Equations, Functions and Logs Previous Year Questions (PYQs) for XAT: 4+ Solved Questions with Step-by-Step Solutions

Solve 4+ Equations, Functions and Logs previous year questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Equations, Functions, and Logs

Chapter Roadmap

1. Linear & Quadratic Equations

Master roots, systems, integer solutions, and sign schemes. (Current Topic)

2. Logarithms, Functions & Trigonometry

Explore domains, log properties, and trigonometric identities.

The Core of Algebra: Linear and Quadratic Equations

The Core of Algebra

Linear Equations

Constant rate of change. Straight lines.

y=mx+c

Quadratic Equations

Changing rate of change. Parabolas.

y=ax2+bx+c

Exam Focus:

Systems of equations (intersections).
Hidden constraints (integer/irrational roots).
Curve geometry (maxima, minima, signs).

Equations, Functions and Logs: Solved Questions with Step-by-Step Explanations (4 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Consider the quadratic function $f (x) = a x^{2} + b x + a$ having two irrational roots, with a and b being two positive integers, such that $a, b \leq 9$ .

If all such permissible pairs(a, b) are equally likely, what is the probability that a+ b is greater than 9?

A.
$\frac{7}{16}$
B.
$\frac{5}{8}$
C.
None of the other answers is correct.
D.
$\frac{7}{15}$
E.
$\frac{2}{3}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a probability problem combined with the nature of roots of a quadratic equation. The condition for "two irrational roots" means the discriminant $D$ must be strictly positive and not a perfect square.

Step 1: Define the condition for irrational roots.

For $f (x) = a x^{2} + b x + a$ , the discriminant is $D = b^{2} - 4 a^{2}$ .

We need $D > 0 ⟹ b^{2} > 4 a^{2} ⟹ b > 2 a$ (since $a, b > 0$ ).

We also need $D$ to not be a perfect square.

Step 2: List all valid pairs $(a, b)$ with $1 \leq a, b \leq 9$ .

If $a = 1$ : $b > 2 ⟹ b \in {3, 4, 5, 6, 7, 8, 9}$ .

$D = b^{2} - 4$ . Values: 5, 12, 21, 32, 45, 60, 77. None are perfect squares. (7 pairs)

If $a = 2$ : $b > 4 ⟹ b \in {5, 6, 7, 8, 9}$ .

$D = b^{2} - 16$ . Values: 9, 20, 33, 48, 65.

$b = 5 ⟹ D = 9$ (perfect square, rational roots). Invalid.

Valid $b$ : 6, 7, 8, 9. (4 pairs)

If $a = 3$ : $b > 6 ⟹ b \in {7, 8, 9}$ .

$D = b^{2} - 36$ . Values: 13, 28, 45. None are perfect squares. (3 pairs)

If $a = 4$ : $b > 8 ⟹ b = 9$ .

$D = 81 - 64 = 17$ . Not a perfect square. (1 pair)

Total valid pairs = $7 + 4 + 3 + 1 = 15$ .

Step 3: Count pairs where $a + b > 9$ .

$a = 1$ : $b = 9$ (sum 10) $⟹ 1$ pair.
$a = 2$ : $b \in {8, 9}$ (sums 10, 11) $⟹ 2$ pairs.
$a = 3$ : $b \in {7, 8, 9}$ (sums 10, 11, 12) $⟹ 3$ pairs.
$a = 4$ : $b = 9$ (sum 13) $⟹ 1$ pair.

Total favorable pairs = $1 + 2 + 3 + 1 = 7$ .

Step 4: Calculate probability.

Probability = Favorable / Total = $7/15$ .

Answer: D

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

How many solutions of the equation $x + y + 2 z = 350$ exist, where x, y and z are positive integers?

A.
17
B.
15
C.
16
D.
None of the other options is correct
E.
18

Correct Answer:

Step-by-Step Solution

Key idea: This is a Diophantine equation problem (finding integer solutions). The number of positive integer solutions to $x_{1} + ... + x_{k} = n$ is given by $(k - 1 n - 1)$ .

Step 1: Identify Parameters.

Equation: $x + y + 2 z = 350$

Variables: $x, y, z$ are positive integers ( $\geq 1$ ).

Step 2: Handle the Coefficient of z.

The standard formula applies when coefficients are 1. Here, $z$ has a coefficient of 2.

Let $x + y = 350 - 2 z$ .

Let $K = 350 - 2 z$ .

Since $x, y \geq 1$ , the minimum value of $x + y$ is 2.

So, $K \geq 2 ⟹ 350 - 2 z \geq 2 ⟹ 2 z \leq 348 ⟹ z \leq 174$ .

Also $z \geq 1$ .

So $z$ can range from $1$ to $174$ .

Step 3: Count Solutions for Fixed z.

For a fixed $z$ , we have $x + y = K$ .

The number of positive integer solutions for $x + y = K$ is $(2 - 1 K - 1) = K - 1$ .

Here $K = 350 - 2 z$ .

So, number of solutions for a given $z$ is $(350 - 2 z) - 1 = 349 - 2 z$ .

Step 4: Sum over all valid z.

Total solutions = $\sum_{z = 1}^{174} (349 - 2 z)$ .

This is an arithmetic progression.

First term ( $z = 1$ ): $349 - 2 (1) = 347$ .

