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Came across this problem in my assignment... Not sure what to do.

Last year, Evelin's invested 10,000 dollars, part at 6% annual interest, and the rest at 8% annual interest. If she recieves $760 at the end of the year, how much did he invest at each rate?
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2 Answers

 X - The part invests in 6%
 
10,000 - X -- The rest invests in 8 %
 
   X ( 0.06) --gain in 6% investment
 
   (10,000 - X ) 0.08 -- gain on 8%
 
   Solve for X , if:
 
     X ( 0.06 ) + ( 10,000 - X ) ( 0.08) = 760
 
       X ( 0.06) - X ( 0.08 ) = -10,000( 0.08)  + 760
 
        X ( - 0.02 ) = - -800 + 760 = -40
 
         X = -40/ -0.02 = 2,000 / invested in 6% Plan
 
         10,000 - 2,000 = 8,000 / invested in 8% plan
This can be solved with a system of equations. Let x be the amount invested at 6% and let y be the amount invested at 8%.
 
You know that the total invested is 10,000. This yields one equation:
x + y = 10,000
 
You also know that the interest earned on x plus the interest earned on y equals $760. The interest earned on x is 0.06x and the interest earned on y is 0.08y. Now you have a second equation:
0.06x + 0.08y = 760
 
There are a variety of ways to solve this system. I will use elimination and back substitution. I will multiply the first equation by -0.06, then when I add that to the second equation I will eliminate the x variable.
 
-0.06x - 0.06y = -600
0.06x + 0.08y = 760  
             .02y = 160
                  y = 8000
 
Now I will substitute 8000 for y in the original first equation and solve for x.
x + 8000 = 10,000
x = 2000
 
Evelin invested $2000 at 6% and $8000 at 8%.