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John purchased large bottles of water for $2 each and small [#permalink] ### Show Tags 23 Apr 2012, 04:12 7 This post was BOOKMARKED 00:00 Difficulty: 75% (hard) Question Stats: 50% (02:21) correct 50% (01:27) wrong based on 232 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 [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 36549 Followers: 7078 Kudos [?]: 93145 [0], given: 10553 Re: John purchased large bottles of water for$2 each and small [#permalink]

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23 Apr 2012, 04:29
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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
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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: 47 Location: Tajikistan Concentration: General Management, Technology GPA: 3.86 Followers: 2 Kudos [?]: 26 [0], given: 4 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. GMAT Club Legend Joined: 09 Sep 2013 Posts: 13441 Followers: 575 Kudos [?]: 163 [0], given: 0 Re: 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 Re: John purchased large bottles of water for$2 each and small   [#permalink] 19 Jun 2016, 03:03
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