Linear Systems, Integer Solutions and Algebraic Expressions Previous Year Questions (PYQs) for CAT: 13+ Solved Questions with Step-by-Step Solutions

Solve 13+ Linear Systems, Integer Solutions and Algebraic Expressions previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Linear Systems, Integer Solutions and Algebraic Expressions

2×2

Chapter roadmap

Linear Systems, Integer Solutions and Algebraic Expressions

⚖️ Linear Equations and Systems

Solve useful combinations, check no-solution conditions, and use infinite-solution proportionality.

3 direct CAT PYQs | selected topic

🔢 Integer Solutions and Natural Number Constraints

Use divisibility, factorization, bounds, and integer feasibility to reduce possibilities.

4 direct CAT PYQs

🧩 Algebraic Identities and Quadratic Forms

Transform quadratic-looking conditions into squares, differences, and identity-based shortcuts.

6 direct CAT PYQs | strongest topic in chapter

🧺 Word Equations and Cost Relations

Convert price or quantity stories into equations and eliminate unnecessary variables.

1 direct CAT PYQ

By the end, you should know whether the question needs solving, elimination, consistency checking, or expression transformation.

Topic Hero: Linear Equations and Systems

Algebra → Linear Systems, Integer Solutions and Algebraic Expressions → Topic 1

ax+by

Combine equations, don’t over-solve

Linear Equations and Systems

CAT systems often hide one clean combination behind several variables.

✅ Combine equations to find target expressions

✅ Know unique, no-solution, and infinite-solution cases

✅ Use determinant logic for two equations

✅ Convert infinite solutions into proportional coefficients

Linear Systems, Integer Solutions and Algebraic Expressions: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

Let

x, y,

and

z

be real numbers satisfying

4 (x^{2} + y^{2} + z^{2}) = a,

4 (x - y - z) = 3 + a

The a equals

A.
$3$
B.
$1 \frac{1}{3}$
C.
$4$
D.
$1$

Correct Answer:

Step-by-Step Solution

Key idea: This is an algebraic identities question. It is recognisable because a square-sum expression and a linear expression are linked through the same parameter $a$ . The clean route is to eliminate $a$ and complete squares.

Step 1: Write the two given equations.

$4 (x^{2} + y^{2} + z^{2}) = a$

$4 (x - y - z) = 3 + a$

Step 2: Substitute $a$ from the first equation into the second.

Since $a = 4 (x^{2} + y^{2} + z^{2})$ , the second equation becomes:

$4 (x - y - z) = 3 + 4 (x^{2} + y^{2} + z^{2})$

Step 3: Bring everything to one side.

$4 (x^{2} + y^{2} + z^{2}) - 4 (x - y - z) + 3 = 0$

Step 4: Expand the linear part.

$- 4 (x - y - z) = - 4 x + 4 y + 4 z$

So the equation is:

$4 x^{2} + 4 y^{2} + 4 z^{2} - 4 x + 4 y + 4 z + 3 = 0$

Step 5: Complete squares separately for $x$ , $y$ , and $z$ .

For $x$ :

$4 x^{2} - 4 x = (2 x - 1)^{2} - 1$

For $y$ :

$4 y^{2} + 4 y = (2 y + 1)^{2} - 1$

For $z$ :

$4 z^{2} + 4 z = (2 z + 1)^{2} - 1$

Step 6: Substitute these square forms back.

$[(2 x - 1)^{2} - 1] + [(2 y + 1)^{2} - 1] + [(2 z + 1)^{2} - 1] + 3 = 0$

The constants cancel:

$- 1 - 1 - 1 + 3 = 0$

Hence,

$(2 x - 1)^{2} + (2 y + 1)^{2} + (2 z + 1)^{2} = 0$

Step 7: Use the sum-of-squares rule.

