Multiple choice

A district has $64000$ inhabitants. If the population increases the rate of $2\cfrac{1}{2}$% per annum, the number of inhabitants at the end of $3$ years will be:

  1. $70000$
  2. $69200$
  3. $68921$
  4. $68911$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Population = 64000 * (1 + 2.5/100)^3 = 64000 * (1.025)^3 = 64000 * 1.076890625 = 68921.

AI explanation

Using the compound interest formula for population growth, the final population is 64000 multiplied by the quantity (1 plus 0.025) cubed. This evaluates to 64000 multiplied by 1.025 cubed, which is 64000 multiplied by 1.076890625. The number of inhabitants at the end of 3 years is 68921.

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