4900km^2. Suppose that each year this area decreases by 8.5%. What will the area be after 6 years? round your answer to the nearest square kilometer

2 Answers

hkj1385's profile pic

hkj1385 | (Level 1) Assistant Educator

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Area at the beginning of the time:-

A0 = 4900 km^2

Area after the end of 1st year, A1 = 4900 - (8.5/100)*4900 = 4483.5 km^2

Area after the end of 2nd year, A2 = 4483.5 - (8.5/100)*4483.5 = 4102.4025 km^2

Area after the end of 3rd year, A3 = 4102.4025 - (8.5/100)*4102.4025 = 3753.698 km^2

Area after the end of 4th year, A4 = 3753.698 - (8.5/100)*3753.698 = 3434.634 km^2

Area after the end of 5th year, A5 = 3434.634 - (8.5/100)*3434.634 = 3142.690 km^2

Area after the end of 6th year, A6 = 3142.690 - (8.5/100)*3142.690 = 2875.561 km^2 = 2876 km^2 approx. after rounding off

t-rashmi's profile pic

t-rashmi | College Teacher | eNotes Newbie

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SOLUTION: Let the final area be x.

By the formula of compound interest in decreasing quantities, we get

x = A*(1-r/100)^n

   = 4900*(1-8.5/100)^6

   = 2875.56139461 km^2

   = 2876 km^2 

THEORY: This is a question which can be solved by the concept of Compound Interest. There, a fixed amount increases by some percent annually. So we use the formula A = P*(1+r/100)^n ,where A is the final amount, P is the original (or Principal) amount, r is the rate of interest and n is the number of years. In the case of the above problem, the quantity is decreasing at a given rate annually. So we shall replace 'r' with '- r'. Thus the formula becomes A = P*(1-r/100)^n .

Bonus: The graph of  y = 4900*(1-8.5/100)^x is attached with the answer, so that you can see how fast the area decreases over the years. Notice how sharply the value of y decreases in the first 65 years.

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