Lorna invests $6,000, some at a 6% annual interest rate and some at 11

Can you further elaborate on this explanation? The setup is not clear to me. Thanks.

Weighted average
The amounts invested at different rates sum to \($6,000\)
Let \(x\) = amount invested at \(6\)%
\((6,000-x)\) = amount invested at \(11\)%

The dollar amounts of interest earned by each portion at each rate sum to $580. Different rates carry different weights.
Weighted average:
\(.06(x) +.11(6,000-x)=580\)
\(.06x + 660 -.11x = 580\)
\(80=.05x\)
\(x=\frac{80}{.05}=\frac{8000}{5}=1,600\)

ANSWER B

Answer Choices

Start with C) $2200

Amt invested at 6%: \($2200\)
Amt invested at 11%: \($(6000-2200)=$3,800\)

Interest earned at 6%:\(($2200*.06)=$132\)
Interest earned at 11%:\(($3800*.11)=$418\)

Total interest earned?\($(132+418)=$550\)

Too low. The 6% rate has too great a portion of the $6,000. If 11% has a greater portion of the $6,000, total interest earned will increase.

Eliminate Answers D and E.
Their portion for the 6% rate is higher than C, which is already too high. We need a number lower than C, but not by much.

Try B) $1600

Amt invested at 6%: \($1600\)
Amt invested at 11%:\(($6000-1600)=$4400\)

Interest earned at 6%: \(($1600*.06)=$96\)
Interest earned at 11%: \(($4400*.11)=$484\)

Total interest earned:
\($(96+484)=$580\)
That's correct

ANSWER B

We can let n = the amount invested at 6% and (6000 - n) = the amount invested at 11%; thus:

0.06n + 0.11(6,000 - n) = 580

0.06n + 660 - 0.11n = 580

80 = 0.05n

1,600 = n

Answer: B

Make it as easy as possible. This is a mixture (weighted average) problem.

580/6000=29/300=weighted average interest from both

6/100=18/300

11/100=33/300

that means the ratio of the two investments was

18---29---33

4:11

hence invested @6%: 4/(11+4)*6000