SOLUTION: A broker invested $13,000 in two different stocks. One earned dividends at a rate of 8% and the other at 7%. If a dividend of $1370 was earned on both stocks, how much was invested

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Question 172780: A broker invested $13,000 in two different stocks. One earned dividends at a rate of 8% and the other at 7%. If a dividend of $1370 was earned on both stocks, how much was invested at each rate?
I set up the equation as follows:
Quantity(amount Invested)::
StockA = A
StockB = $13000 – A
Value(Interest rate):
StockA = 0.08
StockB = 0.07
Quantity * Value:
StockA = .08A
StockB = 0.07($13000 – A)= 910 -.07A [WHERE DID YOU GET THE 910 TERM?????]
Therefore:
0.08A + 910 – 0.07A = $1370
910 + 0.01A = $13000 = $1370
0.01A = $460
A = $46,000 ????
Here is where I am lost because the broker only had $13,000 to invest. $46,000 > $13,000. What’s wrong???

Found 2 solutions by vleith, checkley77:
Answer by vleith(2983)   (Show Source): You can put this solution on YOUR website!
I am guessing that there must be some missing info (like maybe investing for more than a single year).
It "doesn't work" because given the numbers you have - "you can't get there from here".
Let's say the broker put the entire amount at 8% for a year. In that case, the interest is 13000*08 = 1040. Which is less than the value the problem stated was returned.
The math does not lie. The reason 46000>13000 is because the math, as provided, doesn't "work".
When you get an answer that doesn't make sense, like 5 = 7 or somesuch, then that means "you can't do whatever was being suggested could be done".
The fact that you were aware enough to realize that the answer is not 46000 and -33000 shows you are on the right track mathematically. Good for you!

Answer by checkley77(12844)   (Show Source): You can put this solution on YOUR website!
.08x+.07(18,000-x)=1,370
.08x+1,260-.07x=1,370
.01x=1,370-1,260
.01x=110
x=110/.01
x=11,000 invested @ 8%.
18,000-11,000=7,000 invested @ 7%.
Proof:
.08*11,000+.07*7,000=1,370
880+490=1,370
1,370=1,370

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