Bunuel wrote:

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)

Attachment:

2017-10-29_1222.png

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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