A worker invested $21,000 in two accounts, one yielding 6% interest and the other yielding 10%. If he received a total of $1580 in interest at the end of the year, how much did he invest at 6% ? The amount invested at 6% was

Answer 1

Let x be the amount the worker invested at 6%.
Therefore, the amount he invested at 10% is 21000 - x.
The interest he received from the 6% account is x * 0.06.
The interest he received from the 10% account is (21000 - x) * 0.1.
The total interest he received is x * 0.06 + (21000 - x) * 0.1 = 1580.
0.06x + 2100 - 0.1x = 1580.
0.04x = 520.
x = <<13000=13000>>13000. Answer: \boxed{13000}.