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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 percent 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 percent per year. What is the difference in the number of chickens between these two situations two years from now?
Choose 1 answer:
(A) 0
(B) 
189
(C) 405
(D) 11

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 6464. The number increased by 50%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%50\% per year. What is the difference in the number of chickens between these two situations two years from now?\newlineChoose 11 answer:\newline(A) 00\newline(B) 189189\newline(C) 405405\newline(D) 1111

Full solution

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 6464. The number increased by 50%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%50\% per year. What is the difference in the number of chickens between these two situations two years from now?\newlineChoose 11 answer:\newline(A) 00\newline(B) 189189\newline(C) 405405\newline(D) 1111
  1. Calculate Chickens After 44 Years: First, let's calculate the number of chickens at the end of the four-year period with a 50%50\% increase each year.\newlineInitial number of chickens: 6464\newlineYearly increase: 50%50\%\newlineWe 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.\newlineNumber of chickens after nn years = Initial number ×(1+rate)n\times (1 + \text{rate})^n
  2. Calculate Chickens After 22 Years: Now, let's calculate the number of chickens after four years.\newlineNumber of chickens after 44 years =64×(1+0.50)4= 64 \times (1 + 0.50)^4\newline=64×(1.50)4= 64 \times (1.50)^4\newline=64×5.0625= 64 \times 5.0625\newline=324= 324 (rounded to the nearest whole number)
  3. 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.\newlineThe rate of the most recent year is 50%50\% of the number of chickens at the end of the fourth year.\newlineConstant increase = 50%50\% of 324324\newline= 0.50×3240.50 \times 324\newline= 162162\newlineSo, the number of chickens will increase by 162162 each year for the next two years.
  4. Calculate Chickens After 22 More Years: Now, let's calculate the number of chickens after two more years with the constant increase.\newlineNumber of chickens after 22 more years (constant increase) = 324+162×2324 + 162 \times 2\newline= 324+324324 + 324\newline= 648648
  5. 5050% 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 5050 percent per year.\newlineNumber of chickens after 11 more year (5050% increase) = 324×(1+0.50)324 \times (1 + 0.50)\newline= 324×1.50324 \times 1.50\newline= 486486\newlineNumber of chickens after 22 more years (5050% increase) = 486×(1+0.50)486 \times (1 + 0.50)\newline= 486×1.50486 \times 1.50\newline= 729729
  6. Calculate Difference: Finally, we calculate the difference in the number of chickens between the two scenarios two years from now.\newlineDifference = Number of chickens after 22 more years (5050% increase) - Number of chickens after 22 more years (constant increase)\newlineDifference = 729648729 - 648\newlineDifference = 8181

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