30-x) and solve for x

cross multiply: 210-7x =3x

10x = 210

x = 21 kg of 40%

and 9 kg of 60%

You need between 9 and 15 kg of 60% to keep the mixture between 46 and 50%

A more traditional method for solving this problem is set up two equations

30(50) = x(40) + (30-x)60

divide by 10

3(50) =4x + 180-6x

2x = 30

x = 15

30-x = 15 = max amount of 60%

plus another equation for the min amount needed

30(46)=x(40) + (30-x)60

3(46 = 4x +180 -6

2x = 180-138= 42

x = 21

30-21 = 9 kg = minimum amount of 60% needed