Multiple choice

In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town’s population during this twenty-year period?

  1. 42

  2. 47

  3. 57

  4. 62
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D Correct answer
Explanation

Let population be x^2. Then (x^2 + 150) = y^2 + 9, so x^2 + 141 = y^2. Also (y^2 + 150) = z^2. Solving these integer equations: x=20 (400), y=23 (529), z=26 (676). Growth = (676-400)/400 = 276/400 = 69%. The closest option is 62%.

AI explanation

Let the initial population be a^2, so after 10 years it is a^2 + 150 and this equals b^2 + 9, which means a^2 + 141 = b^2. After 20 years the population is a^2 + 300, which must also be a perfect square, c^2, meaning c^2 minus a^2 equals 300 and factors to (c-a)(c+a) = 300. The factor pairs of 300 that work are 6 and 50, giving c = 28 and a = 22, so the initial population is 22^2 or 484 and the final is 28^2 or 784. The percent growth is ((784 - 484)/484) * 100, which is about 62%.

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