Algebra, Equations, Functions and Sequences Practice Questions for GATE CS

Solve 197+ Algebra, Equations, Functions and Sequences practice questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Algebra, Equations, Functions and Sequences

Chapter Journey

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 MCQ

How many pairs of positive integers $(x, y)$ satisfy the equation $lo g_{10} (\frac{x + y}{2}) = \frac{1}{2} (lo g_{10} x + lo g_{10} y)$ and the condition $x + y \leq 12$ ?

A.
4
B.
5
C.
6
D.
11

Correct Answer:

Step-by-Step Solution

Key idea: Simplify the right side to recognize the AM-GM equality condition.

Step 1: Simplify the right side using log rules.

$\frac{1}{2} (lo g_{10} x + lo g_{10} y) = \frac{1}{2} lo g_{10} (x y) = lo g_{10} ((x y)^{1/2}) = lo g_{10} (x y)$

Step 2: Equate the arguments since the logs are equal.

$\frac{x + y}{2} = x y$

Step 3: Recognize this as the Arithmetic Mean (AM) equals Geometric Mean (GM) condition.

For positive real numbers, AM = GM if and only if $x = y$ .

Step 4: Apply the condition $x + y \leq 12$ .

Since $x = y$ , we have $2 x \leq 12 ⟹ x \leq 6$ .

Step 5: Count the positive integer pairs.

$x$ can be $1, 2, 3, 4, 5, 6$ . Since $y = x$ , the pairs are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ .

There are 6 such pairs.

Answer: 6

Question 2 · Quantitative Aptitude MCQ

Consider the following statements:

Assertion (A): If $\frac{x + y}{x - y} = \frac{3}{1}$ for $x \neq = y$ , then $\frac{x}{y} = 2$ .

Reason (R): The componendo and dividendo rule states that if $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ , which yields $\frac{( x + y ) + ( x - y )}{( x + y ) - ( x - y )} = \frac{3 + 1}{3 - 1} ⟹ \frac{x}{y} = 2$ .

Which of the following is correct?

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is NOT the correct explanation of A
C.
A is true but R is false
D.
Both A and R are false

Correct Answer:

Step-by-Step Solution

Insight: The componendo and dividendo rule directly transforms the given ratio into the target ratio.

Exam route: Apply the rule to $\frac{x + y}{x - y} = \frac{3}{1}$ to get $\frac{2 x}{2 y} = \frac{4}{2} ⟹ \frac{x}{y} = 2$ . Both A and R are true, and R explains A.

Learning route:

Step 1: Evaluate Assertion (A). We are given $\frac{x + y}{x - y} = 3$ . Cross-multiplying gives $x + y = 3 x - 3 y ⟹ 4 y = 2 x ⟹ \frac{x}{y} = 2$ . So, A is true.

Step 2: Evaluate Reason (R). The componendo and dividendo rule states that if $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ .

Step 3: Apply the rule to the given equation: $\frac{( x + y ) + ( x - y )}{( x + y ) - ( x - y )} = \frac{3 + 1}{3 - 1}$ .

Step 4: Simplify: $\frac{2 x}{2 y} = \frac{4}{2} ⟹ \frac{x}{y} = 2$ . This matches A, and R is the direct method used to derive it. So, R is true and correctly explains A.

Trap warning: Do not assume the rule is invalid or miscalculate the sum/difference in the numerator and denominator.

Verification: Let $x = 2, y = 1$ . Then $\frac{2 + 1}{2 - 1} = \frac{3}{1}$ . And $\frac{x}{y} = \frac{2}{1} = 2$ . The rule holds perfectly.

Question 3 · Quantitative Aptitude MCQ

If $ln (\frac{x + y}{2}) = \frac{1}{2} (ln x + ln y)$ for positive real numbers $x$ and $y$ , what is the value of $\frac{x}{y} + \frac{y}{x}$ ?

A.
1
B.
4
C.
It cannot be determined.
D.
2

Correct Answer:

Step-by-Step Solution

Key idea: The right side is the log of the geometric mean, the left side is the log of the arithmetic mean.

Step 1: Simplify the right side: $\frac{1}{2} (ln x + ln y) = \frac{1}{2} ln (x y) = ln (x y)$ .

Step 2: Equate arguments: $\frac{x + y}{2} = x y$ .

