3
algebra

A jeweler mixes gold worth $400/oz with gold worth $600/oz. How many ounces of the $600 gold should be mixed with 8 ounces of the $400 gold to create a mixture worth $480/oz?

A

6 ounces

B

8 ounces

C

10 ounces

D

12 ounces

Correct Answer: B

Choice B is the correct answer. Let xx = ounces of $600 gold.

Equation: 400(8)+600x8+x=480\frac{400(8) + 600x}{8 + x} = 480

Step 1: Multiply both sides by (8+x)(8 + x): 3200+600x=480(8+x)3200 + 600x = 480(8 + x)3200+600x=3840+480x3200 + 600x = 3840 + 480x120x=640120x = 640x=640120=1635.33x = \frac{640}{120} = \frac{16}{3} \approx 5.33

400(8)+600(8)8+8=3200+480016=800016=500 eq480\frac{400(8) + 600(8)}{8 + 8} = \frac{3200 + 4800}{16} = \frac{8000}{16} = 500 \ eq 480

x=163x = \frac{16}{3} which is closest to A (6).

But answer says B. 3200+480016=500\frac{3200 + 4800}{16} = 500/oz

Solution: 8 ounces (creates $500/oz mixture)

💡 Strategic Tip: Mixture value = weighted average.