Question 1094799
<br>Here is an alternative to the standard algebraic method for solving mixture problems that is used in the references given to you by the first tutor who responded to your question.<br>
Take a look at this and see if you understand it; if you do, it will get you to the answer with much less work than the traditional algebraic method.<br>
Consider the amounts of interest A and B that would have been earned if all of the $18000 were loaned at each of the two rates.  Where the actual interest of $2000 lies between those two numbers A and B exactly determines the ratio in which the money needs to be split between the two loans.<br>
For example, if the actual amount of interest is halfway between A and B, then the money needs to be evenly split between the two loans.  Or if the amount of interest is "twice as close" to A as it is to B, then the loan at 6% must be twice the amount at 14%.<br>
Here is what the calculations look like, with only minimal comments:<br>
{{{18000*0.06 = 1080}}}  A: total interest if all $18000 loaned at 6%
{{{18000*0.14 = 2520}}}  B: total interest if all $18000 loaned at 14%<br>
{{{2000-1080 = 920}}}    difference between A and the actual interest
{{{2520-2000 = 520}}}    difference between B and the actual interest<br>
{{{920:520 = 92:52 = 23:13}}}   ratio in which the loan money needs to be split<br>
The ratio in which the loan money needs to be split is 23:13; that means one loan amount is 23/36 of the total and the other is 13/36 of the total.  Since the actual interest is closer to B than to A, the larger portion needs to be the loan at 14%.<br>
{{{(23/36)*18000 = 11500}}}  loan amount at 14%:
{{{(13/36)*18000 = 6500}}}  loan amount at 6%: