Rellie invests 12,000 pesos. A part of it earns 10% interest annually and the rest earns 9% interest annually?

Question

Rellie invests 12,000 pesos. A part of it earns 10% interest annually and the rest earns 9% interest annually. His annual income from his investments is 1,130 pesos. How much is invested at each rate?

Answer 1

Let x be the number of pesos invested at 10%.
Then the number at 9% is 12000 - x.
The interest is:
0.10x + 0.09(12000 - x) = 1130
0.10x + 1080 - 0.09x = 1130
0.01x = 50
x = 5000
He invested 5000 pesos at 10% and 7000 at 9%.

Answer 2

Let x and y be the amounts invested at10% and 9% respectively.

10x/100 + 9y/100 = 1130

x + y = 12000

10x + 9y = 113000

10x + 10y = 120000

y = 7000 (Subtract the upper equation from the lower one)

x = 5000 pesos

Answer 3

let x be the amount invested at 10%

1130 = x*10% + (12000-x)*9%

solve for x. the answer is:

5000 pesos at 10% interest
7000 pesos at 9%

Answer 4

Thi depends, are you taking into account compound interest and taxation, as this would affect the result.

And are you offsetting the 10% against the 9%, or vice versa, as this would affect the final result.

Answer 5

6000 each