CAT 2021 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 13 : If \(3 x+2|y|+y=7\) and \(x+|x|+3 y=1\), then \(x+2 y\) is

- 0
- 1
- \(-\frac{4}{3}\)
- \(\frac{8}{3}\)

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We are given two equations: 3x + 2|y| + y = 7 and x + |x| + 3y = 1

|x| or the modulus of x, is a function of x, that gives the magnitude of x.

|x| = -(x); if x is negative, and

|x| = x; if x is non-negative.

Therefore, depending on whether ‘x’ and ‘y’ are positive or negative, we assume the following cases for the two equations given.

Case (i): ‘x’ and ‘y’ are both positive.

|x| = x and |y| = y

3x + 2|y| + y = 7

x + |x| + 3y = 1

3x + 2y + y = 7

x + x + 3y = 1

3x + 3y = 7

2x + 3y = 1

Solving the two equations, we get x = 6 and y = -11/3.

Since this is contradictory to the assumption that y is positive, we discard this case.

Case (ii): ‘x’ and ‘y’ are both negative.

|x| = -x and |y| = -y

3x + 2|y| + y = 7

x + |x| + 3y = 1

3x - 2y + y = 7

x - x + 3y = 1

3x - y = 7

3y = 1

Solving the two equations, we get x = 22/9 and y = 1/3.

Since this is contradictory to the assumption that both x and y are both negative, we discard this case.

Case (iii): ‘x’ is negative and ‘y’ is positive.

|x| = -x and |y| = y

3x + 2|y| + y = 7

x + |x| + 3y = 1

3x + 2y + y = 7

x - x + 3y = 1

3x + 3y = 7

3y = 1

Solving the two equations, we get x = 2 and y = 1/3.

Since this is contradictory to the assumption that both x is negative, we discard this case.

Case (iv): ‘x’ is positive and ‘y’ is negative.

|x| = x and |y| = -y

3x + 2|y| + y = 7

x + |x| + 3y = 1

3x - 2y + y = 7

x + x + 3y = 1

3x - y = 7

2x + 3y = 1

Solving the two equations, we get x = 2 and y = -1.

This satisfies the assumption that x is positive and y is negative.

Hence x = 2 and y = -1

Therefore, x + 2y = 2 + 2(-1) = 0

Hence, x + 2y = 0.

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

Choice A is the correct answer.

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