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Question

Tom invested $11,000, part at 9% and part at 10% simple interest. How much money was invested at each interest rate if the total interest earned from both is $1050?

Answer 1

Call the part at 9% P9 and the part at 10%P10.
The total is P9+P10 = 11000

THe interest is .10P10 + .09P9 = 1050.

Now you have two equations and can solve for the two variables.

Multiply both sides of that second equation by 100 to get rid of the decimal points and you have 10P10 + 9P9 = 105000

Get rid of P9 by subtracting it from both sides of the first equation so you get P9 = 11000 - P10.

Substitute that value for P9 into the big equation. You get
10P10 + 9(11000-P10) = 105000

10P10+99000-9P10 = 105000
P10 = 105000 - 99000 so P10 = 6000

Going back to the first equation with P9 and P10 subtract that known value of P10 from the total 11000 invested and get 11000-6000 = 5000 for P9

The two answers are $6000 invested at 10% and 5000 invested at 9%.

Answer 2

We have two sums of money which means we need to have two equations. We'll call the first sum - the one at 9% - x, which means the second sum - the one at 10% - can be called y. So the two things we know are this:

1) The first sum (x) plus the second sum (y) is equal to the total he has to invest ($11,000). Thus, our first equation is:

x + y = 11000

2) The interest paid to the first sum (9% times x, or .09x) plus the interest paid to the second sum (10% times y, or .10y) equals the total interest paid out (which is $1,050). Thus:

.09x + .10y = 1050

So given the two equations, we can use substitution to solve:

x + y = 11000
.09x + .10y = 1050

We want to solve for one of our variables. The easier way to do this is to use the first equation:

x + y = 11000, which means that
x = 11000 - y.

Now we can plug the new x equivalent into the second equation.

.09x + .10y = 1050
.09(11000 - y) + .10y = 1050
990 - .09y + .10y = 1050
990 + .01y = 1050
.01y = 60
y = 6000.

Therefore, he invested $6,000 at 10%. Since the total invested was $11,000, that means the other $5,000 was invested at 9%.

Does that make sense?

Answer 3

9%=0.09 and 10% =0.1%

lets x fraction invested at 9%

so 11000*x*0.09 +11000*0.1 *(1-x) =1050
990x [phone]x =1050
-110x= -50
11x=5
x=5/11
(5/11)*11000=5000

so tom invested 5000$ at 9% and 6000$ at 10%