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There are three varieties of sugar in a store. Sugar (S1)

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There are three varieties of sugar in a store. Sugar (S1) [#permalink] New post 27 Oct 2012, 00:21
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There are three varieties of sugar in a store. Sugar (S1) costs $6 per kilogram and sugar (S2) costs $9 per kilogram. Sugar (S1) is mixed with Sugar (S2) and a new variety of Sugar (S3) is formed. If 20 kilograms of Sugar (S3) consists of A kilogram of Sugar (S1) and B kilogram of Sugar (S2), is A > B ?

(1) A > 9
(2) The cost of 20 kilogram of Sugar (S3) is more than $150.
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Oct 2012, 04:51, edited 1 time in total.
Renamed the topic and edited the question.
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Re: There are three varieties of sugar in a store [#permalink] New post 27 Oct 2012, 16:02
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(1) Insufficient. A could be, for example, 10 or 11. If A = 10, then B=10 and A is not bigger than B. However, if A = 11, then B = 9 and A>B.

(2) Sufficient. If you would have to pay $150 for 20kg, then 1kg costs 150/20 = $7.50. Note that this would be exactly the price for 1kg if S3 consisted of equal parts of S1 and S2, since (6+9)/2 = 7.5. Since the statement tells us that the price per kg for S3 is greater than $7.50 we know that there has to be more of the more expensive sugar (= S2) in the mixture. This means A<B and we have our answer.
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Re: There are three varieties of sugar in a store. Sugar (S1) [#permalink] New post 06 Nov 2012, 01:15
GMATBaumgartner wrote:
There are three varieties of sugar in a store. Sugar (S1) costs $6 per kilogram and sugar (S2) costs $9 per kilogram. Sugar (S1) is mixed with Sugar (S2) and a new variety of Sugar (S3) is formed. If 20 kilograms of Sugar (S3) consists of A kilogram of Sugar (S1) and B kilogram of Sugar (S2), is A > B ?

(1) A > 9
(2) The cost of 20 kilogram of Sugar (S3) is more than $150.


(1) Given: A+B = 20 kg
When A=10kg , B= 10kg
When A = 11 kg, B = 9 kg.
Hence, not sufficient.

(2) Assume cost of 20 kg = $ 150.
Then, cost of 1 kg = \frac{150}{20}= $7.5
Hence the cost of 1 kg of S3 is $7.5.

The cost of 1 kg can also be expressed as \frac{Total Cost}{Total Weight}
or, \frac{6A+9B}{A+B}= 7.5

7.5 is the weighted average between 6 and 9, and is equidistant from both 6 and 9. This means that when the cost is equal to $150, then A = B. But if it is greater than $ 150, then the balance of weighted average will shift towards B, and hence, B>A. Sufficient.
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Re: There are three varieties of sugar in a store. Sugar (S1) [#permalink] New post 19 Feb 2013, 00:17
There are three varieties of sugar in a store. Sugar (S1) costs $6 per kilogram and sugar (S2) costs $9 per kilogram. Sugar (S1) is mixed with Sugar (S2) and a new variety of Sugar (S3) is formed. If 20 kilograms of Sugar (S3) consists of A kilogram of Sugar (S1) and B kilogram of Sugar (S2), is A > B ?

(1) A > 9. A=B if A=10 or A>B
(2) The cost of 20 kilogram of Sugar (S3) is more than $150. IF A=B=10 then , we can get cost = 150 . But cost is more than 150, so B>A.
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Re: There are three varieties of sugar in a store. Sugar (S1) [#permalink] New post 04 Jul 2014, 00:33
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Re: There are three varieties of sugar in a store. Sugar (S1)   [#permalink] 04 Jul 2014, 00:33
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