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Algebra word problem:How many pounds of hamburger that costs \$1.10 per pound must be mixed with 50 pounds of hamburger that costs \$1.80 per pound to make a mixture that costs \$1.30 per pound.

Relevance

Let h = amount of \$1.10 hamburger (cheap) to use

50 = amount of \$1.80 hamburger (expensive) to use

50+h = amount of \$1.30 hamburger (medium) to make

cost of cheap HB + cost of expensive HB = cost of medium HB

1.10h + 1.8*50 = 1.3(50+h)

11h + 18*50 = 13(50+h) <== multiply by 10

11h + 900 = 650 + 13h <== distribute

11h + 900 - 11h - 650 = 650 + 13h - 11h - 650 <== subtract 11h and 650

250 = 2h <== simplify

h=125 <== divide by 2

Therefore, add 125 lbs of cheap hamburger to 50 lbs of expensive to get 175 lbs of medium priced (\$1.30/lb) hamburger.

Source(s): Me, a math teacher.
• Let x pounds of hamburger @\$1.10 /pound be mixed with 50 pounds of hamburger @1.80/pound to make mixture of 50+x pounds of mixture that costs \$1.30/pound

According to the problem,

1.10x+50*1.80=1.30(50+x)

or,1.10x+90=65+1.30x

or, 1.10x-1.30x=65-90

or, -0.20x= -25

or x= -25/-0.20=125

Therefore 125 pounds of hamburger of \$1.10 is to be mixed.

• let x be pound of hambergur that cost 1.10

1.10x + 1.80(50) = 1.30 (x + 50)

1.10x + 90 = 1.30x + 65

-25 = -.2x

x = 125 pounds

• When in doubt, let X and Y be equal to something.

X = no of lbs of 1.10 meat

50 = no of lbs of 1.80 meat

Total lbs = X + 50

Total cost = 1.1*X + 1.8*50

Thus cost per lb = (1.1*X + 90) / (X+50) = 1.3

1.1*X + 90 = 1.3*X + 65

0.2*X = 25

X = 125 lbs