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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. 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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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.
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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.

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.
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GMATBaumgartner
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

B
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This doesnt seem to be a mixtures Question. Algebra/ word problems / inequalities is better

Posted from my mobile device
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GMAT Data Sufficiency
Do not forget.

We are asked if A is greater than B?

Let's analyze the information 1):

1) Assert that A is greater than 9.

What do we know from the statement?

A + B = 20

If A is greater than 9, A could be 10, fulfilling condition 1), and in this case A = B
But it could also be A = 11, fulfilling 1), and in this case: A = 11 and B = 9, fulfilling A is greater than B.
SINCE THERE ARE TWO SITUATIONS, in which one does not satisfy that A is greater than B and in the other it satisfies that A is greater than B.

With 1) ambiguity occurs.

Then 1) It is not enough.

Let's analyze situation 2)

In 2) we are told that:
The cost of producing the 20Kg of S3 is greater than $150.

Again, let's remember what the statement says:

S1 + S2 = S3

Cost of each kilogram of S1 is $6
Cost of each kilogram of S2 is $9
A kilograms + B kilograms = 20 Kg


Then:
At this point, it is convenient to place yourself at a point that allows us to draw quick conclusions: Do not continue advancing and state what that point would be?

Very well:
If A=B=10, the cost of producing the 20Kg of S3 would be:
10x6 + 10x9= $150

2) It tells us that the cost of producing 20Kg of S3 is greater than $150, so the only way for the latter to be true is that the amount of B (kilograms supplied by S2) is greater than the amount A (kilograms supplied by S1).

For the cost of producing 20 Kg to be greater than $150, B is always greater than A. We can affirm that A will never be greater than B, a categorical statement.
2) is enough.

Answer B
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1) If A>9, and A+B=20 => B<11. This is not enough to say whether A>B, because (A,B) = (9.5,10.5) and (A,B) = (11,9) both satisfy the conditions by the first gives A<B and the former A>B. Insufficient
2) 6A+9B > 150 => 2A+3B>50 . Since A+B=20, we can say 2A+3(20-B)>50 <=> A<10 <=> (20-B)<10 <=> B>10 <=> A<10<B <=> A<B. Sufficient
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Quantity of S1 = X
Quantity of S2 = 20-X

For Statement 2
6(X) + 9(20-X) > 150
10>X (S1 quantity is less than 10 means automatically B's quantity will be higher)

Hence St.2 is Suff.
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