Step 3: Recognize this as Arithmetic Mean = Geometric Mean. This equality holds only when $x = y$ .

Step 4: Substitute $x = y$ into the target expression: $\frac{x}{x} + \frac{x}{x} = 1 + 1 = 2$ .

Answer: D

Question 4 · Quantitative Aptitude MCQ

Consider the following:

Assertion (A): If $lo g (x^{2} + y^{2}) = lo g x + lo g y + lo g 4$ for positive real numbers $x$ and $y$ , then $\frac{x}{y} + \frac{y}{x} = 4$ .

Reason (R): $lo g x + lo g y = lo g (x + y)$ .

Which of the following is correct?

A.
Both A and R are true, and R is the correct explanation of A
B.
Both A and R are true, but R is NOT the correct explanation of A
C.
A is true, but R is false
D.
A is false, but R is true

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate the assertion and reason separately.

Step 1: Check Reason (R)

The statement $lo g x + lo g y = lo g (x + y)$ is FALSE.

The correct identity is $lo g x + lo g y = lo g (x y)$ .

Step 2: Check Assertion (A)

Given: $lo g (x^{2} + y^{2}) = lo g x + lo g y + lo g 4$

Using the correct product rule on the right side:

$lo g x + lo g y + lo g 4 = lo g (x y) + lo g 4 = lo g (4 x y)$

Since logs are equal, arguments are equal:

$x^{2} + y^{2} = 4 x y$

Divide both sides by $x y$ :

$\frac{x ^{2}}{x y} + \frac{y ^{2}}{x y} = 4$

$\frac{x}{y} + \frac{y}{x} = 4$

So Assertion (A) is TRUE.

Step 3: Conclusion

A is true, R is false.

Answer: A is true, but R is false

Question 5 · Quantitative Aptitude MCQ

Match the logarithmic equations in List I with their simplified algebraic forms in List II.

List I

P. $lo g (a^{2} + b^{2}) = lo g a + lo g b + lo g 8$

Q. $lo g (a^{2} + b^{2}) = lo g a + lo g b + 2 lo g 2$

R. $lo g (a^{3} + b^{3}) = lo g a + lo g b + 2 lo g 3$

List II

$a^{2} + b^{2} = 8 ab$
$a^{2} + b^{2} = 4 ab$
$a^{3} + b^{3} = 9 ab$

Options:

A.
P-2, Q-1, R-3
B.
P-1, Q-2, R-3
C.
P-2, Q-3, R-1
D.
P-3, Q-1, R-2

Correct Answer:

Step-by-Step Solution

Key idea: Use log rules to combine the right side into a single log, then equate arguments.

Step 1: Simplify P.

$lo g a + lo g b + lo g 8 = lo g (ab) + lo g (8) = lo g (8 ab)$ .

Equating arguments: $a^{2} + b^{2} = 8 ab$ . (Matches 1)

Step 2: Simplify Q.

$lo g a + lo g b + 2 lo g 2 = lo g (ab) + lo g (2^{2}) = lo g (ab) + lo g (4) = lo g (4 ab)$ .

Equating arguments: $a^{2} + b^{2} = 4 ab$ . (Matches 2)

Step 3: Simplify R.

$lo g a + lo g b + 2 lo g 3 = lo g (ab) + lo g (3^{2}) = lo g (ab) + lo g (9) = lo g (9 ab)$ .

Equating arguments: $a^{3} + b^{3} = 9 ab$ . (Matches 3)

Step 4: Match the pairs.

P-1, Q-2, R-3.

Answer: P-1, Q-2, R-3

Question 6 · Quantitative Aptitude MCQ

Consider the following Assertion and Reason:

Assertion (A): If $f (x) = g (y)$ for all real values of $x$ and $y$ , then $f (x)$ must be a constant function.

Reason (R): Because $x$ and $y$ are independent variables, changing $x$ would change $f (x)$ unless $f (x)$ is constant, which would violate the equality since $g (y)$ would remain unchanged.

Which of the following is correct?

A.
Both A and R are true and R is the correct explanation of A.
B.
Both A and R are true but R is NOT the correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: If an equation holds for all independent values of two separate variables, both sides must be constant.

Step 1: Analyze Assertion (A). If $f (x) = g (y)$ for all $x, y$ , fix $y = y_{0}$ . Then $f (x) = g (y_{0})$ for all $x$ , meaning $f (x)$ is a constant. Thus, A is true.

