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CAT 2022 Slot 2 — QA Question 6

Number TheoryEasy

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

Answer & solution

  • A

    4

  • 4.5

  • C

    3.5

  • D

    3

Solution

Easy

Express the target a+ba+b using the constraint so it depends on a single variable, then push that variable to its allowed extremes. Both aa and bb are non-negative, which caps how large each can grow.

1

Reduce to one variable. From a+2b=6a+2b=6, a=6−2ba=6-2b, so

a+b=(6−2b)+b=6−b.a+b=(6-2b)+b=6-b.
2

Find the feasible range of bb. Need a≥0a\ge0 and b≥0b\ge0:

a=6−2b≥0  ⇒  b≤3,b≥0  ⇒  0≤b≤3.a=6-2b\ge0 \;\Rightarrow\; b\le3, \qquad b\ge0 \;\Rightarrow\; 0\le b\le3.
3

Extremes of a+b=6−ba+b=6-b (decreasing in bb):

max at b=0: a+b=6min at b=3: a+b=3\begin{aligned} &\text{max at }b=0:\ a+b=6\\ &\text{min at }b=3:\ a+b=3 \end{aligned}

Average =6+32=4.5=\dfrac{6+3}{2}=\mathbf{4.5} — option (b).

CAT 2022 Slot 2 QA Q6: If a and b are non-negative real numbers such that a + 2b = 6, then the average of the maximum and minimum val — Solution | TheCATExam