MATH word problem help?

Question

An investor plans to invest 1 million dollars to provide an income of 54,000 per year. There are two possible investments available. One pays 4% per year, the other pays 6% per year. How much should be invested at each rate to exactly total 54, 000 per year?

Answer 1

i wasn't quick enough and the question was answered but i'll still tell you my answer.

it's a simple case of simultaneous equation,the answer is as follows

you should invest 300,000 dollars of the one million on the 4% per annum (brit style) interest rate and the rest 700,000 on the 6% per annum interest rate.

i'm wondering if it's you who's investing such a principal,if yes fantastic!,if not

God bless

and finally thanks for asking a real question but don't ask every home work question you get, that's just cheating

Answer 2

You should NOT be asking everyone to answer your question for you.  You get those answers to do them yourself to learn from your mistakes

God Bless

Answer 3

This is a place where you have 2 equations and 2 unknowns which is a solvable system.

X & Y are your two investment options

X+Y=1,000,000

0.04X * 0.06Y=54,000

Now solve the two equations and that is your solution.

Answer 4

let X be the amount invested with 4% rate.

let Y be the amount invested with 6% rate

X + Y = 1000000

X = 1000000 - Y

.04X + .06Y = 54000

substitute X

.04(1000000 - Y) + .06Y = 54000

40000 - .04Y + .06Y = 54000

.02Y = 54000 - 40000

.02Y = 14000

Y = 14000 / .02

Y = 700000

X = 1000000 - 700000

X = 300000

proof:

300000 x 4% = 12000

700000 x 6% = 42000

12000 + 42000 = 54000

Answer 5

It sounds time consuming but what you do is you multiply .06 or .04 by random numbers under 1000000 until you get 54000

Answer 6

investing x $ in first and y $ in second one

x+y= 1000000

0.04x + 0.06y = 54000

..........................

0.01x + 0.01y = 10000

0.01x + 0.015y = 13500

-----------------------------

0.005y = 3500

y= 700000

x=300000

Answer 7

i dont know