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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 \(\frac{y}{x+y}=\frac{35}{15+35}=\frac{70}{100}\).

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

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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.
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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 \(y=3\) 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 \(y=2\)

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

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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

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

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19 Jun 2014, 04:33

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

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12 Jul 2015, 08:59

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John purchased large bottles of water for $2 each and small [#permalink]

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31 Dec 2015, 19:16

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

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

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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