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

(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.

(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.

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.

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.

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

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

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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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