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GMAT Algebra (Data Sufficiency)
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Author:  questioner [ Thu Nov 04, 2010 3:20 am ]
Post subject:  GMAT Algebra (Data Sufficiency)

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 John purchased was $1.65.

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

(B) Statement (1) alone is not sufficient. For instance, John could have purchased 15 large bottles and two small bottles, spending $33 in all, or he could have bought 12 large bottles and 6 small ones. In these two cases, the percent of bottles that were small is different.

Statement (2), however, is sufficient. If we know that the average price per bottle is $1.65, we can solve for the percent of bottles that are small. Using s for small bottles and L for large bottles:
$1.65 = (Total $ spent on water)/(Number of bottles)
$1.65 = ($1.5s + $2L)/(s + L)
$1.65s + $1.65L = $1.5s + 2L
$0.15s = $0.35L
s/L = $0.35/$0.15 = 7/3.

Therefore, the ratio of small to large bottles is 7 : 3, which means that 7/10 of the bottles are small and 3/10 of the bottles are large.

Using only Statement (2), we can determine that 70% of the bottles are small. Since Statement (1) is insufficient alone, and Statement (2) is sufficient, the correct answer is choice (B).
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The question does not seem to make sense to me. Instead of saying L = 2 and S = 1.5, why not say L = 2 and S = L – .5 then solve for both variables in statement 1?

I understand why conceptually you cannot use statement 1 to solve for L and S but don't see why the math doesn't work. Is this just poor wording?

Author:  Gennadiy [ Thu Nov 04, 2010 3:32 am ]
Post subject:  Re: math (test 2, question 4): algebra, data sufficiency

We denote the NUMBER of bottles, NOT the prices. We know the prices: $2 and $1.5.

If John purchased L large bottles, he paid $(2L) for those.
If John purchased s small bottles, he paid $(1.5L) for those.

So, if we use the first statement then we can write down the following equation:
1.5L + 2s = 33
s = (33 – 1.5L)/2

therefore

s/L = (16.5 – 0.75L)/L = 16.5/L – 0.75

BUT we see that the ratio of s to L depends on the value of L, which is unknown. It was shown in the explanation that different values of L yield different percentage.

Author:  questioner [ Tue May 29, 2012 8:55 am ]
Post subject:  Re: math (test 2, question 4): algebra, data sufficiency

Hi,
my final ratio is 3/7, can you please revise?
s/L is 0.15 /0.35 .
Thank you.

Author:  Gennadiy [ Tue May 29, 2012 8:59 am ]
Post subject:  Re: math (test 2, question 4): algebra, data sufficiency

Quote:
s/L is 0.15 /0.35
We have $0.15s = $0.35L . Cross-multiply, which is the same as to divide by 0.15L , to get

s/L = 0.35/0.15


However you do not need to find the exact value once you are sure that the information is sufficient. Though it helps to avoid a mistake if you have enough time to do so.

Author:  fedana [ Tue Nov 06, 2012 4:21 pm ]
Post subject:  Re: GMAT Algebra (Data Sufficiency)

The result of the comparison is incorrect: s/L is 0.15 / 0.35, not the other way around. Herewith 30% of the bottles are small.

Author:  Gennadiy [ Tue Nov 06, 2012 4:25 pm ]
Post subject:  Re: GMAT Algebra (Data Sufficiency)

fedana wrote:
The result of the comparison is incorrect: s/L is 0.15 / 0.35, not the other way around. Herewith 30% of the bottles are small.
If there were 30% small bottles and 70% large bottles, then the average price would be closer to $2 than to $1.5 .

Be careful, when getting a proportion. Take more smaller steps, if you need.
$0.15s = $0.35L |divide by L
$0.15s / L = $0.35 |divide by $0.15
s / L = 0.35 / 0.15

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