Algebra, Equations, Functions and Sequences PYQs for GATE CS

Solve 12+ Algebra, Equations, Functions and Sequences previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Algebra, Equations, Functions and Sequences

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Your path through Algebra, Equations, Functions and Sequences.

1. Number Properties and Proportional Relationships

Current topic · Foundation of variables

2. Exponential and Logarithmic Identities

Simplifying complex expressions

3. Recursive Sequences

Jumping between terms

4. Functions, Growth and Piecewise Models

Mapping inputs to outputs

5. Algebraic Equations and Constraint Problems

Solving for unknowns

Number Properties and Proportional Relationships

Quantitative Aptitude · Algebra

Number Properties and Proportional Relationships

Decode the hidden rules of numbers and the true meaning of proportionality.

Prime number parity Direct proportionality Componendo and dividendo

Chapter context: Algebra, Equations, Functions and Sequences

Algebra, Equations, Functions and Sequences: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Quantitative Aptitude · 2026_Set1 MCQ

A student needs to enroll for a minimum of 60 credits. A student cannot enroll for more than 70 credits. The credits are divided amongst project and three distinct sets of courses namely, core courses, specialization courses, and elective courses. It is compulsory for a student to enroll for exactly 15 credits of core courses and exactly 20 credits of project. In addition, a student has to enroll for a minimum of 10 credits of specialization courses. The maximum credits of elective courses that a student can enroll for is ______

A.
10
B.
15
C.
20
D.
25

Correct Answer:

Step-by-Step Solution

Insight: To maximize one variable in a sum with a fixed upper bound, you must minimize all other variables to their lowest allowed values.

Exam route: Total = Core (15) + Project (20) + Specialization (S) + Elective (E) = 35 + S + E. We know Total $\leq$ 70 and S $\geq$ 10. To maximize E, set S to its minimum (10) and Total to its maximum (70). Thus, 35 + 10 + E = 70 $⟹$ E = 25.

Learning route:

Step 1: Formulate the total credits equation: $Total = 15 + 20 + S + E = 35 + S + E$ .

Step 2: Apply the given constraints: $60 \leq Total \leq 70$ and $S \geq 10$ .

Step 3: We want to maximize $E$ . From the total equation, $E = Total - 35 - S$ .

Step 4: To make $E$ as large as possible, we must maximize $Total$ and minimize $S$ .

Step 5: The maximum allowed $Total$ is 70, and the minimum allowed $S$ is 10.

Step 6: Substitute these values: $E = 70 - 35 - 10 = 25$ .

Step 7: Verify this satisfies the minimum total constraint: $35 + 10 + 25 = 70$ , which is $\geq 60$ . The solution is valid.

Question 2 · Quantitative Aptitude · 2026_Set1 MCQ

For positive real numbers

S

and

K

, the function

H_{K} (S)

is defined as:

H_{K} (S) = max (S - K, 0)

. The max function is defined as:

max (a, b) = {a, when a > b \b, when a \leq b

The graph below shows the plot of a function

N (S)

versus

S

N (S)

can be expressed as _____.

A.
$H_{10} (S) - H_{20} (S)$
B.
$H_{10} (S) - 2 H_{20} (S)$
C.
$- H_{10} (S) + H_{20} (S)$
D.
$H_{15} (S) - H_{20} (S)$

Correct Answer:

Step-by-Step Solution

Insight: The graph is a piecewise linear function that is 0 for $S \leq 10$ , rises with slope 1 for $10 < S \leq 20$ , and is constant at 10 for $S > 20$ . This matches the difference of two max functions.

Exam route: Test the options at key points. At $S = 15$ , the graph shows $N (15) = 5$ . Option A: $H_{10} (15) - H_{20} (15) = max (5, 0) - max (- 5, 0) = 5 - 0 = 5$ . Matches. At $S = 25$ , graph is 10. Option A: $H_{10} (25) - H_{20} (25) = 15 - 5 = 10$ . Matches perfectly.

Learning route:

Step 1: Analyze the graph to identify the corner points and slopes. The function is 0 until $S = 10$ .

