ruplun wrote:[/img]

From the question stem, all we know is that there are 2 types of pencils, and we want to know how many of the expensive ones Marta bought. Let's call L the "# of low cost 21-cent pencils she buys" and H the "# of high cost 23-cent pencils she buys". Therefore, the question becomes, what is H?

Statement (1)
H + L = 6.
- we could have {H=2,L=4} or {H=5,L=1} and so on - Insufficient

Statement (2)
21*L + 23*H = 130
- This is harder to list the possible combinations BUT we should actually assume that with a fixed set of prices, there might only be 1 combination of integer L and H that make this equation work. So we start with the higher value (the more strict constraint) and see the possibilities:

1 H = 23, so 107 left to spend on L...this is not divisible by 21 so NOT POSSIBLE
2 H = 46, so 84 left to spend on L...this IS DIVISIBLE by 21
3 H = 69, so 61 left to spend on L...this is not divisible by 21 so NOT POSSIBLE
4 H = 92, so 38 left to spend on L...this is not divisible by 21 so NOT POSSIBLE
5 H = 115, so 15 left to spend on L...this is not divisible by 21 so NOT POSSIBLE
6 H = 138, overspent so NOT POSSIBLE

The only combination of integer H and L that makes the equation true is {H=2, and L=4}. SUFFICIENT

The correct answer is B (statement 2 alone).


Whit