<PRE><B>Question 1162082</B>
&lt;br&gt;
A setup using the traditional algebraic approach would look something like this:&lt;br&gt;
let x = amount invested at 4%
then 29000-x = amount invested at 7%&lt;br&gt;
The interest from one investment is .04(x); from the other is .07(29000-x).&lt;br&gt;
The total interest was $1160:&lt;br&gt;
{{{.04(x)+.07(29000-x) = 1160}}}&lt;br&gt;
Solve using basic algebra.  I leave that to you.&lt;br&gt;
I find this alternative method faster and easier....&lt;br&gt;
(1) Determine how the actual interest compares to the interest that would have been earned if the whole $29000 had been invested at each rate.
all at 4%: .04(29000) = 1160
actual interest: 1550
all at 7%: .07(29000) = 2030&lt;br&gt;
(2) Where the actual interest lies between the two extremes exactly determines what fraction of the total was invested at each rate.&lt;br&gt;
The difference between 1160 and 2030 is 870
The difference between 1160 and 1550 is 390
The actual interest 1550 is 390/870 = 39/87 = 13/29 of the way from 1160 to 2030.&lt;br&gt;
That means 13/29 of the total was invested at the higher rate.&lt;br&gt;
ANSWER: 13/29 of $29,000, or $13,000, at 7%; the other $16,000 at 4%.&lt;br&gt;
CHECK: .07(13000)+.04(16000) = 910+640 = 1550&lt;br&gt;
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