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23 Apr 2012, 05:12
7
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Difficulty:

75% (hard)

Question Stats:

50% (01:20) correct 50% (01:27) wrong based on 240 sessions

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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?

(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 23 Apr 2012, 05:29 Expert's post 1 This post was BOOKMARKED 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 $$\frac{y}{x+y}=\frac{35}{15+35}=\frac{70}{100}$$. Answer: B. _________________ Kudos [?]: 132526 [0], given: 12324 Manager Joined: 28 Jul 2011 Posts: 233 Kudos [?]: 161 [0], given: 16 Re: John purchased large bottles of water for$2 each and small [#permalink]

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27 Apr 2012, 06:10
1
This post was
BOOKMARKED
got B

(A)

2(l) + 1.5(s) = 33

not sufficient

(B)

[2(l) + 1.5(s)] / (l+s) = 1.65

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

2l - 1.65l = 1.65s - 1.5s

0.35(l) = 0.15 (s)

35/15 = s/l

% of small bottles

=35/ (35+15)*100
= 70%

sufficient

- bookmarking for future reference

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Re: John purchased large bottles of water for $2 each and small [#permalink] ### Show Tags 11 Sep 2012, 07:16 1 This post received KUDOS 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. _________________ If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth -Game Theory If you have any question regarding my post, kindly pm me or else I won't be able to reply Kudos [?]: 867 [1], given: 276 Math Expert Joined: 02 Sep 2009 Posts: 42248 Kudos [?]: 132526 [0], given: 12324 Re: John purchased large bottles of water for$2 each and small [#permalink]

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11 Sep 2012, 08:38
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 $$\frac{y}{x+y}=\frac{35}{15+35}=\frac{70}{100}$$.

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.
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Re: John purchased large bottles of water for $2 each and small [#permalink] ### Show Tags 11 Sep 2012, 09:43 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 _________________ God loves the steadfast. Kudos [?]: 33 [0], given: 4 Director Joined: 22 Mar 2011 Posts: 610 Kudos [?]: 1072 [0], given: 43 WE: Science (Education) Re: John purchased large bottles of water for$2 each and small [#permalink]

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11 Sep 2012, 14:00
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 $$\frac{N_2}{N_1+N_2}$$.
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12 Jul 2015, 08:59
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19 Jun 2016, 04:03
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

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