You have one type of nut that sells for $2.50/lb and another type of nut sells for $9.70/ltain the desired mixture?

Question

You have one type of nut that sells for $2.50/lb and another type of nut that sells for $9.70/lb. You would like to have 14.4 lbs of a nut mixture that sells for $2.60/lb. How much of each nut will you need to obtain the desired mixture?

lbs of the cheaper nut =

 lbs of the expensive nut.

Dilbert needs to mix a 20% saline solution with a 60% saline solution to create 100 millileters of a 20% solution. How many millileters of each solution must Dilbert use?

Answer: Dilbert must mix 

 millileters of 20% solution =

 millileters of 60% solution=

A chemist has 20% and 40% solutions of acid available. How many liters of each solution should be mixed to obtain 62.5 liters of 28% acid solution?

 liters of 20% acid=

 liters of 40% acid=

Answer 1

2.5C + 9.7E = 14.4(2.6)

2.5(14.4-E) + 9.7E = 37.44

36- 2.5E + 9.7E = 37.44

7.2E = 37.44 -36 = 1.44

E = 1.44/7.2 = .2 lbs of the expensive $9.70 per pound nuts

C = 14.4 - 0.2 = 14.2 lbs of the cheaper $2.50 per pound nuts

to get 14.4 lbs of $2.60 per pound nuts

2.5 is within .1 of 2.6

9.7 is within 7.1 of 2.6

14.2 is 71 times .2

7.1 is 71 times .1

you need far more of the nuts that are closer to 2.6

71 times more of those nuts

20x + 60y = 28(62.5) = 1750

20(62.5-y) + 60y = 1750

1250 - 20y + 60y = 1750

40y = 500

y = 500/40 = 12.5 liters of 60%

x = 62.5-12.5 = 50 liters of 20%

20 is within 8 of 28

60 is within 32 of 28

60 is 4 times further away 32 = 8x4

you need 4 times more of 20 than 60

4 x 12.5 = 50