SOLUTION: Question content area top Part 1 A person invested ​$6900 for 1​ year, part at 8​%, part at 11​%, and the remainder at 14​%. The total annual income from these investme

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Question content area top Part 1 A person invested ​$6900 for 1​ year, part at 8​%, part at 11​%, and the remainder at 14​%. The total annual income from these investme      Log On

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Question 1207173: Question content area top
Part 1
A person invested ​$6900 for 1​ year, part at 8​%, part at 11​%, and the remainder at 14​%. The total annual income from these investments was ​$834. The amount of money invested at 14% was ​$900 more than the amounts invested at 8​% and 11​% combined. Find the amount invested at each rate.
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Try tabulating the information and form your needed equations.
PART AT RATE         AMOUNT          INTEREST IN YEAR
       8%             x               0.08x         
       11%            y              0.11y
       14%           x+y+900         0.14(x+y+900)                
TOTAL                6900                   834

system%282x%2B2y%2B900=6900%2C0.08x%2B0.11y%2B0.14x%2B0.14y%2B%280.14%29%28900%29=834%29

system%28x%2By=3000%2C0.22x%2B0.25y=834-%280.14%29%28900%29%29

system%28x%2By=3000%2C0.22x%2B0.25y=708%29

0.22%283000-y%29%2B0.25y=708

660-0.22y%2B0.25y=708

0.03y=708-660

highlight%28y=1600%29--------for 11% rate

3000-1600=highlight%28x=1400%29-------for 8% rate

highlight%283900%29----------for 14% rate

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There are always different ways to set up a problem for solving.

The method shown by the other tutor basically translates the given information directly into equations that are then solved to find the answer.

I think a very different approach, analyzing the given information to find a way to set up the problem that makes solving the problem easier, is better.

Given that the total invested was $6900 and that the amount invested at 14% was $900 more than the combined amounts invested at 8% and 11%, we can quickly determine that $3900 was invested at 14% and $3000 combined at 8% and 11%. I leave the details for finding that result to the student, if formal algebra is required.

$3900 invested at 14% yields $546 income; that leaves $834-$546 = $288 income from the other two investments.

A standard formal solution for that part of the problem would start something like this:

x = amount invested at 8%
3000-x = amount invested at 11%

The total income from those two investments was $288:

.08(x)+.11(3000-x)=288

Solving that equation then finishes the problem.

I use a different, less formal method for solving 2-part "mixture" problems like this. Here is my method for solving this part of the problem.

All $3000 invested at 8% would yield $240 income; all invested at 11% would yield $330 income; the actual income from those two investments was $288.
Look at those three numbers 240, 288, and 330 (on a number line, if it helps), and observe/calculate that 288 is 48/90 = 8/15 of the way from 240 to 330.
That means 8/15 of the $3000 was invested at the higher rate.

8/15 of $3000 is 8*$200 = $1600, so $1600 was invested at 11% and $1400 at 8%.

ANSWERS: $3900 at 14%, $1600 at 11%, and $1400 at 8%

CHECK: .14(3900)+.11(1600)+.08(1400) = 546+176+112 = 834