Step 2: Analyze Reason (R). The reasoning correctly explains that since $x$ and $y$ vary independently, a non-constant $f (x)$ would change while $g (y)$ stays fixed, breaking the equality. Thus, R is true and correctly explains A.

Answer: A

Question 7 · Quantitative Aptitude MCQ

If $f (x)$ and $g (y)$ are functions such that $f (x) = g (y)$ for all real values of $x$ and $y$ , and $f (1) = 7$ , what is the value of $g (10) - f (- 5)$ ?

A.
15
B.
0
C.
-15
D.
12

Correct Answer:

Step-by-Step Solution

Key idea: If $f (x) = g (y)$ for all independent $x$ and $y$ , both functions must be the same constant.

Step 1: Since $f (x) = g (y)$ for all $x, y$ , $f (x)$ cannot depend on $x$ . Thus, $f (x) = c$ for all $x$ .

Step 2: Given $f (1) = 7$ , the constant $c = 7$ . So $f (x) = 7$ and $g (y) = 7$ for all inputs.

Step 3: Calculate $g (10) - f (- 5) = 7 - 7 = 0$ .

Answer: B

Question 8 · Quantitative Aptitude MCQ

Consider the functions $f (t) = t^{2}$ and $g (t) = 100 t$ for $t > 0$ . Which of the following statements is true?

A.
$f (t) > g (t)$ for all $t > 0$
B.
$f (t) < g (t)$ for all $t > 0$
C.
$f (t) > g (t)$ for all $t > 100$
D.
$f (t) < g (t)$ for all $t > 100$

Correct Answer:

Step-by-Step Solution

Key idea: Quadratic growth eventually dominates linear growth, but you must find the exact crossover point.

Step 1: Set $f (t) = g (t)$ to find the crossover: $t^{2} = 100 t ⟹ t (t - 100) = 0$ .

Step 2: The positive crossover is at $t = 100$ .

Step 3: For $t > 100$ , $t^{2} > 100 t$ , so $f (t) > g (t)$ .

Answer: C

Question 9 · Quantitative Aptitude MCQ

Let $r$ be a root of the equation $x^{2} - 3 x - 4 = 0$ . What is the maximum possible value of the expression $r^{3} - 3 r^{2} + 2 r$ ?

A.
24
B.
-6
C.
4
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: Since $r$ is a root of $x^{2} - 3 x - 4 = 0$ , we have $r^{2} - 3 r - 4 = 0$ , which means $r^{2} - 3 r = 4$ . We can factor the target expression to use this substitution.

Step 1: Factor the target expression: $r^{3} - 3 r^{2} + 2 r = r (r^{2} - 3 r) + 2 r$ .

Step 2: Substitute $r^{2} - 3 r = 4$ : $r (4) + 2 r = 6 r$ .

Step 3: Find the roots of $x^{2} - 3 x - 4 = 0$ . Factoring gives $(x - 4) (x + 1) = 0$ , so the roots are $r = 4$ and $r = - 1$ .

Step 4: Evaluate $6 r$ for both roots to find the maximum:

For $r = 4$ : $6 (4) = 24$ .
For $r = - 1$ : $6 (- 1) = - 6$ .

The maximum possible value is 24.

Answer: A

Question 10 · Quantitative Aptitude MCQ

If $2 e^{x} = 8 e^{- x}$ , which of the following statements about $x$ is true?

A.
$x > 1$
B.
$x = 0.5$
C.
$0 < x < 1$
D.
$x < 0$

Correct Answer:

Step-by-Step Solution

Key idea: Isolate the exponential term by multiplying both sides by $e^{x}$ .

Step 1: Multiply both sides by $e^{x}$ :

$2 e^{x} \cdot e^{x} = 8 e^{- x} \cdot e^{x}$

$2 e^{2 x} = 8$

Step 2: Isolate $e^{2 x}$ :

$e^{2 x} = 4$

Step 3: Take natural log of both sides:

$2 x = ln 4$

$x = \frac{l n 4}{2} = \frac{2 l n 2}{2} = ln 2$

Step 4: Estimate $ln 2 \approx 0.693$ , so $0 < x < 1$ .

Answer: $0 < x < 1$

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Algebra, Equations, Functions and Sequences Practice Questions 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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