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Firstly, for my own sanity, I multiplied every number by 10 so I was working with integers.

From the problem stem we can conclude that the number of Pencils will be a multiple of 3 and the number of Erasers will be a multiple of 5. We also are given the number of pencils purchased cannot equal 4, or in other words the total cost of the pencils cannot equal 12.

1) This provides us a total of 42 for the combined cost of pencils and erasers. From here I simply ran through the combinations of multiples of 5 subtracted from 42 in order to find a multiple of 3.
Visually It looked a bit like this
42=5(1)+37. Incorrect, 37 is not a multiple of 3.
42=5(2)+32. Incorrect, 32 is not a multiple of 3.
42=5(3)+27. Possibly correct, 27 is 3x9.
42=5(4)+22. Incorrect, 22 is not a multiple of 3.
42=5(5)+17. Incorrect, 17 is not a multiple of 3
42=5(6)+12. Incorrect the number of pencils purchased cannot be 4. 3x4=12.
42=5(7)+7. Incorrect, 7 is not a multiple of 3.

This leaves us with the only possible combination of 3 Erasers and 3 Pencils purchased.
Suff

2). Gives us the number of erasers purchased but without an upper limit for the total cost we have nothing to go off here as the number of pencils could be anywhere between 1 and ∞.
Insuff

Answer choice A
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