Numerical computation and estimation PYQs for GATE CS

GATE CS Numerical computation and estimation: 4 chapters, 28 previous year questions (85% of Quantitative Aptitude), 461 practice questions and one solved question from each chapter.

About Numerical computation and estimation Previous Year Questions (PYQs)

28 previous year questions from Numerical computation and estimation in GATE CS, grouped by chapter with the exam year, answer key and step-by-step solution for each.

Numerical computation and estimation Weightage in GATE CS

Numerical computation and estimation accounts for 28 of 33 Quantitative Aptitude previous year questions in our bank (85%), about 2.8 per paper across 10 papers.

Numerical computation and estimation Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Arithmetic, Ratios, Percentages and Commercial Mathematics	Averages, Median and Central Tendency, Ratios, Proportions and Percentages, Profit, Loss and Investment Returns	8	29%	134
Algebra, Equations, Functions and Sequences	Number Properties and Proportional Relationships, Exponential and Logarithmic Identities, Recursive Sequences, Functions, Growth and Piecewise Models, Algebraic Equations and Constraint Problems	12	43%	197
Geometry, Mensuration and Area	Similarity and Area Ratios, Plane Geometry and Area Computation, Solid Geometry, Volumes and Diagonals	5	18%	82
Counting, Sets and Elementary Probability	Sets and Inclusion-Exclusion, Permutations and Arrangements, Elementary Dice Probability	3	11%	48

One Solved Question from Each Numerical computation and estimation Chapter

Question 1 · Arithmetic, Ratios, Percentages and Commercial Mathematics · 2026_Set2 MCQ

The values of Stock A and Stock B on a particular day are Rs. 50 and Rs. 80, respectively. An investor invests Rs. 100 in Stock A and Rs. 80 in Stock B. He sells all the stocks the next day when the value of Stock A is Rs. 55 and Stock B is Rs. 70. The profit made by the investor is Rs. ________

A.
0
B.
5
C.
10
D.
20

Correct Answer:

Step-by-Step Solution

Insight: Calculate the number of shares purchased for each stock using the initial investment and price, then find the total selling value.

Exam route: Shares of A = 100 / 50 = 2. Shares of B = 80 / 80 = 1. Selling value = 2 55 + 1 70 = 110 + 70 = 180. Total cost = 180. Profit = 180 - 180 = 0.

Learning route:

Step 1: Find the number of shares bought for Stock A. Investment = Rs. 100, Price = Rs. 50. Shares of A = 100 / 50 = 2.

Step 2: Find the number of shares bought for Stock B. Investment = Rs. 80, Price = Rs. 80. Shares of B = 80 / 80 = 1.

Step 3: Calculate the total selling value the next day. Price of A = Rs. 55, Price of B = Rs. 70.

Selling value of A = 2 * 55 = 110.

Selling value of B = 1 * 70 = 70.

Total selling value = 110 + 70 = 180.

Step 4: Calculate profit. Total cost = 100 + 80 = 180. Profit = Total selling value - Total cost = 180 - 180 = 0.

Trap warning: A common mistake is to just average the percentage changes or add the price differences (55 - 50 + 70 - 80 = -5) without weighting by the number of shares.

Verification: Cost = 180. Final value = 180. Profit = 0. Matches option A.

Question 2 · Algebra, Equations, Functions and Sequences · 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 3 · Geometry, Mensuration and Area · 2025_Set2 MCQ

In the given figure, PQRS is a square of side 2 cm and PLMN is a rectangle. The corner L of the rectangle is on the side QR. Side MN of the rectangle passes through the corner S of the square.

What is the area (in cm

^{2}

) of the rectangle PLMN?

Note: The figure shown is representative.

A.
$22$
B.
$2$
C.
$8$
D.
$4$

Correct Answer:

Step-by-Step Solution

Insight: The area of the rectangle PLMN is invariant and equals the area of the square PQRS due to a shared geometric relationship through triangle PLS.

Exam route: Area of rectangle = Area of square = $2^{2} = 4$ .

