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John purchased large bottles of water for $2 each and small

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John purchased large bottles of water for $2 each and small [#permalink] New post 23 Apr 2012, 04:12
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Question Stats:

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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] New post 23 Apr 2012, 04:29
Expert's post
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.
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Re: John purchased large bottles of water for $2 each and small [#permalink] New post 27 Apr 2012, 05:10
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

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Re: John purchased large bottles of water for $2 each and small [#permalink] New post 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}.

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.
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Re: John purchased large bottles of water for $2 each and small [#permalink] New post 11 Sep 2012, 07:38
Expert's post
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.
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Re: John purchased large bottles of water for $2 each and small [#permalink] New post 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] New post 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}.
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Re: John purchased large bottles of water for $2 each and small [#permalink] New post 19 Jun 2014, 03:33
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Re: John purchased large bottles of water for $2 each and small   [#permalink] 19 Jun 2014, 03:33
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