Question 1205484: Ethan invested $19,000 for 1 year, part at 11% simple interest and part at 13% simple interest. If he earned a total of $2,290 in interest, how much was invested at each rate?
Found 4 solutions by ikleyn, mananth, Edwin McCravy, greenestamps:
Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website! .
Ethan invested $19,000 for 1 year, part at 11% simple interest and part at 13% simple interest.
If he earned a total of $2,290 in interest, how much was invested at each rate?
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Let x be the investment at 13%.
Then the investment at 11% is (19000-x) dollars.
Write the total annual interest equation
0.13x + 0.11*(19000-x) = 2290 dollars.
Simplify this equation and find x
0.13x + 0.11*19000 - 0.11x = 2290
0.13x - 0.11x = 2290 - 0.11*19000
0.02x = 200
x = 200/0.02 = 10000.
ANSWER. $10,000 invested at 13% and the rest, $19000 - $10000 = $9000 invested at 11%.
CHECK. 0.13*10000 + 0.11*9000 = 2290 dollars, the total annual interest. ! correct !
Solved.
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To see many other similar solved problems using one equation in one unknown, look into the lesson
- Typical investment problems
in this site.
To see many other similar solved problems using systems of two equations in two unknowns, look into the lesson
- Using systems of equations to solve problems on investment
in this site.
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! Ethan invested $19,000 for 1 year, part at 11% simple interest and part at 13% simple interest. If he earned a total of $2,290 in interest, how much was invested at each rate?
let the amount invested at 11% a be x
and the amount invested at 13% awill be 19,000 -x
−
The formula for simple interest is I=prt
The interest earned from the amount invested at 11%
is given by x*0.11*1 = 0.11x
the amount invested at 13% =19000-x
interest earned is given by
(19,000−x)*0.13*1
2470-0.13x
total interest =2290
0.11x+2470+0.13x = 2290
-0.02x = 2290-2470
-0.02x =-180
x = 9000
19000-x =10000
Ethan invested $9,000 at 11% and $10,000 (since $ at 13%.
Answer by Edwin McCravy(20064)  (Show Source):
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
Here is a solution by a method completely different from the standard algebraic solutions shown by the other tutors. This method can be use to solve any 2-part "mixture" problem like this one.
All $19,000 invested at 11% would yield $2090 interest; all invested at 13% would yield $2470 interest.
The amounts invested at the two rates to yield $2290 in interest is exactly determined by where the actual interest of $2290 lies between $2090 and $2470.
Use a number line, if it helps, to observe/calculate that $2290 is $2000/$3800 = 20/38 = 10/19 of the way from $2090 to $2470. That means 10/19 of the total was invested at the higher rate.
10/19 of $19,000 is, trivially, $10,000.
ANSWER: $10,000 was invested at 13%, the other $9000 at 11%
CHECK: .13(10000)+.11(9000) = 1300+990 = 2290
For me, at least, that solution is much easier than a formal algebraic solution using either one or two variables.
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