SOLUTION: susan invest 2 times as much money at 9% as she does at 6%. If her total interest after 1 year is $1440, how much does she invest at each rate?
Question 1116363: susan invest 2 times as much money at 9% as she does at 6%. If her total interest after 1 year is $1440, how much does she invest at each rate? Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52770) (Show Source):
Let x be an amount invested at 6% (the lesser amount).
Then the amount invested at 9% is 2x, according to the condition.
Your equation to find x is
interest + interest = total interest, or
0.06x + 0.09*(2x) = 1440
0.06x + 0.18x = 1440 ====> 0.24x = 1440 ====> x = = 6000.
Answer. $6000 was invested at 6%. $12000 was invested at 9%.
Check. 0.06*6000 + 0.09*12000 = 1440 dollars. ! Correct !
Solved.
--------------------
It is a typical and standard problem on investment.
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
...or here is a way to solve the problem with logical analysis instead of formal algebra
The amount invested at 9% is 2 times the amount invested at 6%; and the 9% interest rate is 1.5 times the 6% interest rate. Together those two things mean the interest earned at 9% is 2*1.5 = 3 times the interest earned at 6%.
So 1/4 of the interest earned is from the 6% investment and 3/4 is from the 9% investment.
1/4 of the total interest of $1440 is $360, so the amount of interest from the 6% investment is $360.
And then the amount invested at 6% is $360/.06 = $6000.
Then the amount invested at 9% is 2*$6000 = $12000.
