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frosja888 [35]
4 years ago
10

A grocer sells milk chocolate at $2.90 per pound, dark chocolate at $4.90 per pound, and dark chocolate with almonds at $5.50 pe

r pound. He wants to make a mixture of 50 pounds of mixed chocolates to sell at $4.80 per pound. The mixture is to contain as many pounds of dark chocolate with almonds as milk chocolate and dark chocolate combined. How many pounds of each type must he use in this mixture?
Mathematics
1 answer:
daser333 [38]4 years ago
6 0

Answer: milk chocolate = 10 pounds

Dark chocolate =15 pounds

Dark chocolate with almonds = 25 pounds.

Step-by-step explanation:

What we have here is a proportion problem.

Milk chocolate = $2.90

Dark chocolate = $4.90

Dark chocolate with almonds = $5.50

We are to have 50 pounds of mixed chocolate

Amount the grocer makes from this

= $4.80 × 50 = $240

However, we must consider how much the grocer would have made if he had sold them separately

Weight of milk chocolate = x

Weight of dark chocolate = y

Weight of dark chocolate with almond = z

x+y = z (according to the question)

x+y+z= 50

Equating the collective amount of each type of chocolate with the amount they all cost together

2.90x + 4.90y + 5.50z = 50(4.80)

2.90x + 4.90y + 5.50z = $240

But z = x + y

2.90x + 4.90y + 5.50 (x+y) = $240

2.90x + 4.90y + 5.50x + 5.50y = 240

8.40x + 10.40y = 240 .....eqn 1

Since x + y + z = 50

x + y = 50 - z

But x+ y = z

x + y = 50 - (x+y)

x + y = 50 - x - y

2x + 2y = 50

(Divide through by 2)

x+y = 25 ...... Eqn 2

y = 25 - x

Substitute for y in eqn2

8.4x + 10.4 (25-x) = 240

8.4x + 260 - 10.4x = 240

8.4x - 10.4x = 240 - 260

-2x = -20

x = 10 pounds

y = 25-x

y = 25-10

y=15pounds

x+y+z= 50

10+15+z= 50

25+z=50

z= 50-25

z = 25

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I'll do problem 1 to get you started.

The answer to problem 1 is <u>3400 miles</u>

============================================================

Explanation for problem 1:

We're asked to find the perimeter, which is the total distance around the exterior or outside. In other words, we need to add up the four side lengths of this quadrilateral.

We'll need the distance formula to find the lengths of the slanted sides AB and BC

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Let's first find the length of segment AB

A = (x1,y1) = (0,0)

B = (x2,y2) = (400,300)

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d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(0-400)^2 + (0-300)^2}\\\\d = \sqrt{(-400)^2 + (-300)^2}\\\\d = \sqrt{160000 + 90000}\\\\d = \sqrt{250000}\\\\d = 500\\\\

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Now find the length of segment BC

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d = length of BC

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Segment DA is a similar story. But we subtract the x coordinates. This only works for horizontal lines. DA is 800 miles long because |-800-0| = 800. Absolute value is used to make sure the distance is never negative.

----------------------------------------

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