Each square is non-negative. If their sum is zero, each square must be zero:

$2 x - 1 = 0, 2 y + 1 = 0, 2 z + 1 = 0$

So,

$x = \frac{1}{2}, y = - \frac{1}{2}, z = - \frac{1}{2}$

Step 8: Find $a$ .

$a = 4 (x^{2} + y^{2} + z^{2})$

$a = 4 (\frac{1}{4} + \frac{1}{4} + \frac{1}{4})$

$a = 4 \cdot \frac{3}{4} = 3$

Answer: 3

Question 2 · Quantitative Ability MCQ

A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

A.
11
B.
13
C.
10
D.
12

Correct Answer:

Step-by-Step Solution

Key idea: This is a cost-relation word equation. We translate the baskets into algebraic expressions and use equality to eliminate unnecessary variables.

Step 1: Assign variables to the prices.

Let $A$ , $O$ , and $M$ be the prices of one apple, one orange, and one mango respectively.

The three baskets have the following cost expressions:

Basket 1: $2 A + 4 O + 6 M$

Basket 2: $A + 4 O + 8 M$

Basket 3: $8 O + 7 M$

Step 2: Equate Basket 1 and Basket 2.

$2 A + 4 O + 6 M = A + 4 O + 8 M$

Subtract $A + 4 O + 6 M$ from both sides:

$A = 2 M$

Step 3: Equate Basket 2 and Basket 3.

$A + 4 O + 8 M = 8 O + 7 M$

Subtract $4 O + 7 M$ from both sides:

$A + M = 4 O$

Substitute $A = 2 M$ into this equation:

$2 M + M = 4 O ⟹ 3 M = 4 O ⟹ O = \frac{3}{4} M$

Step 4: Find the cost of a basket in terms of $M$ .

Using Basket 2:

Cost = $A + 4 O + 8 M$

Substitute $A = 2 M$ and $O = \frac{3}{4} M$ :

Cost = $2 M + 4 (\frac{3}{4} M) + 8 M = 2 M + 3 M + 8 M = 13 M$

Since the cost is $13 M$ , a basket containing only mangoes with the same cost must contain 13 mangoes.

Answer: 13

Question 3 · Quantitative Ability MCQ

If $p^{2} + q^{2} - 29 = 2 pq - 20 = 52 - 2 pq$ , then the difference between the maximum and minimum possible value of $(p^{3} - q^{3})$

A.
243
B.
486
C.
378
D.
189

Correct Answer:

Step-by-Step Solution

Key idea: This is a chain-equality algebraic identities question. It is recognisable because three expressions are declared equal, so equating pairs gives clean relations among $p$ and $q$ .

Step 1: Understand the chain equality.

$p^{2} + q^{2} - 29 = 2 pq - 20 = 52 - 2 pq$

Let the common value be $k$ . Then we can equate any two parts.

Step 2: Equate the second and third expressions.

$2 pq - 20 = 52 - 2 pq$

Add $2 pq$ to both sides:

$4 pq - 20 = 52$

Add 20:

$4 pq = 72$

So,

$pq = 18$

Step 3: Find the common value.

Using $2 pq - 20$ :

$2 (18) - 20 = 36 - 20 = 16$

So the common value is 16.

Step 4: Use the first expression.

$p^{2} + q^{2} - 29 = 16$

Hence,

$p^{2} + q^{2} = 45$

Step 5: Find $p - q$ .

Use the identity:

$(p - q)^{2} = p^{2} + q^{2} - 2 pq$

Substitute:

$(p - q)^{2} = 45 - 36 = 9$

Therefore,

$p - q = \pm 3$

Step 6: Express $p^{3} - q^{3}$ .

Use:

$p^{3} - q^{3} = (p - q) (p^{2} + pq + q^{2})$

We know:

$p^{2} + q^{2} = 45, pq = 18$

So,

$p^{2} + pq + q^{2} = 45 + 18 = 63$

Therefore,

$p^{3} - q^{3} = 63 (p - q)$

Step 7: Find maximum and minimum values.

