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Manager
Joined: 06 Jul 2011
Posts: 207
Location: Accra, Ghana
John purchased large bottles of water for $2 each and small [#permalink] ### Show Tags 23 Apr 2012, 04:12 6 00:00 Difficulty: 75% (hard) Question Stats: 51% (01:52) correct 49% (02:05) wrong based on 257 sessions ### HideShow timer Statistics John purchased large bottles of water for$2 each and small bottles of water for $1.50 each. What percent of the bottles purchased were small bottles? (1) John spent$33 on the bottles of water
(2) The average price of bottles purchased was $1.65 Math Expert Joined: 02 Sep 2009 Posts: 50572 Re: John purchased large bottles of water for$2 each and small  [#permalink]

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23 Apr 2012, 04:29
John purchased large bottles of water for $2 each and small bottles of water for$1.50 each. What percent of the bottles purchased were small bottles?

Say John purchased x large bottles and y small bottles.

(1) John spent $33 on the bottles of water --> $$2x+1.5y=33$$ --> $$4x+3y=66$$. Several integer solutions possible to satisfy this equation, for example $$x=15$$ and $$y=2$$ OR $$x=12$$ and $$y=6$$. Not sufficient. (2) The average price of bottles purchased was$1.65 --> $$\frac{2x+1.5y}{x+y}=1.65$$ --> $$2x+1.5y=1.65x+1.65y$$ --> $$0.35x=0.15y$$ --> $$\frac{y}{x}=\frac{35}{15}$$, we have the ratio, which is sufficient to get the percentage.

Just to illustrate $$\frac{y}{x+y}=\frac{35}{15+35}=\frac{70}{100}$$.

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11 Sep 2012, 06:16
1
Bunuel wrote:
John purchased large bottles of water for $2 each and small bottles of water for$1.50 each. What percent of the bottles purchased were small bottles?

Say John purchased x large bottles and y small bottles.

(1) John spent $33 on the bottles of water --> $$2x+1.5y=33$$ --> $$4x+3y=66$$. Several integer solutions possible to satisfy this equation, for example $$x=15$$ and $$y=3$$ OR $$x=12$$ and $$y=6$$. Not sufficient. (2) The average price of bottles purchased was$1.65 --> $$\frac{2x+1.5y}{x+y}=1.65$$ --> $$2x+1.5y=1.65x+1.65y$$ --> $$0.35x=0.15y$$ --> $$\frac{y}{x}=\frac{35}{15}$$, we have the ratio, which is sufficient to get the percentage.

Just to illustrate $$\frac{y}{x+y}=\frac{35}{15+35}=\frac{70}{100}$$.

Hi Bunuel,

There is small error in one of the calculations.
$$x=15$$ and $$y=3$$
should be
$$x=15$$ and $$y=2$$

Kindly correct me if i am wrong.
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Re: John purchased large bottles of water for $2 each and small [#permalink] ### Show Tags 11 Sep 2012, 07:38 fameatop wrote: Bunuel wrote: John purchased large bottles of water for$2 each and small bottles of water for $1.50 each. What percent of the bottles purchased were small bottles? Say John purchased x large bottles and y small bottles. (1) John spent$33 on the bottles of water --> $$2x+1.5y=33$$ --> $$4x+3y=66$$. Several integer solutions possible to satisfy this equation, for example $$x=15$$ and $$y=3$$ OR $$x=12$$ and $$y=6$$. Not sufficient.

(2) The average price of bottles purchased was $1.65 --> $$\frac{2x+1.5y}{x+y}=1.65$$ --> $$2x+1.5y=1.65x+1.65y$$ --> $$0.35x=0.15y$$ --> $$\frac{y}{x}=\frac{35}{15}$$, we have the ratio, which is sufficient to get the percentage. Just to illustrate $$\frac{y}{x+y}=\frac{35}{15+35}=\frac{70}{100}$$. Answer: B. Hi Bunuel, There is small error in one of the calculations. $$x=15$$ and $$y=3$$ should be $$x=15$$ and $$y=2$$ Kindly correct me if i am wrong. Typo edited. Thank you. +1. _________________ Intern Status: Life begins at the End of your Comfort Zone Joined: 31 Jul 2011 Posts: 43 Location: Tajikistan Concentration: General Management, Technology GPA: 3.86 Re: John purchased large bottles of water for$2 each and small  [#permalink]

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11 Sep 2012, 08:43
Let x be the number of large bottles of water and y be the number of small bottles of water, from the question stem we get:
2*x+1.5*y=33, thus 1 is INSUFFICIENT.
Mowing to 2 condition:
(2*x+1.5*y)/(x+y) = 1.65 ------>>>> 2x+1.5y = 1.65x+1.65y, from here we easily get that 7x=3y, OR x = 3y/7 now we now x we can easily find the ratio of y in total of bottles: y/(y+3y/7) = 7/10 or 70%

Please correct me, if I went awry. Actually we do not need the solution, as it is data sufficiency so 2 is SUFFICIENT

dzodzo85 wrote:
John purchased large bottles of water for $2 each and small bottles of water for$1.50 each. What percent of the bottles purchased were small bottles?

