Question

James invested 20,000 for one year and earned 1470 interest. If part of the money is invested at 10% and the remainder is invested at 6% how much is the invested at each rate

Linear equation.

Answer 1

Let's represent the amount invested at 10% as x and the amount invested at 6% as y. Then we can set up a system of two equations to represent the given information:

x + y = 20,000 (since the total amount invested is 20,000)

0.10x + 0.06y = 1,470 (since the interest earned is 1,470 and the interest rate at which x is invested is 10% and the interest rate at which y is invested is 6%)

We can use the first equation to solve for one of the variables in terms of the other:

x = 20,000 - y

Now we can substitute this expression for x into the second equation and solve for y:

0.10(20,000 - y) + 0.06y = 1,470

2,000 - 0.10y + 0.06y = 1,470

-0.04y = -530

y = 13,250

So $13,250 was invested at 6%. We can find the amount invested at 10% by plugging in this value of y into the first equation:

x + 13,250 = 20,000

x = 6,750

So $6,750 was invested at 10%.