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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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If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink]

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New post 29 Oct 2017, 00:25
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Difficulty:

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Question Stats:

47% (02:47) correct 53% (01:16) wrong based on 43 sessions

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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)


[Reveal] Spoiler:
Attachment:
2017-10-29_1222.png
2017-10-29_1222.png [ 21.48 KiB | Viewed 838 times ]
[Reveal] Spoiler: OA

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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink]

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New post 29 Oct 2017, 14:35
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This post received
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Bunuel wrote:
Image
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)


[Reveal] Spoiler:
Attachment:
2017-10-29_1222.png

Bunuel, Answer B is missing text.

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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink]

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New post 29 Oct 2017, 18:04
The Cost will be .018x+.03y+.0275z

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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink]

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New post 29 Oct 2017, 19:32
Bunuel wrote:
Image
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)


[Reveal] Spoiler:
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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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink]

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New post 29 Oct 2017, 20:18
genxer123 wrote:
Bunuel wrote:
Image
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)


[Reveal] Spoiler:
Attachment:
2017-10-29_1222.png

Bunuel, Answer B is missing text.

_______________
Edited. Thank you.
_________________

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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink]

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New post 29 Oct 2017, 20:33
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This post received
KUDOS
Bunuel wrote:
Image
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)


[Reveal] Spoiler:
Attachment:
2017-10-29_1222.png


$(1.8x+3y+2.75z)/100
$(9x/500 + 3y/100 + 11z/400)
¢(9x/5 + 3y + 11z/4)
A
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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink]

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New post 29 Oct 2017, 21:00
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.
---

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

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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink]

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New post 31 Oct 2017, 15:17
Bunuel wrote:
Image
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)


[Reveal] Spoiler:
Attachment:
2017-10-29_1222.png


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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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F   [#permalink] 31 Oct 2017, 15:17
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