In a sale promotional campaign by Assamadiya frozen foods on 02/01/2025,80% of first new year Monday (02/1/2015),slaughtered cows is given out to customer as free beef prizes based on on previous years patronage. There are free beef prizes altogether, the first prize winner gets 10,950 kgs of beef,the second gets 3,645 kgs the third 1,215 kgs as so on. how much beef(kgs) does the eight person get? how much free prizes is there altogether how much beef (kgs) was realized on first new year Monday (]2/01/2025) Assuming each kilogram of beef cost 800 naira,what is the total value of beef realized from the slaughtered cows on the new year? if the inflationary trend increasing by 2% monthly,what would have been the total value of beef realized from the slaughtered cows as at 01/12/2025?

The prizes form a geometric progression.
The first term is $a_1 = 10,950 \text{ kgs}$.
The common ratio $r$ is found by dividing the second term by the first term:

$$r = \frac{3,645}{10,950} = \frac{1}{3}$$

The formula for the $n$-th term of a geometric progression is $a_n = a_1 \cdot r^{n-1}$.
For the eighth person ($n=8$):

$$a_8 = 10,950 \cdot \left(\frac{1}{3}\right)^{8-1}$$ $$a_8 = 10,950 \cdot \left(\frac{1}{3}\right)^7$$ $$a_8 = 10,950 \cdot \frac{1}{2187}$$ $$a_8 = \frac{10,950}{2187}$$ $$a_8 = 5 \text{ kgs}$$

The eighth person gets $\boxed{\text{5 kgs}}$ of beef.

We need to find the number of terms $n$ for which the prize amount $a_n$ is at least 1 kg (a reasonable minimum for a "prize").

$$a_n = 10,950 \cdot \left(\frac{1}{3}\right)^{n-1} \ge 1$$ $$\left(\frac{1}{3}\right)^{n-1} \ge \frac{1}{10,950}$$ $$3^{n-1} \le 10,950$$

To solve for $n-1$, we take the logarithm base 3 of both sides:

$$n-1 \le \log_3(10,950)$$

Using the change of base formula, $\log_3(10,950) = \frac{\ln(10,950)}{\ln(3)}$:

$$n-1 \le \frac{9.301}{1.0986} \approx 8.466$$ $$n \le 9.466$$

Since $n$ must be an integer, the maximum number of prizes is 9.
There are $\boxed{\text{9}}$ free prizes altogether.

First, calculate the total amount of beef given as prizes ($S_9$), which is the sum of the first 9 terms of the geometric progression.
The formula for the sum of a geometric series is $S_n = \frac{a_1(1-r^n)}{1-r}$.

$$S_9 = \frac{10,950 \left(1 - \left(\frac{1}{3}\right)^9\right)}{1 - \frac{1}{3}}$$ $$S_9 = \frac{10,950 \left(1 - \frac{1}{19683}\right)}{\frac{2}{3}}$$ $$S_9 = \frac{10,950 \left(\frac{19682}{19683}\right)}{\frac{2}{3}}$$ $$S_9 = 10,950 \cdot \frac{19682}{19683} \cdot \frac{3}{2}$$ $$S_9 = 10,950 \cdot \frac{9841}{6561}$$ $$S_9 = 16,425 \text{ kgs}$$

This amount ($16,425 \text{ kgs}$) represents 80% of the total beef realized from slaughtered cows. Let $X$ be the total beef realized.

$$0.80 \cdot X = 16,425 \text{ kgs}$$ $$X = \frac{16,425}{0.80}$$ $$X = 20,531.25 \text{ kgs}$$

The total beef realized on 02/01/2025 was $\boxed{\text{20,531.25 kgs}}$.

The total beef realized is $20,531.25 \text{ kgs}$ (from Step 3).
The cost per kilogram is 800 naira.
Total value = Total beef $\times$ Cost per kg

$$\text{Total Value} = 20,531.25 \text{ kgs} \times 800 \text{ naira/kg}$$ $$\text{Total Value} = 16,425,000 \text{ naira}$$

The total value of beef realized was $\boxed{\text{16,425,000 naira}}$.

The initial value (PV) on 02/01/2025 is $16,425,000 \text{ naira}$ (from Step 4).
The monthly inflation rate is $r = 2\% = 0.02$.
The number of months ($n$) from 02/01/2025 to 01/12/2025 is 11 months (February, March, April, May, June, July, August, September, October, November, December).
The future value (FV) with compound inflation is given by $FV = PV \cdot (1 + r)^n$.

$$FV = 16,425,000 \cdot (1 + 0.02)^{11}$$ $$FV = 16,425,000 \cdot (1.02)^{11}$$ $$FV = 16,425,000 \cdot 1.24337385...$$ $$FV \approx 20,429,999.25 \text{ naira}$$

The total value of beef realized as at 01/12/2025 would have been $\boxed{\text{20,429,999.25 naira}}$.