Question

At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added \(p\%\) of the goats at the beginning of the year and sold \(q\%\) of the goats at the end of the year where \(p > 0\) and \(q > 0\). If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?

• A. \(p = q\)
• B. \(p < q\)
• C. \(p > q\)
• D. \(p = q/2\)

Hint:

Each year the herd is multiplied by the same factor. For the herd to be unchanged after four years, think about what that factor must be.

Answer 1

The Correct Option is C

Step 1: Understand the yearly change:
Nine dozen goats is \(9 \times 12 = 108\) goats. This is the stock at the end of 1998, after that year's sales.
Each later year has two moves in a fixed order. First \(p\%\) of the goats at the start of the year are added. Then \(q\%\) of the goats at the end of the year are sold.
The years 1999, 2000, 2001 and 2002 are four full years, so the two moves repeat four times.

Step 2: Write the yearly multiplier:
Adding \(p\%\) multiplies the stock by \(\left(1 + \frac{p}{100}\right)\). Selling \(q\%\) of what is left multiplies it by \(\left(1 - \frac{q}{100}\right)\).
Let \(f = \left(1 + \frac{p}{100}\right)\left(1 - \frac{q}{100}\right)\). Then one full year changes the stock from \(N\) to \(fN\).

Step 3: Use the stock at the end of 2002:
After four years the stock is \(108 f^{4}\). The question says this is again 108.
\[ 108 f^{4} = 108 \Rightarrow f^{4} = 1 \]
Since \(f\) is a positive number, the only possible value is \(f = 1\).

Step 4: Turn f = 1 into a relation between p and q:
\[ \left(1 + \frac{p}{100}\right)\left(1 - \frac{q}{100}\right) = 1 \]
\[ 1 + \frac{p}{100} = \frac{100}{100 - q} \]
\[ \frac{p}{100} = \frac{100}{100 - q} - 1 = \frac{q}{100 - q} \]
\[ p = \frac{100 q}{100 - q} \]

Step 5: Compare p with q:
Because \(q > 0\), the number \(100 - q\) is less than 100 (and it must be positive, since we cannot sell more than 100 percent). So \(\frac{100}{100 - q} > 1\).
Multiply both sides by \(q\): \(p = q \cdot \frac{100}{100 - q} > q\).

Step 6: Check the options:
Option 1, \(p = q\), would give \(f = 1 - \frac{q^{2}}{10000} < 1\), so the herd would shrink. It is false.
Option 2, \(p < q\), makes \(f\) even smaller than 1, so the herd shrinks. It is false.
Option 4, \(p = q/2\), also gives \(f < 1\). It is false.
Only option 3, \(p > q\), matches \(p = \frac{100q}{100 - q}\).

Final Answer:
Since the stock returns to 108 after four years, the yearly multiplier is 1, which forces \(p > q\). This is option 3. \[ \boxed{\text{Option 3: } p > q} \]