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If you purchase x grams of Food A, y grams of Food B, and z grams of F 29 Oct 2017, 01:25
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If you purchase x grams of Food A, y grams of Food B, and z grams of Food C, the cost will be

(A) (9x/5 + 3y + 11z/4)¢
(B) $(9x/5 + 3y + 11z/4)
(C) $(1.8x + 3z + 2.75y)
(D) (3x + 1.8y + 2.75z)¢
(E) $(x + y + z)


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A
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 29 Oct 2017, 15:35
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Bunuel

If you purchase x grams of Food A, y grams of Food B, and z grams of Food C, the cost will be

(A) (9x/5 + 3y + 11z/4)¢
(B) $
(C) $(1.8x + 3z + 2.75y)
(D) (3x + 1.8y + 2.75z)¢
(E) $(x + y + z)


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Bunuel, Answer B is missing text.
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 29 Oct 2017, 19:04
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The Cost will be .018x+.03y+.0275z
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 29 Oct 2017, 20:32
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Bunuel

If you purchase x grams of Food A, y grams of Food B, and z grams of Food C, the cost will be

(A) (9x/5 + 3y + 11z/4)¢
(B) $
(C) $(1.8x + 3z + 2.75y)
(D) (3x + 1.8y + 2.75z)¢
(E) $(x + y + z)


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I think Answer A.

Let \(x = 10, y = 50, and z = 100\) grams

\(\frac{$1.80}{100g}=\frac{$??}{10g}=\) $0.18 for 10g = 18 cents = cost of x

\(\frac{$3.00}{100g}=\frac{$??}{50g}=\) $1.50 for 50g = 150 cents = cost of y

\(\frac{$2.75}{100g}=\frac{$??}{100g}=\) $2.75 for 100g = 275 cents = cost of z

Total cost in cents: 443
Total cost in dollars: $4.43

Using Answer A, converting to cents/grams and multiplying by x, y, or z grams, I get the same result:

\(\frac{$1.80}{100g} = \frac{180 cents}{100g} = \frac{9 cents}{5 g} * 10 g(=x) =\) 18 cents (x cost)

\(\frac{$3.00}{100g}= \frac{300 cents}{100g} =\frac{3 cents}{1g} * 50 g(=y) =\) 150 cents (y cost)

\(\frac{$2.75}{100g} = \frac{275 cents}{100g} = \frac{11 cents}{4g} * 100g(=z) =\) 275 cents (z cost)

Unless I am missing something, latter parts of equations immediately above, summed, are equivalent to

\((\frac{9x}{5}\) cents + \(3y\) cents + \(\frac{11z}{4}\) cents) or

\((\frac{9x}{5} + 3y + \frac{11z}{4})\)¢

I think Answer A.
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 29 Oct 2017, 21:18
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Bunuel

If you purchase x grams of Food A, y grams of Food B, and z grams of Food C, the cost will be

(A) (9x/5 + 3y + 11z/4)¢
(B) $
(C) $(1.8x + 3z + 2.75y)
(D) (3x + 1.8y + 2.75z)¢
(E) $(x + y + z)


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Bunuel, Answer B is missing text.
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Edited. Thank you.
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 29 Oct 2017, 21:33
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Bunuel

If you purchase x grams of Food A, y grams of Food B, and z grams of Food C, the cost will be

(A) (9x/5 + 3y + 11z/4)¢
(B) $(9x/5 + 3y + 11z/4)
(C) $(1.8x + 3z + 2.75y)
(D) (3x + 1.8y + 2.75z)¢
(E) $(x + y + z)


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$(1.8x+3y+2.75z)/100
$(9x/500 + 3y/100 + 11z/400)
¢(9x/5 + 3y + 11z/4)
A
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 29 Oct 2017, 22:00
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Cost per 100g of food A is: $1.8 => Cost of x grams of Food A = $1.8 * x/100 = $(9/5 * x/100)
Cost of y grams of Food B = $3 * y/100 = $(3 * y/100)
Cost of z grams of Food C = $2.75 * z/100 = $(11/4 * z/100)
=> total cost = $(9x/5 + 3y + 11z/4)/100 = ¢(9x/5 + 3y + 11z/4)

Hence, the answer is A.
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 31 Oct 2017, 16:17
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Bunuel

If you purchase x grams of Food A, y grams of Food B, and z grams of Food C, the cost will be

(A) (9x/5 + 3y + 11z/4)¢
(B) $(9x/5 + 3y + 11z/4)
(C) $(1.8x + 3z + 2.75y)
(D) (3x + 1.8y + 2.75z)¢
(E) $(x + y + z)


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Note that since 100 grams of Food A cost $1.80, 1 gram of Food A will cost 1.8¢. Similarly, each gram of Food B and Food C will cost 3¢ and 2.75¢, respectively.

We can thus create an expression for the total price, in cents, as follows:

1.8x + 3y + 2.75z

This answer is not one of the choices, so we must re-express it to match one of them:

18x/10 + 3y + 275z/100

9x/5 + 3y + 11z/4

Answer: A
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 06 Dec 2023, 03:55
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I found that the easiest way to solve the problem was by creating a table as shown below:

Item.....$....¢....¢ per 100g....¢ per g....$ per 100g....$ per g

A..........1.8...180......180...........1.8...........1.8...............0.018
B...........3.....300......300...........3..............3.................0.03
C........2.75....275......275...........2.75.........2.75.............0.0275

From the table above, one can see that the only equation that makes sense is eqn. A.
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If you purchase x grams of Food A, y grams of Food B, and z grams of F 06 Dec 2023, 06:24
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It may be easier just substituting a simple number, such as 200 grams each, so that cost equals:

200/100(1.8+3+2.75) = $15.10=1510 cents

And then comparing to answer choices.

So, Answer A:

200(9/5 + 3 + 11/4)=

(200/20)(36+60+55)=10*151=

1510 cents

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If you purchase x grams of Food A, y grams of Food B, and z grams of F 25 Jan 2025, 07:54
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