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2016/17 AHSMC Part 1 Problem 3

Problem 3 of 16

A ring of 1010 grams is 60%60\% gold and 40%40\% silver. A jeweller wants to melt it down, add 22 grams of silver and add enough gold to make it 70%70\% gold. How many grams of gold should be added?

Answer choices

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Solution

Let xx be the grams of gold which should be added. Then 710=10×610+x10+2+x. \frac{7}{10} = \frac{10 \times \dfrac{6}{10} + x}{10 + 2 + x}. Solving the equation one obtains x=8x = 8.

Alternative solution: The original ring contains 44 g of silver and 66 g of gold. The new ring will contain 4+2=64+2 = 6 g of silver, which must account for 30%30\% of the total. Thus the new ring must weigh 6×103=206 \times \dfrac{10}{3} = 20 g, of which therefore 20−10−2=820-10-2=8 g must be added gold. The answer is (C).

University of Alberta