SOLUTION: sheryl won $60000 on slot machine in oxford. She invested part of the money at 2% and the rest at 3%. in one year she earned a total of $1600 in interest. How much was invested at

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Question 1018957: sheryl won $60000 on slot machine in oxford. She invested part of the money at 2% and the rest at 3%. in one year she earned a total of $1600 in interest. How much was invested at each rate?
Found 2 solutions by Marz157, MathTherapy:
Answer by Marz157(7) (Show Source):
You can put this solution on YOUR website!
Lets call the percent of the money invested in the 2% account "x" which means the percent invested in the 3% account is "1-x" (this makes sense because if we add together x and 1-x, we just get 1 meaning the total 100% of the money).
Depending on different values of p, we get different amounts of money at the end. If we find the value of p that makes $1600 we solve the problem.
If we invest $A at a interest rate of B per year, we make A*B money in interest.
So, the formula for the money we make in interest is

Solving this for x can be done like this
Dividing by 60000
Expand (1-x) * .03 by multiplying both by .03
Add the x terms together and subract the .03 from the other side.
divide both sides by -.01
So 33.33% (one third) of the money was invested in the 2% interest account. To double check, if we plug this number in for x in the first equation we should see that we get $1600.

Answer by MathTherapy(10551) (Show Source):
You can put this solution on YOUR website!

sheryl won $60000 on slot machine in oxford. She invested part of the money at 2% and the rest at 3%. in one year she earned a total of $1600 in interest. How much was invested at each rate?

Let amount invested at 2%, be T
Then amount invested at 3% is: 60,000 - T
We then get: .02T + .03(60,000 - T) = 1,600
.02T + 1,800 - .03T = 1,600
.02T - .03T = 1,600 - 1,800
- .01T = - 200
T, or amount invested at 2% = , or
Amount invested at 3% = $60,000 - $20,000, or