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There are three varieties of sugar in a store. Sugar (S1) [#permalink]

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27 Oct 2012, 01:21

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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.

Re: There are three varieties of sugar in a store [#permalink]

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27 Oct 2012, 17:02

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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.

Re: There are three varieties of sugar in a store. Sugar (S1) [#permalink]

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06 Nov 2012, 02:15

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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]

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19 Feb 2013, 01:17

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. _________________

Re: There are three varieties of sugar in a store. Sugar (S1) [#permalink]

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04 Jul 2014, 01:33

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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There are three varieties of sugar in a store. Sugar (S1) [#permalink]

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23 May 2015, 09:20

2nd statement: 6x + 9x > 150 15x >150 x > 10

Therefore, because S3 consists of only 20 kgs you have sufficient information to prove that one of the quantities (S1 or S2, it does not matter which one) is bigger than the other.

Re: There are three varieties of sugar in a store. Sugar (S1) [#permalink]

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26 Jun 2015, 11:41

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.

State 1: is clearly insufficient

State 2: Total cost is > 150, implies that: 6a+9b > 150 & we know that a + b = 20 2a + 3b >50 ; substituting for a and simplifying will give us a>10 & since a + b is 20, b has to be less than a.

Re: There are three varieties of sugar in a store. Sugar (S1) [#permalink]

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29 Oct 2016, 12:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: There are three varieties of sugar in a store. Sugar (S1) [#permalink]

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13 Nov 2016, 10:31

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.

IF A>B therefore the price per kilogram of mixture will be closer to 6 i.e. 6<= price per kg of mixture<7.5 or the weight of A in the mixture >10

from 1

1) clearly insuff

from 2

the cost /kg of mix = 150/20 = 7.5 thus A=B.... SUFF

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