Question 1205918: 2000 shares of stocks and bonds sold for $200,000. Bonds were sold for $400 each. Preferred stock sold at 2 for $100. Common stock (employees only) sold at 4 for $100. Determine number of common stock sold.
Not sure how to solve.
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13196)   (Show Source):
You can put this solution on YOUR website!
Required information is missing from your post.
With only the given information, it is impossible to find the answer. There are many possible combinations of bonds, preferred stock, and common stock that give a total of 2000 shares at a total cost of $200,000.
Here is what we can do with the problem as stated....
x = # bonds
y = # shares of preferred stock
z = # shares of common stock
The given conditions give us these two equations:
(1)  the total number of shares was 2000
(2)  the total cost was $200,000
With three variables and only two equations, we can only find a family of solutions. The number of solutions is limited by the fact that x, y, and z must be whole numbers.
Simplify (2) (divide everything by 25):
(3) 
Use (1) and (3) to eliminate z; then solve the resulting equation for one of the variables:
(4) 
That gives us an expression for y in terms of x. Now use (1) and (4) to find an expression for z in terms of x:
Now we have expressions for two of the variables in terms of the third, so we can define parametric equations to find different solutions to the problem.
The expression for y in terms of x tells us that any whole number value of x will yield an integer number value for y; and since y must be a whole number (not a negative integer), the expression also tells us the maximum value for x is 400.
The expression for z will not yield a whole number value for any whole number value for x. To find all the whole number solutions to the problem, we could do some further mathematics; I won't go that path, since we already know we can't solve the problem as posted. Instead we can find a few possible answers to the problem by trying different values for x and finding ones that yield whole number values for both y and z.
x y=15(400-x) z=14x-4000
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400 15(0)=0 5600-4000=1600
300 15(100)=1500 4200-4000=200
350 15(50)=750 4900-4000=900
There are 3 combinations that satisfy the given conditions; there are others.
To find a single solution, we would have to have another piece of information. For example, seeing the second possible solution in the list above, we could find that single solution if the given information said either (a) the number of shares of preferred stock was 5 times the number of bonds, or (b) the number of bonds was 1.5 times the number of shares of common stock.
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The student replied in a "thank you" note to me that the answer was 200 shares of common stock, with 1500 shares of preferred stock and 300 bonds. He also stated that he didn't know where they cam up with 200 shares of common stock.
As I showed in my response, that is one of many possible solutions; it is in fact one of the specific solutions I showed.
The fact remains that, with the problem as presented in the student's post, a single solution to the problem is not possible.
Perhaps the student will see this note and look at the given information again, finding that a required piece of information was not provided in his post.
Answer by MathTherapy(10549)  (Show Source):
You can put this solution on YOUR website!
2000 shares of stocks and bonds sold for $200,000. Bonds were sold for $400 each. Preferred stock sold at 2 for $100. Common stock (employees only) sold at 4 for $100. Determine number of common stock sold.
Not sure how to solve.
I found that the least number of bonds sold would've been greater than 285, and the greatest number would've been
400. I also found that more than 400 (401) bonds could not have been sold, as that number would've resulted in 1,614
shares of common stock and - 15 shares of preferred stock. In addition, the sale of 285 bonds would also yield
nonsensical results. So, the number of bonds that could've been sold would've been between 286 and 400, inclusive, a
TOTAL of 115 different possibilities. However, because bonds come in $50 increments/denominations, the least and
greatest amount that could've been sold would've been 300 and 400, respectively.
So, we now have 3 $50-increment possibilities for bonds, from 300 - 400, as follows:
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|Instrument | No. | No. | No. |
|BONDS | 300| 350| 400|
|Shares of COMMON STOCK | 200| 900| 0|
|Shares of PREFERRED STOCK |1,500| 750|1,600|
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|Total |2,000|2,000|2,000|
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Now, since 400 bonds lead to 0 common stocks being sold - and it's stated that common and preferred were sold - I
would rule out the 400-bond, 1,600 preferred-stock sale.
This now leaves the following 2 possibilities:
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|Instrument | No. | No. |
|BONDS | 300| 350|
|Shares of COMMON STOCK | 200| 900|
|Shares of PREFERRED STOCK |1,500| 750|
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|Total |2,000|2,000|
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You now have 2 LEGIT possibilities, in my opinion.
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