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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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Joined: 02 Sep 2009
Posts: 47112
If you purchase x grams of Food A, y grams of Food B, and z grams of F  [#permalink]

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29 Oct 2017, 01:25
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Difficulty:

65% (hard)

Question Stats:

50% (02:08) correct 50% (01:09) wrong based on 88 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) $(x + y + z) Attachment: 2017-10-29_1222.png [ 21.48 KiB | Viewed 1364 times ] _________________ SC Moderator Joined: 22 May 2016 Posts: 1830 Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink] ### Show Tags 29 Oct 2017, 15:35 1 Bunuel wrote: If you purchase x grams of Food A, y grams of Food B, and z grams of Food C, the cost will be (A) (9x/5 + 3y + 11z/4)¢ (B)$
(C) $(1.8x + 3z + 2.75y) (D) (3x + 1.8y + 2.75z)¢ (E)$(x + y + z)

Attachment:
2017-10-29_1222.png

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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29 Oct 2017, 19:04
The Cost will be .018x+.03y+.0275z
SC Moderator
Joined: 22 May 2016
Posts: 1830
Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F  [#permalink]

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29 Oct 2017, 20:32
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. _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. -- Albert Camus, "Return to Tipasa" Math Expert Joined: 02 Sep 2009 Posts: 47112 Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink] ### Show Tags 29 Oct 2017, 21:18 genxer123 wrote: 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

Bunuel, Answer B is missing text.

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Edited. 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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29 Oct 2017, 21:33
1
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) $(9x/5 + 3y + 11z/4) (C)$(1.8x + 3z + 2.75y)
(D) (3x + 1.8y + 2.75z)¢
(E) $(x + y + z) Attachment: 2017-10-29_1222.png$(1.8x+3y+2.75z)/100
$(9x/500 + 3y/100 + 11z/400) ¢(9x/5 + 3y + 11z/4) A _________________ We must try to achieve the best within us Thanks Luckisnoexcuse Intern Joined: 12 Oct 2017 Posts: 39 Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink] ### Show Tags 29 Oct 2017, 22:00 1 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 = $(9x/5 + 3y + 11z/4)/100 = ¢(9x/5 + 3y + 11z/4) Hence, the answer is A. --- Kindly press +1 Kudos if the explanation is clear. Thank you! Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 2949 Location: United States (CA) Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F [#permalink] ### Show Tags 31 Oct 2017, 16:17 1 Bunuel wrote: If you purchase x grams of Food A, y grams of Food B, and z grams of Food C, the cost will be (A) (9x/5 + 3y + 11z/4)¢ (B)$(9x/5 + 3y + 11z/4)
(C) $(1.8x + 3z + 2.75y) (D) (3x + 1.8y + 2.75z)¢ (E)$(x + y + z)

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

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Re: If you purchase x grams of Food A, y grams of Food B, and z grams of F &nbs [#permalink] 31 Oct 2017, 16:17
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