Last term ( $z = 174$ ): $349 - 2 (174) = 349 - 348 = 1$ .

Number of terms = 174.

Sum = $\frac{n}{2} (F i r s t + L a s t)$

Sum = $\frac{174}{2} (347 + 1) = 87 \times 348 = 30276$ .

Step 5: Check Options.

Options: 17, 15, 16, None, 18.

The calculated value 30,276 is not 17, 15, 16, or 18.

Therefore, "None of the other options is correct".

Answer: D

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

For how many distinct real values of $x$ does the equation below hold true?(Consider $x > 0$ .)

$lo g_{a} (32) - x lo g_{a} (16) + lo g_{a} (32) - x lo g_{a} (64) = 0$

A.
1
B.
Infinitely many
C.
2
D.
0
E.
Depends on the value of $a$

Correct Answer:

Step-by-Step Solution

Key idea: This is a logarithmic equation problem, recognizable because it involves terms with $lo g_{a}$ and an unknown variable $x$ multiplied by logarithms. The method involves grouping terms and using log properties to solve for $x$ .

Step 1: Group the terms with and without $x$ .

The equation is $lo g_{a} (32) - x lo g_{a} (16) + lo g_{a} (32) - x lo g_{a} (64) = 0$ .

Combine the constant terms and the terms with $x$ :

$2 lo g_{a} (32) - x (lo g_{a} (16) + lo g_{a} (64)) = 0$ .

Step 2: Simplify using logarithm properties.

We know that $32 = 2^{5}$ , $16 = 2^{4}$ , and $64 = 2^{6}$ .

So, $lo g_{a} (32) = 5 lo g_{a} (2)$ , $lo g_{a} (16) = 4 lo g_{a} (2)$ , and $lo g_{a} (64) = 6 lo g_{a} (2)$ .

Substitute these into the equation:

$2 (5 lo g_{a} (2)) - x (4 lo g_{a} (2) + 6 lo g_{a} (2)) = 0$ .

$10 lo g_{a} (2) - x (10 lo g_{a} (2)) = 0$ .

Step 3: Solve for $x$ .

Factor out $10 lo g_{a} (2)$ :

$10 lo g_{a} (2) (1 - x) = 0$ .

Since $a$ is a valid base ( $a > 0, a \neq = 1$ ), $lo g_{a} (2) \neq = 0$ .

Therefore, $1 - x = 0 ⟹ x = 1$ .

There is exactly 1 distinct real value of $x$ .

Answer: A

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Consider the system of two linear equations as follows: 3x+ 21y+ p= 0; and qx+ ry −7= 0, where p, q, and r are real numbers.
Which of the following statements DEFINITELY CONTRADICTS the fact that the lines represented by the two equations are coinciding?

A.
p and q must have opposite signs
B.
The smallest among p, q, and r is r
C.
The largest among p, q, and r is q
D.
r and q must have same signs
E.
p cannot be 0

Correct Answer:

Step-by-Step Solution

Key idea: Two lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ are coincident if their coefficients are proportional: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}} = k$ .

Step 1: Set up the proportionality.

Line 1: $3 x + 21 y + p = 0$

Line 2: $q x + r y - 7 = 0$

Condition: $\frac{3}{q} = \frac{21}{r} = \frac{p}{- 7} = k$

Step 2: Express variables in terms of k.

$3 = k q ⟹ q = 3/ k$

$21 = k r ⟹ r = 21/ k$

$p = - 7 k$

Step 3: Analyze the signs and magnitudes.

From $q = 3/ k$ and $r = 21/ k$ , $q$ and $r$ always have the same sign (same as $1/ k$ ).

From $p = - 7 k$ , $p$ has the opposite sign of $k$ .

Therefore:

$q$ and $r$ have the same sign. (Option D is True).
$p$ has the opposite sign to $q$ and $r$ . (Option A is True).
$p \neq = 0$ because $k \neq = 0$ . (Option E is True).

Step 4: Evaluate Options B and C.

We need to find which statement DEFINITELY CONTRADICTS (i.e., is ALWAYS FALSE).

Can $q$ ever be the largest among $p, q, r$ ? (Option C)

We need $q > r$ and $q > p$ .

$q > r ⟹ 3/ k > 21/ k$ .

If $k > 0$ , $3 > 21$ (False).

If $k < 0$ , $3 < 21$ (True, since multiplying by $k$ flips inequality).

So we must have $k < 0$ for $q > r$ .

Now we need $q > p ⟹ 3/ k > - 7 k$ .

Since $k < 0$ , multiply by $k$ (flips inequality): $3 < - 7 k^{2} ⟹ 7 k^{2} < - 3$ .

This is impossible for real $k$ .

So $q$ can NEVER be strictly greater than $p$ when $k < 0$ .

Therefore, $q$ can NEVER be the largest among $p, q, r$ .

Statement C is ALWAYS FALSE. It definitely contradicts the condition.

Answer: C

Equations, Functions and Logs Previous Year Questions (PYQs) for XAT: 4+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Equations, Functions, and Logs

Chapter Roadmap

The Core of Algebra: Linear and Quadratic Equations

The Core of Algebra

Equations, Functions and Logs: Solved Questions with Step-by-Step Explanations (4 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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