CAT 2021 Slot 3 QA Q13 Inequalities & Modulus

Question 13 of the Quantitative Ability (QA) section, CAT 2021 Slot 3, with CAT's official answer and a worked solution.

If \(3 x+2|y|+y=7\) and \(x+|x|+3 y=1\), then \(x+2 y\) is

A
0
B
1
C
\(-\frac{4}{3}\)
D
\(\frac{8}{3}\)

Answer: (A) 0 Inequalities & Modulus MCQ

Solution

If \(x \lt 0\), the second equation gives \(3y = 1\), then the first gives \(3x + 1 = 7\), \(x = 2\): contradiction. So \(x \ge 0\) and \(2x + 3y = 1\).
If \(y \ge 0\): \(3x + 3y = 7\), which with \(2x + 3y = 1\) gives \(x = 6, y = -\frac{11}{3}\): contradiction. So \(y \lt 0\) and \(3x - y = 7\).
Solving \(3x - y = 7\) and \(2x + 3y = 1\): \(x = 2, y = -1\).
\(x + 2y = 0\).