If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?
(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.
This is a MAX/MIN problem.
Try to MAXIMIZE and MINIMIZE the price of each item.
Nothing in the problem restrict the prices to integers.
The price of each notebook can be INFINITELY small, as can the price of each pencil.
Thus, we can simplify the problem by trying one case in which the price of each notebook = 0, and one case in which the price of each pencil = 0.
Statement 1:
Since 5 pencils can be purchased for 20 francs, the maximum possible price of each pencil = 20/5 = 4 francs.
Since 9 notebooks can be purchased for 20 francs, the maximum possible price of each notebook = 20/9 francs.
Case 1: Price of each notebook = 0 francs, price of each pencil = 4 francs
Total cost for 12 notebooks and 12 pencils = 12*(0) + 12(4) = 48 francs, which is MORE than 40 francs.
Case 2: Price of each notebook = 20/9 francs, price of each pencil = 0 francs
Total cost for 12 notebooks and 12 pencils = 12*(20/9) + 12(0) ≈ 24 francs, which is LESS than 40 francs.
Since 40 francs is NOT enough in Case 1 but IS enough in Case 2, INSUFFICIENT.
Statement 2:
Since 8 pencils can be purchased for 20 francs, the maximum possible price of each pencil = 20/8 = 5/2 francs.
Since 9 notebooks can be purchased for 20 francs, the maximum possible price of each notebook = 20/9 francs.
Case 1: Price of each notebook = 0 francs, price of each pencil = 5/2 francs
Total cost for 12 notebooks and 12 pencils = 12*(0) + 12(5/2) = 30 francs, which is LESS than 40 francs.
Case 2: Price of each notebook = 20/9 francs, price of each pencil = 0 francs
Total cost for 12 notebooks and 12 pencils = 12*(20/9) + 12(0) ≈ 24 francs, which is LESS than 40 francs.
Since 40 francs is enough in each case, SUFFICIENT.
The correct answer is
B.
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