amt invested 25,000 @ 6% for 1 year and 25,000 @ 9% for 1 year, how much was invested @ each rate if the total annual percentage from both investments was 2070?

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Answer 1

i think what the question asks for is as follows..

 

if a total amount of 25000 is invested at two different rates, 6% and 9%. the sum of the interest at these rates after one year is 2070. how much was invested at 6% , and how much was invested at 9%?

 

if this is the question , we can form two equations.

 

say 25000= X+Y, where X is invested at 6% and Y is invested at 9%------------(equation 1)

 

then we have the data from the question

 

Interest at 6% rate on amount X= 0.06X

 

Interest at 9% rate on amount Y= 0.09X

 

0.06*X + 0.09*Y = 2070------------------(equation 2)

 

Solving these two equations, can yeild the value for X and Y as follows:

X=6000

Y=19000

Hope this helps.

 

Answer 2

there are two investments, x and y, that total $25,000......that is, x + y = 25,000

let x be the amount invested at 6% and let y be the amount invested at 9%, the total annual percentage from these two investments equals $2,070......that is, 0.06x + 0.09y = 2,070

notice that you must convert from percent form to decimal form because a percentage is out of 100 parts, (for example, 6 out of 100 is 6% so 6%/100% = 6/100 = 0.06)

thus, the system contains the following 2 equations:

   (1.)             x +        y = 25,000

   (2.)       0.06x + 0.09y = 2,070

first, you want to eliminate one of the variables and solve for the other, then plug in the number for the variable you solved for to find the other variable...

let's solve for x first, so we want to eliminate y to do that.....multiply equation (1.) by -0.09 then add it to equation (2.):

(1.)       x + y = 25,000   ===>   -0.09 (x + y = 25,000)   ===>   -0.09x - 0.09y = -2,250

(2.)                                                                                            0.06x + 0.09y = 2,070

                                                                                           ___________________________

                                                                                                 -0.03x + 0y = -180

so,     -0.03x = -180    (divide both sides by -0.03) ===>  (-0.03/-0.03)x = -180/-0.03 ===> x = 6,000

plug x back into one of the main equations (the first one would be the easiest) and solve for y:

   x + y = 25,000    ===>     6,000 + y = 25,000      (subtract 6,000 from both sides) 

                                         -6,000          -6,000

                                    _____________________

                                               0  +  y  = 19,000     ===>    y = 19,000

so, the amount invested at 6% (x) was $6,000 and the amount invested at 9% (y) was $19,000

Plug these #s back into equation (2.) to check:

    0.06x + 0.09y = 2,070    ===>    0.06(6,000) + 0.09(19,000) = 360 + 1,710 = 2,070

Answer 3

Hi Annette, I’m thinking you may have copied the problem wrong here, because it doesn’t quite make sense. You say that $25,000 is invested at each amount, but then you ask how much was invested at each amount. Also, I’m thinking that perhaps the word “percentage” might be “interest” in the original problem? Would you mind checking how this problem was worded? In general, Interest = Principal x Rate x Time so, for example $25,000 at 6% interest for 1 year would yield 25,000 x .06 x 1 = $1500 in interest. I’m not sure how to relate this to your question until we get the above clarified.