Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.9 AMC 10/12 Find the answer South Africa

If $x = \sqrt[3]{900}$ then
(A) $7 < x < 8$ (B) $9 < x < 10$ (C) $11 < x < 12$ (D) $10 < x < 11$ (E) $12 < x < 13$

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is B, and the solution below works it through.

Solution

Answer B.
Since $9^3 = 729 < 900$ and $10^3 = 1000 > 900$ , it follows that $9 < \sqrt[3]{900} < 10$ .

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