Step 2: From $S = 10$ to $S = 20$ , the function rises linearly from 0 to 10. The slope is $\frac{10 - 0}{20 - 10} = 1$ .

Step 3: For $S > 20$ , the function remains constant at 10.

Step 4: Recall that $H_{K} (S) = max (S - K, 0)$ is 0 for $S \leq K$ and rises with slope 1 for $S > K$ .

Step 5: To get a slope of 1 starting at $S = 10$ , we add $H_{10} (S)$ .

Step 6: To flatten the slope back to 0 at $S = 20$ , we must subtract a function that starts rising with slope 1 at $S = 20$ , which is $H_{20} (S)$ .

Step 7: Therefore, $N (S) = H_{10} (S) - H_{20} (S)$ .

Question 3 · Quantitative Aptitude · 2025_Set2 MCQ

If $P e^{x} = Q e^{- x}$ for all real values of $x$ , which one of the following statements is true?

A.
$P = Q = 0$
B.
$P = Q = 1$
C.
$P = 1; Q = - 1$
D.
$\frac{P}{Q} = 0$

Correct Answer:

Step-by-Step Solution

Insight: An equation involving a variable exponent, like $P e^{2 x} = Q$ , can only hold for <i>all</i> real values of $x$ if the coefficient of the variable term is zero.

Exam route: Multiply both sides by $e^{x}$ to get $P e^{2 x} = Q$ . Since $e^{2 x}$ varies with $x$ and is not constant, the only way this equality holds for all $x$ is if $P = 0$ . If $P = 0$ , then $Q$ must also be 0.

Learning route:

Step 1: Start with the given equation: $P e^{x} = Q e^{- x}$ .

Step 2: Multiply both sides by $e^{x}$ (which is never zero) to eliminate the negative exponent: $P e^{2 x} = Q$ .

Step 3: Analyze the condition "for all real values of $x$ ". The term $e^{2 x}$ is a strictly increasing function that takes all positive real values.

Step 4: If $P \neq = 0$ , then $e^{2 x} = Q / P$ , which implies $e^{2 x}$ is a constant. This is a contradiction because $e^{2 x}$ varies with $x$ .

Step 5: Therefore, we must have $P = 0$ .

Step 6: Substitute $P = 0$ back into the equation: $0 \cdot e^{2 x} = Q ⟹ 0 = Q$ .

Step 7: Thus, the only solution that satisfies the condition for all $x$ is $P = 0$ and $Q = 0$ .

Question 4 · Quantitative Aptitude · 2025_Set2 MCQ

Let $p_{1}$ and $p_{2}$ denote two arbitrary prime numbers. Which one of the following statements is correct for all values of $p_{1}$ and $p_{2}$ ?

A.
$p_{1} + p_{2}$ is not a prime number.
B.
$p_{1} p_{2}$ is not a prime number.
C.
$p_{1} + p_{2} + 1$ is a prime number.
D.
$p_{1} p_{2} + 1$ is a prime number.

Correct Answer:

Step-by-Step Solution

Insight: The product of any two prime numbers is always a composite number because it has at least four distinct factors: 1, $p_{1}$ , $p_{2}$ , and $p_{1} p_{2}$ .

Exam route: Test the edge case where one prime is 2 (the only even prime). For $p_{1} = 2, p_{2} = 3$ : $p_{1} + p_{2} = 5$ (prime, so A is false). $p_{1} + p_{2} + 1 = 6$ (not prime, so C is false). $p_{1} p_{2} + 1 = 7$ (prime, but test $p_{1} = 3, p_{2} = 5 ⟹ 16$ , not prime, so D is false). Option B is universally true by definition of primes.

Learning route:

Step 1: Analyze Option A: $p_{1} + p_{2}$ . If $p_{1} = 2$ and $p_{2} = 3$ , the sum is 5, which is prime. Thus, A is not true for all values.

Step 2: Analyze Option B: $p_{1} p_{2}$ . By definition, a prime number has exactly two distinct positive divisors: 1 and itself. The product $p_{1} p_{2}$ has at least the divisors 1, $p_{1}$ , $p_{2}$ , and $p_{1} p_{2}$ . Since $p_{1}, p_{2} \geq 2$ , these are at least three distinct divisors (four if $p_{1} \neq = p_{2}$ ). Therefore, $p_{1} p_{2}$ is always composite, never prime.

