# Harry has \$4000 invested in two savings accounts. One account earns 6% interest per year, and the other pays 7% per year. If his total interest for the year is \$264, how much is invested at 7%?

There are several different ways you can work this kind of mixture problem. I'll show 3 of them.
1. Figure out what it takes to get where you are from one (usually the lower) extreme.
If all the money were invested at 6%, the interest would be .06*4000 = 240. The alternate investment (7%) pays an additional dollar of interest for each 100 dollars invested. Harry's return is 264 - 240 = 24 dollars more than the lower extreme, so he must have invested 24*100 = 2400 dollars at 7%. (You can do that in your head.)

2. Figure out the proportion invested at each rate. A convenient way to do this is to draw a sort of X diagram. For it to work, you need to know the actual average rate achieved, or you need to know what would have been returned had the entire investment been done at one rate. The first of these first. The actual average rate is \$264/\$4000*100% = 6.6%. Our X diagram looks like this, where the numbers on the right are the diagonal differences between the center and the numbers on the left.
7%          .6%
6.6%
6%          .4%
These tell us that the proportion of the investment at 7% is (.6%)/(.6% + .4%) = 6/(6+4) = 6/10.
The amount associated with that proportion is \$4000*(6/10) = \$2400. (same as above)

I have left the units in the difference numbers .6% and .4%. All you need to recognize is their relative size: 6 to 4, with the 6 (top number) being associated with the 7% investment (top interest rate).

The alternate way to draw the X diagram uses the interest that would have been achieved at each rate.
\$4000*7% = \$280                      24
\$264
\$4000*6% = \$240                      16
The proportion at 7% is (24)/(24+16) = 24/40 = 6/10, as above.

3. Write the equation. Usually, you will form the equation in terms of the amount you are looking for, and the result of doing the mix. We will let "x" represent our seven percent investment amount, and write an equation telling us the total interest received.
7%*x + 6%*(4000 - x) = 264
x(7% - 6%) + 240 = 264
x/100 = 264 - 240
x = 24*100 = 2400
\$2400 was invested at the 7% rate.
Note that this is very similar to our word description of how to work the problem as given in solution number 1. That's the way is will work out if the variable is associated with the amount invested at the higher interest rate.
thanked the writer.
2400\$
thanked the writer.

Harry has 1600 invested @ 6% and 2400 @ 7%

Explanation:

\$x invested at 6% pa and (4000-x)@ 7% pa
interest on each investment
6x/100 and 7(4000-x)/100 adds up to \$264
hence 6x/100 +(2800-7)x/100=264
hence 6x+2800-7x=26400

hencex=1600

thanked the writer. 