A metal alloy weighing 7oz and containing 70% iron is melted and mixed with 8oz of pure iron, what percent of the resulting alloy is iron?

Question

this is for algebra 2 but an algebra 1 problem.

Answer 1

Pure iron in 70% of 7 oz metal is a = 7·0.7 = 4.9 oz.

After mixer we have 8 + 4.9 = 12.9 oz of pure iron in total 7 + 8 oz of metal.

To get percent of iron in new alloy we have 12.9/15·100% = 86%.

Answer: 86%

Answer 2

70% of 7 oz is 0.7*7=4.9 oz

8+4.9=12.9 oz of iron

the resulting alloy is 8 oz + 7 oz=15 oz

12.9/15=

129/150=

simplify

43/50=

86/100=86%