Step 3: Analyze Option C: $p_{1} + p_{2} + 1$ . If $p_{1} = 2, p_{2} = 3$ , the result is 6, which is not prime. Thus, C is false.

Step 4: Analyze Option D: $p_{1} p_{2} + 1$ . If $p_{1} = 2, p_{2} = 3$ , the result is 7 (prime). However, if $p_{1} = 3, p_{2} = 5$ , the result is 16 (not prime). Thus, D is not true for all values.

Step 5: Conclude that Option B is the only statement that holds for all arbitrary prime numbers.

Question 5 · Quantitative Aptitude · 2024_Set2 MCQ

For positive non-zero real variables $x$ and $y$ , if

$ln (\frac{x + y}{2}) = \frac{1}{2} [ln (x) + ln (y)]$

then, the value of

$\frac{x}{y} + \frac{y}{x}$

A.
1
B.
$\frac{1}{2}$
C.
2
D.
4

Correct Answer:

Step-by-Step Solution

Insight: The equation $ln (\frac{x + y}{2}) = \frac{1}{2} [ln (x) + ln (y)]$ is the logarithmic form of the AM-GM equality condition.

Exam route: Simplify the RHS to $ln (x y)$ . Equate arguments: $\frac{x + y}{2} = x y$ . This implies $x = y$ . Thus, $\frac{x}{y} + \frac{y}{x} = 1 + 1 = 2$ .

Learning route:

Step 1: Use logarithm properties on the right side: $\frac{1}{2} [ln (x) + ln (y)] = \frac{1}{2} ln (x y) = ln ((x y)^{1/2}) = ln (x y)$ .

Step 2: The equation becomes $ln (\frac{x + y}{2}) = ln (x y)$ .

Step 3: Since the natural logarithm is a one-to-one function, we can equate the arguments: $\frac{x + y}{2} = x y$ .

Step 4: Recognize this as the condition where the Arithmetic Mean (AM) equals the Geometric Mean (GM).

Step 5: AM = GM holds for positive real numbers if and only if the variables are equal, so $x = y$ .

Step 6: Substitute $x = y$ into the target expression: $\frac{x}{x} + \frac{x}{x} = 1 + 1 = 2$ .

Question 6 · Quantitative Aptitude · 2024_Set1 MCQ

For positive non-zero real variables

p

and

q

, if

lo g (p^{2} + q^{2}) = lo g p + lo g q + 2 lo g 3,

then, the value of

\frac{p ^{4} + q ^{4}}{p ^{2} q ^{2}}

A.
$79$
B.
$81$
C.
$9$
D.
$83$

Correct Answer:

Step-by-Step Solution

Insight: Use logarithmic identities to combine the right side into a single logarithm, then equate the arguments to get an algebraic equation relating $p$ and $q$ .

Exam route: $lo g (p^{2} + q^{2}) = lo g (pq) + lo g (9) = lo g (9 pq)$ . Equating arguments gives $p^{2} + q^{2} = 9 pq$ . Square both sides: $(p^{2} + q^{2})^{2} = 81 p^{2} q^{2} ⟹ p^{4} + 2 p^{2} q^{2} + q^{4} = 81 p^{2} q^{2} ⟹ p^{4} + q^{4} = 79 p^{2} q^{2}$ . Divide by $p^{2} q^{2}$ to get 79.

Learning route:

Step 1: Simplify the right side of the given equation using logarithmic identities.

Step 2: $lo g p + lo g q = lo g (pq)$ (Product rule).

Step 3: $2 lo g 3 = lo g (3^{2}) = lo g 9$ (Power rule).

Step 4: Combine them: $lo g (pq) + lo g 9 = lo g (9 pq)$ .

Step 5: The equation is now $lo g (p^{2} + q^{2}) = lo g (9 pq)$ .

Step 6: Since the logarithm function is one-to-one, we can equate the arguments: $p^{2} + q^{2} = 9 pq$ .

