Question 1125192
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The traditional algebraic solution method....<br>
Let x be the amount invested at 6%; then (30000-x) is the amount invested at 8%.<br>
The interest earned is then 6% of x plus 8% of (30000-x).  Since the earned interest was 2260,<br>
{{{.06(x)+.08(30000-x) = 2260}}}<br>
Perhaps a bit awkward to solve because of the decimals; still basic algebra that you should be able to solve in a minute or two.<br>
But here is a method which, if you understand it, will get you to the answer much faster and with much less work.<br>
(1) The interest earned if all $30000 were invested at 6% is $1800; the interest if all were invested at 8% is $2400.  Where the actual interest of $2260 lies between $1800 and $2400 determines the ratio in which the money was split between the two investments.
(2) Think of it on a number line if it helps.  2260-1800=460; 2400-1800=600.  So the actual amount of interest is "460/600 of the way from 1800 to 2400".  That means 460/600 of the total investment was at the higher rate.
(3) 460/600 = 230/300 = 23000/30000.  So $23000 was invested at the higher 8% rate; the other $7000 was invested at 6%.<br>
CHECK: .06(7000)=.08(23000) = 420+1840 = 2260.