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
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}.
Let's denote the amount invested at 6% as x.
The amount invested at 10% would then be (21,000 - x).
The interest earned from the account at 6% would be (x * 0.06), and the interest earned from the account at 10% would be [(21,000 - x) * 0.10].
Given that the total interest earned is $1580, we can write the equation:
(x * 0.06) + [(21,000 - x) * 0.10] = $1580
Now, let's solve for x:
0.06x + (21,000 - x) * 0.10 = 1580
0.06x + 2100 - 0.10x = 1580
-0.04x = 1580 - 2100
-0.04x = -520
x = (-520) / (-0.04)
x = 13,000
Therefore, the worker invested $13,000 at 6%.