Step 7: We need to find the value of $\frac{p ^{4} + q ^{4}}{p ^{2} q ^{2}}$ . To get fourth powers, square the equation from Step 6.

Step 8: $(p^{2} + q^{2})^{2} = (9 pq)^{2} ⟹ p^{4} + 2 p^{2} q^{2} + q^{4} = 81 p^{2} q^{2}$ .

Step 9: Isolate $p^{4} + q^{4}$ by subtracting $2 p^{2} q^{2}$ from both sides: $p^{4} + q^{4} = 79 p^{2} q^{2}$ .

Step 10: Divide both sides by $p^{2} q^{2}$ (which is valid since $p, q$ are positive non-zero): $\frac{p ^{4} + q ^{4}}{p ^{2} q ^{2}} = 79$ .

Question 7 · Quantitative Aptitude · 2024_Set1 MCQ

If two distinct non-zero real variables $x$ and $y$ are such that $(x + y)$ is proportional to $(x - y)$ then the value of $\frac{x}{y}$

A.
depends on $x y$
B.
depends only on $x$ and not on $y$
C.
depends only on $y$ and not on $x$
D.
is a constant

Correct Answer:

Step-by-Step Solution

Insight: Proportionality means the ratio of the two quantities is a constant. We can set up an equation with a constant $k$ and solve for $x / y$ .

Exam route: $(x + y) = k (x - y) ⟹ x + y = k x - k y ⟹ x (1 - k) = - y (1 + k) ⟹ x / y = - (1 + k) / (1 - k)$ . Since $k$ is constant, $x / y$ is constant.

Learning route:

Step 1: Translate "proportional to" into an equation: $x + y = k (x - y)$ for some constant $k$ .

Step 2: Expand the right side: $x + y = k x - k y$ .

Step 3: Group $x$ terms on one side and $y$ terms on the other: $x - k x = - y - k y$ .

Step 4: Factor out $x$ and $y$ : $x (1 - k) = - y (1 + k)$ .

Step 5: Solve for the ratio $x / y$ : $\frac{x}{y} = \frac{- ( 1 + k )}{1 - k} = \frac{k + 1}{k - 1}$ .

Step 6: Since $k$ is a fixed constant of proportionality, the expression $\frac{k + 1}{k - 1}$ is also a fixed constant. Thus, $x / y$ is a constant.

Question 8 · Quantitative Aptitude · 2023 MCQ

Consider two functions of time

(t)

f (t) = 0.01 t^{2}

g (t) = 4 t

where

0&lt;t&lt;\infty

.

Now consider the following two statements:

(i) For some

t&gt;0

g(t)&gt;f(t)

.
(ii) There exists a

T

, such that

f(t)&gt;g(t)

for all

t&gt;T

.

Which one of the following options is TRUE?

A.
only (i) is correct
B.
only (ii) is correct
C.
both (i) and (ii) are correct
D.
neither (i) nor (ii) is correct

Correct Answer:

Step-by-Step Solution

Insight: Compare the growth rates of a quadratic and a linear function; the linear function wins initially, but the quadratic eventually dominates.

Exam route: Set $f (t) = g (t)$ to find the crossover point $t = 400$ . For $0 < t < 400$ , $g (t) > f (t)$ , proving (i). For $t > 400$ , $f (t) > g (t)$ , proving (ii) with $T = 400$ .

Learning route:

Analyze the inequality for statement (i): $g (t) > f (t) ⟹ 4 t > 0.01 t^{2} ⟹ t (400 - t) > 0$ . Since $t > 0$ , this holds for $0 < t < 400$ . Thus, statement (i) is true.

Next, analyze the inequality for statement (ii): $f (t) > g (t) ⟹ 0.01 t^{2} > 4 t ⟹ t (t - 400) > 0$ . For $t > 400$ , this holds. Thus, choosing $T = 400$ satisfies statement (ii).

Both statements are correct.

Question 9 · Quantitative Aptitude · 2023 MCQ

A series of natural numbers

F_{1}, F_{2}, F_{3}, F_{4}, F_{5}, F_{6}, F_{7}, \dots

obeys

F_{n + 1} = F_{n} + F_{n - 1}

for all integers

n \geq 2

.

