A poultry farmer has been keeping track of the number of chickens at his farm over previous years. Four years ago, the number of chickens was $64$ . The number increased by $50\%$ each year for four years until the present day. In the future, the farmer considers the situation where the number of chickens increases at a constant rate equal to the rate of the most recent year or the situation where the number continues to increase by $50\%$ per year. What is the difference in the number of chickens between these two situations two years from now? $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $189$ $\newline$ (C) $405$ $\newline$ (D) $11$

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Grade 6

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Q. A poultry farmer has been keeping track of the number of chickens at his farm over previous years. Four years ago, the number of chickens was

64

. The number increased by

50\%

each year for four years until the present day. In the future, the farmer considers the situation where the number of chickens increases at a constant rate equal to the rate of the most recent year or the situation where the number continues to increase by

50\%

per year. What is the difference in the number of chickens between these two situations two years from now?

\newline

Choose

1

answer:

\newline

(A)

0

\newline

(B)

189

\newline

(C)

405

\newline

(D)

11

Calculate Chickens After $4$ Years: First, let's calculate the number of chickens at the end of the four-year period with a $50\%$ increase each year. $\newline$ Initial number of chickens: $64$ $\newline$ Yearly increase: $50\%$ $\newline$ We will use the formula for compound interest to calculate the number of chickens after four years, which is similar to calculating the future value of an investment. $\newline$ Number of chickens after $n$ years = Initial number $\times (1 + \text{rate})^n$
Calculate Chickens After $2$ Years: Now, let's calculate the number of chickens after four years. $\newline$ Number of chickens after $4$ years $= 64 \times (1 + 0.50)^4$ $\newline$ $= 64 \times (1.50)^4$ $\newline$ $= 64 \times 5.0625$ $\newline$ $= 324$ (rounded to the nearest whole number)
Constant Increase Calculation: Next, we need to calculate the number of chickens two years from now under the first scenario where the number of chickens increases at a constant rate equal to the rate of the most recent year. $\newline$ The rate of the most recent year is $50\%$ of the number of chickens at the end of the fourth year. $\newline$ Constant increase = $50\%$ of $324$ $\newline$ = $0.50 \times 324$ $\newline$ = $162$ $\newline$ So, the number of chickens will increase by $162$ each year for the next two years.
Calculate Chickens After $2$ More Years: Now, let's calculate the number of chickens after two more years with the constant increase. $\newline$ Number of chickens after $2$ more years (constant increase) = $324 + 162 \times 2$ $\newline$ = $324 + 324$ $\newline$ = $648$
$50$ % Increase Calculation: For the second scenario, we need to calculate the number of chickens two years from now if the number continues to increase by $50$ percent per year. $\newline$ Number of chickens after $1$ more year ( $50$ % increase) = $324 \times (1 + 0.50)$ $\newline$ = $324 \times 1.50$ $\newline$ = $486$ $\newline$ Number of chickens after $2$ more years ( $50$ % increase) = $486 \times (1 + 0.50)$ $\newline$ = $486 \times 1.50$ $\newline$ = $729$
Calculate Difference: Finally, we calculate the difference in the number of chickens between the two scenarios two years from now. $\newline$ Difference = Number of chickens after $2$ more years ( $50$ % increase) - Number of chickens after $2$ more years (constant increase) $\newline$ Difference = $729 - 648$ $\newline$ Difference = $81$