Bunuel wrote:

Mr. Williams invested a total of $12,000 for a one-year period. Part of the money was invested at 5% simple interest, and the rest was invested at 12% simple interest. If he earned a total of $880 in interest for the year, how much of the money was invested at 12%?

(A) $1,920

(B) $4,000

(C) $4,800

(D) $7,200

(E) $8,000

Try a weighted average approach, similar to that in mixture problems.

Interest, as a percent, is like concentration.

Let x = amount in dollars invested at 5% (below, equivalent to "weight")

Let y = amount in dollars invested at 12% (below, equivalent to "weight")

\(x + y = $12,000\)

\(x = $12,000 - y\)\((Rate_1)(Weight_1) + (R_2)(W_2)\) = Total Interest

\(.05x + .12y = $880\)Substitute

\(x = ($12,000 - y)\)

\(.05($12,000 - y) + .12y = $880\)

\($600 - .05y + .12y = $880\)

\(.07y = $280\)

\(y = \frac{$280}{.07}=\frac{$28,000}{7}=$4,000\)Answer B