Learning route:

Consider $△ P L S$ . Its base is $P L$ , which is a side of the rectangle PLMN.
The vertex $S$ lies on the line $M N$ , which is parallel to $P L$ . Therefore, the perpendicular height of $△ P L S$ from $S$ to $P L$ is exactly the width of the rectangle, $L M$ .
Area of $△ P L S = \frac{1}{2} \times base \times height = \frac{1}{2} \times P L \times L M = \frac{1}{2} \times Area (P L M N)$ .
Now consider $△ P L S$ with respect to the square PQRS. Let the base be $P S$ (a side of the square, length 2). The vertex $L$ lies on $QR$ . The perpendicular distance from $L$ to the line containing $P S$ is exactly the side length of the square, which is 2.
Area of $△ P L S = \frac{1}{2} \times P S \times (distance from L to P S) = \frac{1}{2} \times 2 \times 2 = 2$ .
Equating the two expressions for the area of $△ P L S$ : $\frac{1}{2} \times Area (P L M N) = 2 ⟹ Area (P L M N) = 4$ .

Wrong path: Trying to assign coordinates to $L$ and $M$ and solving complex algebraic equations, or assuming the rectangle must be a square and guessing $22$ .

Question 4 · Counting, Sets and Elementary Probability · 2026_Set1 MCQ

An unbiased six-faced dice whose faces are marked with numbers 1, 2, 3, 4, 5, and 6 is rolled twice in succession and the number on the top face is recorded each time. The probability that the number appearing in the second roll is an integer multiple of the number appearing in the first roll is __________

A.
$\frac{1}{6}$
B.
$\frac{5}{18}$
C.
$\frac{7}{18}$
D.
$\frac{5}{6}$

Correct Answer:

Step-by-Step Solution

Insight: fix the first roll, count the multiples available to the second roll; "integer multiple" includes equality.

Exam route:

Sample space $∣ S ∣ = 6 \times 6 = 36$ ordered pairs $(a, b)$ .
For each first roll $a$ , count $b \in {1, \dots, 6}$ with $b = k a$ :
$a = 1$ : $b \in {1, 2, 3, 4, 5, 6}$ → 6
$a = 2$ : $b \in {2, 4, 6}$ → 3
$a = 3$ : $b \in {3, 6}$ → 2
$a = 4$ : $b \in {4}$ → 1
$a = 5$ : $b \in {5}$ → 1
$a = 6$ : $b \in {6}$ → 1
Favourable $= 6 + 3 + 2 + 1 + 1 + 1 = 14$ .
$P = 14/36 = 7/18$ .

Learning route:

Let the first roll be $a$ and the second be $b$ . The condition " $b$ is an integer multiple of $a$ " means $b = k a$ for some positive integer $k$ . Since $1 \leq b \leq 6$ , the multiples of $a$ not exceeding $6$ are exactly $a, 2 a, 3 a, \dots$ up to $6$ .

Build the case table:

| $a$ | Allowed $b$ | Count |

|---|---|---|

| 1 | 1, 2, 3, 4, 5, 6 | 6 |

| 2 | 2, 4, 6 | 3 |

| 3 | 3, 6 | 2 |

| 4 | 4 | 1 |

| 5 | 5 | 1 |

| 6 | 6 | 1 |

Total favourable $= 6 + 3 + 2 + 1 + 1 + 1 = 14$ . Since the two rolls are independent and ordered, $∣ S ∣ = 36$ , so

$P = \frac{14}{36} = \frac{7}{18} .$

Verification: the complement (second roll is NOT a multiple of the first) has $36 - 14 = 22$ outcomes; $22/36 = 11/18$ , and $7/18 + 11/18 = 1$ .

Wrong-path autopsy:

$1/6$ (A) counts only the $a = 1$ row (6 outcomes) and forgets the other five rows.
$5/18$ (B) drops equality — counting only <i>strict</i> multiples — giving $5 + 2 + 1 + 0 + 0 + 0 = 8$ or a similar miscount; the problem says "integer multiple", which includes $b = a$ .
$5/6$ (D) is the complement of $1/6$ , i.e. the same miscount promoted to the other side.

Generalisation: for any condition linking the first and second rolls, fix one roll and enumerate the compatible values of the other; never treat the 11 possible sums or the 21 unordered pairs as equally likely.

A question from this chapter

Login to view the complete previous-year questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Numerical computation and estimation PYQs for GATE CS

About Numerical computation and estimation Previous Year Questions (PYQs)

Numerical computation and estimation Weightage in GATE CS

Numerical computation and estimation Chapter Matrix

More from Quantitative Aptitude

One Solved Question from Each Numerical computation and estimation Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution