Question

A brokerage house offers 3 stock portfolios. Portfolio I consists of 2 blocks of common stock and 1 municipal bond. Portfolio II consists of 4 blocks of common stock, 2 municipal bonds and 3 blocks of preferred stock. Portfolio III consists of 2 blocks of common stock, 2 municipal bonds and 3 blocks of preferred stock. A customer wants 12 blocks of common stock, 6 municipal bonds and 6 preferred stocks. How many portfolio III should be offered?

• A. 1
• B. 2
• C. 3
• D. None of the above

Hint:

Write one equation per stock type (common, municipal, preferred) in terms of the number of each portfolio, then eliminate variables to find how many Portfolio III units are needed.

Answer 1

The Correct Option is D

Instead of solving the full system in one go, plug candidate values of Portfolio III (z) into the three requirement equations and see which one actually works. The three needs are 12 common stocks, 6 municipal bonds and 6 preferred stocks, coming from Portfolio I (2 common, 1 bond, 0 preferred), Portfolio II (4 common, 2 bonds, 3 preferred) and Portfolio III (2 common, 2 bonds, 3 preferred).

The preferred stock only comes from Portfolios II and III: $3y + 3z = 6$, so $y = 2 - z$. This already tells us z can only be 0, 1 or 2 for y to stay non-negative, which rules out 3 and 4 right away as candidate values for z.

1. z = 1: then $y = 1$. Bonds needed: $x + 2(1) + 2(1) = 6 \Rightarrow x = 2$. Check common stock: $2(2) + 4(1) + 2(1) = 4+4+2 = 10 \neq 12$. Fails.
2. z = 2: then $y = 0$. Bonds: $x + 0 + 4 = 6 \Rightarrow x = 2$. Common stock: $2(2)+0+4 = 8 \neq 12$. Fails.
3. z = 3: already ruled out, since $y = 2-3 = -1$ would be negative, which is not allowed for a count of portfolios.

Since z = 1 and z = 2 both fail, and z = 3 (or higher) is impossible, try $z=0$ directly: $y = 2$, bonds give $x + 4 + 0 = 6 \Rightarrow x = 2$, and common stock checks as $2(2)+4(2)+2(0) = 4+8+0=12$, which matches exactly.

Let's summarize:

• The preferred stock equation pins $y = 2-z$ immediately, cutting the search down to $z=0,1,2$.
• Testing $z=1$ and $z=2$ against the common stock requirement breaks the equation, so the true value is $z=0$.

So Portfolio III is not offered at all, meaning the correct choice is "None of the above".