If you purchase x grams of Food A, y grams of Food B, and z grams of F

Question

https://gmatclub.com/forum/download/file.php?id=38211 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) 2017-10-29_1222.png

generis · Answer

Bunuel , Answer B is missing text.

Turkish · Answer

The Cost will be .018x+.03y+.0275z

generis · Answer

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.

Bunuel · Answer

_______________
Edited. Thank you.

Luckisnoexcuse · Answer

$(1.8x+3y+2.75z)/100
$(9x/500 + 3y/100 + 11z/400)
¢(9x/5 + 3y + 11z/4)
A

anhht13 · Answer

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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Kindly press +1 Kudos if the explanation is clear.
Thank you!

ScottTargetTestPrep · Answer

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

youcouldsailaway · Answer

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.

Regor60 · Answer

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

Posted from my mobile device

bumpbot · Answer

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If you purchase x grams of Food A, y grams of Food B, and z grams of F

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