If

F_{6} = 37

, and

F_{7} = 60

, then what is

F_{1}

A.
4
B.
5
C.
8
D.
9

Correct Answer:

Step-by-Step Solution

Insight: To find earlier terms in a recursive sequence when later terms are known, rearrange the recurrence relation to step backwards one index at a time.

Exam route: The rule is $F_{n + 1} = F_{n} + F_{n - 1}$ , which rearranges to $F_{n - 1} = F_{n + 1} - F_{n}$ . Given $F_{7} = 60$ and $F_{6} = 37$ : $F_{5} = 60 - 37 = 23$ . $F_{4} = 37 - 23 = 14$ . $F_{3} = 23 - 14 = 9$ . $F_{2} = 14 - 9 = 5$ . $F_{1} = 9 - 5 = 4$ .

Learning route:

Step 1: Identify the forward recurrence relation: $F_{n + 1} = F_{n} + F_{n - 1}$ .

Step 2: Rearrange the formula to solve for the oldest term: $F_{n - 1} = F_{n + 1} - F_{n}$ .

Step 3: Use the given values $F_{7} = 60$ and $F_{6} = 37$ to find $F_{5}$ . Set $n = 6$ : $F_{5} = F_{7} - F_{6} = 60 - 37 = 23$ .

Step 4: Find $F_{4}$ using $F_{6}$ and $F_{5}$ . Set $n = 5$ : $F_{4} = F_{6} - F_{5} = 37 - 23 = 14$ .

Step 5: Find $F_{3}$ using $F_{5}$ and $F_{4}$ . Set $n = 4$ : $F_{3} = F_{5} - F_{4} = 23 - 14 = 9$ .

Step 6: Find $F_{2}$ using $F_{4}$ and $F_{3}$ . Set $n = 3$ : $F_{2} = F_{4} - F_{3} = 14 - 9 = 5$ .

Step 7: Find $F_{1}$ using $F_{3}$ and $F_{2}$ . Set $n = 2$ : $F_{1} = F_{3} - F_{2} = 9 - 5 = 4$ .

Question 10 · Quantitative Aptitude · 2023 MCQ

$f (x)$ and $g (y)$ are functions of $x$ and $y$ , respectively, and $f (x) = g (y)$ for all real values of $x$ and $y$ . Which one of the following options is necessarily TRUE for all $x$ and $y$ ?

A.
$f (x) = 0$ and $g (y) = 0$
B.
$f (x) = g (y) = constant$
C.
$f (x) \neq = constant$ and $g (y) \neq = constant$
D.
$f (x) + g (y) = f (x) - g (y)$

Correct Answer:

Step-by-Step Solution

Insight: If a function of $x$ equals a function of $y$ for all independent real values of $x$ and $y$ , both functions must be equal to the same constant.

Exam route: Fix $y = y_{0}$ . Then $f (x) = g (y_{0})$ for all $x$ , meaning $f (x)$ is constant. Similarly, fix $x = x_{0}$ to show $g (y)$ is constant. Thus, $f (x) = g (y) = constant$ .

Learning route:

Step 1: We are given $f (x) = g (y)$ for all real $x$ and $y$ .

Step 2: Choose an arbitrary but fixed value for $y$ , say $y = c$ .

Step 3: The equation becomes $f (x) = g (c)$ for all $x$ . Since $g (c)$ is just a number, $f (x)$ must be a constant function.

Step 4: Similarly, choose a fixed value for $x$ , say $x = d$ . The equation becomes $g (y) = f (d)$ for all $y$ , meaning $g (y)$ is also a constant function.

Step 5: Since they are equal to each other, they must be the same constant. Therefore, $f (x) = g (y) = constant$ .

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Algebra, Equations, Functions and Sequences PYQs for GATE CS

Chapter Roadmap: Algebra, Equations, Functions and Sequences

Chapter Journey

Number Properties and Proportional Relationships

Number Properties and Proportional Relationships

Algebra, Equations, Functions and Sequences: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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