(1) John spent $33 on the bottles of water (2) The average price of bottles purchased was$1.65

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Re: John purchased large bottles of water for $2 each and small [#permalink] ### Show Tags 11 Sep 2012, 13:00 dzodzo85 wrote: John purchased large bottles of water for$2 each and small bottles of water for $1.50 each. What percent of the bottles purchased were small bottles? (1) John spent$33 on the bottles of water
(2) The average price of bottles purchased was $1.65 Dealing with Statement (2): remember weighted average (also used when dealing with mixture problems). If we have $$N_1$$ numbers with average $$A_1$$, and $$N_2$$ numbers with average $$A_2$$, the final average being A, then the differences between the final average and the initial averages are inversely proportional to the two numbers of numbers (assume $$A_1>A_2$$): $$(A_1-A)N_1=(A-A_2)N_2$$ or $$\frac{A_1-A}{A-A_2}=\frac{N_2}{N_1}$$. This follows from the equality $$\frac{N_1A_1+N_2A_2}{N_1+N_2}=A.$$ In our case we know $$A, A_1,A_2$$ so we can find the ratio $$\frac{N_2}{N_1}$$ and then, obviously $$\frac{N_2}{N_1+N_2}$$. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Board of Directors Joined: 17 Jul 2014 Posts: 2645 Location: United States (IL) Concentration: Finance, Economics GMAT 1: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) John purchased large bottles of water for$2 each and small  [#permalink]

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31 Dec 2015, 18:16
1. small bottles can be 2, and 15 large or 6 small and 12 large. 1 insufficient.
2. 2L+3S/L+S = 1.65
we are given proportions. thus, we can solve the question.
2L+1.5S=1.65L + 1.65S
0.35L=0.15S.
multiply by 100
35L=15S.
S/L = 35/15

S/S+L = 35/15+35 = 35/50 or 7/10. 70%
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Re: John purchased large bottles of water for $2 each and small [#permalink] ### Show Tags 19 Jun 2016, 03:03 dzodzo85 wrote: John purchased large bottles of water for$2 each and small bottles of water for $1.50 each. What percent of the bottles purchased were small bottles? (1) John spent$33 on the bottles of water
(2) The average price of bottles purchased was $1.65 (1)Not suff. (2) I will use weighted avg. method here W1/W2=(A2-Avg.)/(Avg.-A2) W1=No. of large bottles W2=No. of large bottles A1=2 A2=1.5 Avg.=1.65 W1/W2=3/7 %age of small bottles= 7/(3+7)------>7/10----70% Ans B Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2830 Re: John purchased large bottles of water for$2 each and small  [#permalink]

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07 Dec 2017, 10:15
dzodzo85 wrote:
John purchased large bottles of water for $2 each and small bottles of water for$1.50 each. What percent of the bottles purchased were small bottles?

(1) John spent $33 on the bottles of water (2) The average price of bottles purchased was$1.65

We are given that John purchased large bottles of water for $2 each and small bottles of water for$1.50 each. We can let s = the number of small bottles purchased, and b = the number of larger bottles purchased. We need to determine what percentage of the bottles purchased were small bottles, i.e. the value of s/(s+b) x 100.

Statement One Alone:

John spent $33 on the bottles of water. Using the information in statement one, we can create the following equation: 2b + 1.5s = 33 We can multiply the entire equation by 2 and we have: 4b + 3s = 66 4b = 66 - 3s 4b = 3(22 - s) b = [3(22 - s)]/4 Since b must be an integer, 3(22 - s) must be a multiple of 4. 3(22 - s) is a multiple of 4 when s = 2, 6, 10, 14, or 18. Since we have multiple values of s, we will also have multiple values of b, and thus we do not have enough information to answer the question. Statement Two Alone: The average price of bottles purchased was$1.65.

Using the information in statement two, we can create the following equation:

1.65 = (2b + 1.5s)/(b + s)

1.65(b + s) = 2b + 1.5s

165(b + s) = 200b + 150s

165b + 165s = 200b + 150s

15s = 35b

3s = 7b

(3/7)s = b

We can now determine the value of s/(s+b) x 100 by substituting (3/7)s for b:

s/[s+(3/7)s] x 100

s/[(10/7)s] x 100

1/(10/7) x 100

7/10 x 100 = 70

So the small bottles account for 70 percent of the bottles purchased. We have answered the question.

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