Since $p - q = \pm 3$ :

$p^{3} - q^{3} = 63 (3) = 189$

$p^{3} - q^{3} = 63 (- 3) = - 189$

The maximum is $189$ , and the minimum is $- 189$ .

Step 8: Find the difference.

$189 - (- 189) = 378$

Answer: 378

Question 4 · Quantitative Ability NAT

If x and y are real numbers such that $4 x^{2} + 4 y^{2} - 4 x y - 6 y + 3 = 0$ , then the value of $(4 x + 5 y)$ is

Correct Answer:

Step-by-Step Solution

This is a sum-of-squares question, recognisable because a quadratic expression in two variables equals zero and the coefficients suggest a perfect square decomposition.

Step 1: Group terms to form perfect squares.

Given: \ $4x^2 + 4y^2 - 4xy - 6y + 3 = 0\$

Look for a square involving \ $x\$ and \ $y\$ . The terms \ $4x^2 - 4xy + y^2\$ form a perfect square:

\ $4x^2 - 4xy + y^2 = (2x - y)^2\$

Rewrite the equation by splitting \ $4y^2\$ into \ $y^2 + 3y^2\$ :

\ $(4x^2 - 4xy + y^2) + 3y^2 - 6y + 3 = 0\$

Step 2: Factor the remaining terms.

\ $(2x - y)^2 + 3(y^2 - 2y + 1) = 0\$

\ $(2x - y)^2 + 3(y - 1)^2 = 0\$

Step 3: Apply the zero-sum property of squares.

A square is always \ $\geq 0\$ . The sum of two non-negative terms can be zero <b>only if each term is individually zero</b>:

\ $(2x - y)^2 = 0 \implies 2x = y\$

\ $3(y - 1)^2 = 0 \implies y = 1\$

Step 4: Solve for \ $x\$ and \ $y\$ .

From \ $y = 1\$ :

\ $2x = 1 \implies x = \frac{1}{2}\$

Step 5: Compute the required expression.

\ $4x + 5y = 4\left(\frac{1}{2}\right) + 5(1) = 2 + 5 = 7\$

Verification: \ $4(1/4) + 4(1) - 4(1/2)(1) - 6(1) + 3 = 1 + 4 - 2 - 6 + 3 = 0\$ . ✓

Answer: 7

Question 5 · Quantitative Ability MCQ

Consider the pair of equations: $x^{2} - x y - x = 22$ and $y^{2} - x y + y = 34$ . If $x > y$ , then $x - y$ equals

A.
6
B.
4
C.
7
D.
8

Correct Answer:

Step-by-Step Solution

This is a subtract/add equations to reveal structure question, recognisable because two quadratic equations share the term $- x y$ and adding them creates a perfect square pattern.

Step 1: Add the two equations.

$(x^{2} - x y - x) + (y^{2} - x y + y) = 22 + 34 = 56$

$x^{2} + y^{2} - 2 x y - x + y = 56$

Step 2: Recognise the structure.

$x^{2} - 2 x y + y^{2} = (x - y)^{2}$ and $- x + y = - (x - y)$ .

So: $(x - y)^{2} - (x - y) = 56$

Step 3: Substitute u = x - y.

$u^{2} - u = 56$

$u^{2} - u - 56 = 0$

$(u - 8) (u + 7) = 0$

So $u = 8$ or $u = - 7$ .

Step 4: Apply the condition x > y.

Since $x > y$ , we need $x - y > 0$ , so $u = 8$ .

Answer: 8 (Option D)

Linear Systems, Integer Solutions and Algebraic Expressions Previous Year Questions (PYQs) for CAT: 13+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Linear Systems, Integer Solutions and Algebraic Expressions

Linear Systems, Integer Solutions and Algebraic Expressions

Topic Hero: Linear Equations and Systems

Linear Equations and Systems

Linear Systems, Integer Solutions and Algebraic Expressions: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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