Math Problem Statement

. An Arctic ice study found that in the summer, 70% of an ocean region was ice covered. The study also predicted the region’s ice would be reduced by 6% per year. a) What percent of the region will be ice covered after the fi rst year? Express your answer to the nearest tenth of a percent. b) What percent of the region will be ice covered after the third year? Express your answer to the nearest tenth of a percent. c) How many years will pass before less than one half of the region is covered

Solution

Part (a):

We are given that 70% of the ocean region is ice-covered initially and the ice cover reduces by 6% per year.

After the first year, the percentage of the ice remaining will be: Percentage remaining=70%×(10.06)\text{Percentage remaining} = 70\% \times (1 - 0.06) Percentage remaining=70%×0.94=65.8%\text{Percentage remaining} = 70\% \times 0.94 = 65.8\% Thus, after the first year, 65.8% of the region will be ice-covered.

Part (b):

To find the percentage of the region covered by ice after the third year, we need to reduce the ice cover by 6% for each year, compounded annually.

After the first year: Year 1=70%×0.94=65.8%\text{Year 1} = 70\% \times 0.94 = 65.8\%

After the second year: Year 2=65.8%×0.94=61.852%\text{Year 2} = 65.8\% \times 0.94 = 61.852\%

After the third year: Year 3=61.852%×0.94=58.14088%\text{Year 3} = 61.852\% \times 0.94 = 58.14088\% Rounding to the nearest tenth, the percentage of the region ice-covered after the third year is 58.1%.

Part (c):

We need to find how many years will pass before less than half (50%) of the region is covered by ice. This can be modeled by the equation: P(t)=70%×(0.94)tP(t) = 70\% \times (0.94)^t where P(t)P(t) is the percentage of the region covered by ice after tt years. We want to find tt when P(t)<50%P(t) < 50\%.

Solving: 70×(0.94)t=5070 \times (0.94)^t = 50 (0.94)t=50700.7143(0.94)^t = \frac{50}{70} \approx 0.7143 Now, take the natural logarithm of both sides: ln((0.94)t)=ln(0.7143)\ln((0.94)^t) = \ln(0.7143) tln(0.94)=ln(0.7143)t \cdot \ln(0.94) = \ln(0.7143) t=ln(0.7143)ln(0.94)t = \frac{\ln(0.7143)}{\ln(0.94)}

Let me calculate that value for you.

It will take approximately 5.44 years for less than half of the region to be ice-covered. Since the number of years must be a whole number, this implies that after 6 years, the ice cover will drop below 50%.

Summary:

  • (a) After the first year, 65.8% of the region will be ice-covered.
  • (b) After the third year, 58.1% of the region will be ice-covered.
  • (c) It will take approximately 6 years for less than 50% of the region to be ice-covered.

Would you like more details on any part of the solution?

Here are 5 related questions for practice:

  1. What would the ice cover be after 10 years if the same 6% reduction continues annually?
  2. If the annual reduction were 8%, how many years would it take for the ice cover to drop below 50%?
  3. How much ice cover would remain after 5 years with a 5% annual reduction?
  4. What is the compound percentage reduction over 4 years at 6% per year?
  5. How would the result change if the initial ice cover were 80% instead of 70%?

Tip: When dealing with percentages and compounded rates, think of each year as multiplying by a reduction factor, which can be more accurate than subtracting the percentage directly each year.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Percentages
Compounded Reduction

Formulas

Percentage remaining after n years: P(t) = P0 * (1 - r)^t
Natural logarithm to solve for time: t = ln(final percentage / initial percentage) / ln(1 - r)

Theorems

Exponential Decay Model

Suitable Grade Level

Grades 9-11