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Average of Max and Min Values

CAT 2022 Slot 2 · Quantitative Ability · Easy · Algebra

This is an easy Quantitative Ability question from the CAT 2022 Slot 2 paper. It tests Algebra. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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If aa and bb are non-negative real numbers such that a+2b=6a+2b=6, then the average of the maximum and minimum possible values of (a+b)(a+b) is

  1. A.

    4

  2. B.

    4.5

  3. C.

    3.5

  4. D.

    3

Answer

B

Explanation

a+b=a+b=(6−2b)+b=6−ba + b = a + b = (6 - 2b) + b = 6 - b. Since a,b≥0a, b \ge 0: a=6−2b≥0  ⟹  b≤3a = 6 - 2b \ge 0 \implies b \le 3. Also b≥0b \ge 0. So 0≤b≤30 \le b \le 3. Max value of (a+b)=6−0=6(a+b) = 6 - 0 = 6 (when b=0,a=6b = 0, a = 6). Min value of (a+b)=6−3=3(a+b) = 6 - 3 = 3 (when b=3,a=0b = 3, a = 0). Average of max and min = 6+32=4.5\frac{6 + 3}{2} = 4.5.

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