CAT 2022 Question Paper | Quant Slot 2

CAT 2022 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 4 : If \(a\) and \(b\) are non-negative real numbers such that \(a+2 b=6\), then the average of the maximum and minimum possible values of \((a+b)\) is

Video Explanation

Explanatory Answer

a + 2b = 6
a + b = 6 - b
Clearly, the maximum/minimum value of (a + b) depends on the value of b.
Since a and b are positive real numbers,
The maximum value that b can take is 0.
The maximum value that b can take is when a is 0.
0 + 2b = 6
b = 3.
When b = 0; a + b = 6 - b = 6
When b = 3; a + b = 6 - 3 = 3
Therefore, the minimum and maximum values of (a + b) are 3 and 6 respectively.
The average of these extreme values is \( \frac { 3 + 6 } { 2 } \) = 4.5

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

Hence, the answer is '4.5'

Choice B is the correct answer.