Problema Solution
A $200,000 retirement fund was split into two investments, one portion (x) at 10% annual interest and the rest (y) at 6%. If the annual interest is 18,000 how much was invested at each rate?
Answer provided by our tutors
The amount of interest earned each year on amount x would be 10% of x or 0.1x
The amount of interest earned each year on amount y would be 6% of y or 0.06y
The total interest is $18,000 so 0.1x + 0.06y = 18,000
The total invested is $200,000 so x + y = $200,000
So buy subtracting x from both sides of this equation you have y = 200,000 - x
So you can substitute 200,000 - x in the place of y in the first equation
0.1x + 0.06(200,000 - x) = 18,000 Now we mulitply to eliminate the ( ).
0.1x + 12,000 - 0.06x = 18,000 which simplifies to 0.04x + 12,000 = 18,000
Subtracting 12,000 from both sides of the equation shows that 0.04x = 6,000
When we divide 6,000 by 0.04 to get x alone, we find that x = $150,000.
$200,000 - $150,000 = $50,000 = y So $150,000 was invested at 10% and $50,000 was invested at 6%