CAT 2024 Slot 1 QA Q2 Logarithms & Exponents

Question 2 of the Quantitative Ability (QA) section, CAT 2024 Slot 1, with CAT's official answer and a worked solution.

If \((a+b \sqrt{n})\) is the positive square root of \((29-12 \sqrt{5})\), where \(a\) and \(b\) are integers, and \(n\) is a natural number, then the maximum possible value of \((a+b+n)\) is
  1. A
    22
  2. B
    6
  3. C
    18
  4. D
    4

Answer: (C) 18 Logarithms & Exponents MCQ

Solution

  1. Write \((a + b\sqrt{n})^2 = a^2 + nb^2 + 2ab\sqrt{n}\) and match it with \(29 - 12\sqrt{5}\).
  2. \(\sqrt{n}\) must be a multiple of \(\sqrt{5}\): \(n = 5\) or \(n = 20\) (since \(2\sqrt5 = \sqrt{20}\)); larger \(n\) would make \(b\) a fraction.
  3. \(n = 5\): \(ab = -6\), \(a^2 + 5b^2 = 29 \Rightarrow (a, b) = (\pm3, \mp2)\). The positive root is \(-3 + 2\sqrt5 \approx 1.47\), so \(a + b + n = -3 + 2 + 5 = 4\).
  4. \(n = 20\): \(2ab\sqrt{20} = 4ab\sqrt5 = -12\sqrt5 \Rightarrow ab = -3\), and \(a^2 + 20b^2 = 29 \Rightarrow a = -3, b = 1\). Root \(= -3 + \sqrt{20} > 0\) ✓, so \(a + b + n = 18\).
  5. Maximum value \(= 18\).

Shortcut. \(29 - 12\sqrt5 = (2\sqrt5 - 3)^2\). Writing \(2\sqrt5\) as \(\sqrt{20}\) gives the larger sum \(-3 + 1 + 20 = 18\).

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