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## A fast food restaurant has 25% fat hamburger meat and 15% fat hamburger meat.

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A fast food restaurant has 25% fat hamburger meat and 15% fat hamburger meat. How many pounds of each type of meat should be mixed to make 90 lb of hamburger meat that is 18% fat?
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# 3 Answers

X - lb of 25%
90 - X = amount of 15%

X ( 0.25) + ( 90-x) ).15 = 90( 0.18

0.10 X = 90( 0.03)

X = 2.7 / 0.10 = 27 lb  25% Hamburger

90- 27 = 63 lb     15% Hamburger.
A fast food restaurant has 25% fat hamburger meat and 15% fat hamburger meat. How many pounds of each type of meat should be mixed to make 90 lb of hamburger meat that is 18% fat?

The mixture will be 90 lb, and contain 18% fat, or 16.2 lbs of fat, leaving 73.8 lbs meat.

So, we can use these two equations, where x represents the amount of 25% meat, and y represents the amount of 15% meat:

.25x + .15y = 16.2
x + y = 90

Taking the second equation, subtract y from both sides:

x = -y + 90

Substitute this value for x back into the first equation:

.25(-y + 90) + .15y = 16.2

Using distributive property to remove parentheses:
-.25y + 22.5 + .15y = 16.2

-.1y + 22.5 = 16.2

Subtract 22.5 from both sides:

-.1y = 16.2 - 22.5
-.1y = -6.3

Multiply both sides by -10:
y = 63

Plug this value back into the second equation:
x + y = 90
x + 63 = 90
x = 27

So, the mixture should contain 27 pounds of 25%, and 63 pounds of 15%.

Let's check our work:
(27 lbs * .25) + (63 lbs * .15) =  6.75 + 9.45 = 16.2 lbs of fat
The total weight is 90 lbs (27 + 63).
Set the problem up as a proportion where percentages as shown as decimals (18% = 0.18).

We consider two variable sets of unknowns

x = # of LBS
90- x = # of LBS from 90

0.18 * 90 lbs = 0.25* (90-X) + 0.15 * X

Solve for X

0.18 * 90 = 0.25*90 - 0.25 * X + 0.15 * X
16.2 = 22.5 - 0.1*X
16.2 - 22.5 = -0.1*X
-6.3 = -0.1 *X
X = 63 lbs of meat

Since the X value corresponded to the 0.15 (15% meat) then there should be 63 lbs of 15% meat.

90 lbs - 63 lbs (of 15% meat) = 27 lbs remaining

I.E the answer is 27 LBS of 25% Fat Hamburger and 63 LBS of 15% Fat Hamburger.