Question 1102815
<br>
Let x be the amount invested at 7%; then the amount invested at 11% is 3x.<br>
The interest from the first investment is then .07 times x; the interest from the second is .11 times 3x.<br>
Since the total interest is $2400,
{{{.07(x)+.11(3x) = 2400}}}
{{{.07x+.33x = 2400}}}
{{{.40x = 2400}}}
{{{x = 2400/.40 = 6000}}}<br>
She invested x = $6000 at 7% and 3x = $18000 at 11%.<br>
CHECK:
{{{.07(6000)+.11(18000) = 420+1980 = 2400}}}