Averages , Ratio and Proportion Question 7 - CAT 2022
If a and b are non-negative real numbers such that a+2b=6, then the average of the maximum and minimum possible values of (a+b) is:
Step-by-Step Explanation
We are given that: a + 2b = 6, where a ≥ 0 and b ≥ 0.
We want to analyze the objective expression: S = a + b
- Express the objective equation in terms of a single variable:
Substitute a = 6 − 2b into the expression for S:
S = 6 − 2b + b = 6 − b
- Determine the boundary values for b:
Since a ≥ 0:
6 − 2b ≥ 0
2b ≤ 6
b ≤ 3
Since b is non-negative, its valid range is:
0 ≤ b ≤ 3
- Calculate the maximum and minimum values of S:
S is maximized when b is as small as possible (b = 0):
S maximum = 6 − 0 = 6
S is minimized when b is as large as possible (b = 3):
S minimum = 6 − 3 = 3
Average = (S maximum + S minimum)/2
= (6 + 3)/2
= 4.5
Correct Option: D
💡 FundaMakers Tip:
When you have a linear system with non-negative constraints, the maximum and minimum values always occur at the extreme endpoints of the variable ranges. Testing the boundary conditions (b = 0 and b = 3) directly gives you the extreme values for a + b without any deep optimization steps.