Question 498994: Things did not quite go as planned. You invested $20,000, part of it in a stock that paid 12% annual interest. However, the rest of the money suffered a 5% loss. If the total annual income from both investments was $1890, how much was invested at each rate?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = amount in stock.
20,000 - x = the rest of it.
interest earned on x is .12x
money lost on 20,000 - x is .05(20,000-x)
total income from both investments is $1,890.
your income equation is:
.12x - .05(20,000 - x) = 1890
simplify to get:
.12x - .05*20,000 + .05x = 1890
combine like terms and simplify further to get:
.17x - 1000 = 1890
add 1000 to both sides of the equation to get:
.17x = 2890
divide both sides of the equation by .17 to get:
x = 2890/.17 = 17000.00
that's the amount that was invested earning 12%.
the income from that amount is equal to .12*17000 = 2040.
that least 3000 that was invested elsewhere.
the money that was lost was equal to 5% of that which made it equal to 150.
2040 - 150 = equals a net gain of 1890.
the value for x is confirmed as being good because that value, when substituted into the original problem statement showed that statement to be true.
|
|
|
