Another algebra word problem that I don't understand?

Question

please don't think that I am trying to get all of my hw done by getting answers on here. I had 11 problems to do and this is one of the ones I didn't understand

ok here is the problem:

Jack invests $1000 at a certain annual interest rate, and he invests another $2000 at an annual rate that is one-half percent higher. If he receives a total of $190 interest in one year, at what rate is the $1000 invested?

how do i solve this? thank u all

Answer 1

well 2000 * x = 190

thats to find out what percent of interest.

so x = 190/2000   so x= .095 or 9.5%

because the $1000 invested is .5% lower it is invested at 9% which is the answer.

Answer 2

let  assume that Jack invests  $ 1000 an annual rate of  x%.  So that the second amount is invested at (x + 1/2)%.

Now interest earned  on  his First investment ie $ 1000 =  ( 1000. x . 1 ) / 100  = $ 10.x .

Similarly Interest earned on  $ 2000 =  2000. (x + 0.5 ). 1 / 100 =     $ ( 20x  +  10 )

Hence the total Interest  (on both the investments ) earned  =      10x  +  20x  +  10  =  30x  +  10

But according to question –

30 x   +   10   =   190

=>    30x   =   180

=>    x  =  180/30  =  6.0  %  Ã¢Â€Â¦Ã¢Â€Â¦Ã¢Â€Â¦Ã¢Â€Â¦Ã¢Â€Â¦Ã¢Â€Â¦Ã¢Â€Â¦.  Answer

Dr. PKT

Answer 3

Jesse, the total interest is $190 for the $1000 and the $2000 investment.  The correct calculation would be 1000(x-.5) + 2000x = 190

SOlve for x

Answer 4

wow i dont know either i would just try to do others and ask the teacher on monday

Answer 5

  10*x + 20*(x+0.5) = 190

=> 30x -10 = 190

=> 30x = 200

=> x = 6.67%