Question 104079: Janet invested $29,000, part at 12% and part at 14%. If the total interest at the end of the year is $3,800, how much did she invest at 12%?
I am having a terrible time setting up this problem...
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! First assign a variable to the unknown.
Let x = the amount that Janet invested at 12% per annum. Then ($29,000-x), the remainder, is the amount Janet invested at at 14% per annum.
We can express the interest earned on these two amounts as follows:
First, we'll need to change the percentages to their decimal equivalents: (12% = 0.12 and 14% = 0.14)
The amount of interest earned by (x) at 12% per annum can be expressed algebraically as:
0.12x
and the amount of interest earned by ($29,000-x) at 14% per annum can be expressed algebraically as: 0.14($29,000-x)
Now the problem tells us that the sum of these two = $3,800
So we can set up the necessary equation to solve for x.
0.12x + 0.14($29,000-x) = $3,800 This is the equation you will need to compute x, the amount invested at 12% per annum. Let's solve this for x.
0.12x + 0.14($29,000-x) = $3,800 Simplifying this, we get:
0.12x + $4,060 - 0.14x = $3,800 Combining like-terms (the x-terms), gives us:
-0.02x + $4,060 = $3,800 Now subtract $4,060 from both sides of the equation.
-0.02x = -$260 Dividing both sides by -0.02, we have:
x = $13,000 This is the amount Janet invested at 12% per annum.
If you need to know the amount she invested at 14%, you would subtract $13,000 from $29,000 to get $16,000 invested at 14% per annum.
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