Quantitative Aptitude PYQs for GATE CS

GATE CS Quantitative Aptitude: 2 units and 6 chapters, weightage from 33 previous year questions across 10 papers, a study order by exam weight and 540 practice questions.

About Quantitative Aptitude Previous Year Questions (PYQs)

33 previous year questions from Quantitative Aptitude in GATE CS, grouped by chapter with the exam year, answer key and step-by-step solution for each.

GATE CS Quantitative Aptitude Unit-wise Weightage from Past Papers

We counted every GATE CS Quantitative Aptitude previous year question in our bank (33 questions from 10 papers) and grouped them by unit.

Unit	Chapters	PYQs	Share of section	Avg per paper
Numerical computation and estimation	4	28	85%	2.8
Data interpretation	2	5	15%	0.5

Suggested Quantitative Aptitude Study Order for GATE CS

Numerical computation and estimation: 85% of past Quantitative Aptitude questions, about 2.8 per paper.
Data interpretation: 15% of past Quantitative Aptitude questions, about 0.5 per paper.

Start where the marks are. Units at the top of this list have appeared most often in past GATE CS papers.

Units in GATE CS Quantitative Aptitude

All Quantitative Aptitude chapters

One Solved Question from Each Quantitative Aptitude 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.

Question 5 · Tables and Structured Data · 2025_Set2 MCQ

Which one of the following options is correct for the given data in the table?

i

)0123Input (

I

)20−41015Output (

X

)20162641Output (

Y

)20−80−800−12000

A.
$X (i) = X (i - 1) + I (i); Y (i) = Y (i - 1) I (i); i > 0$
B.
$X (i) = X (i - 1) I (i); Y (i) = Y (i - 1) + I (i); i > 0$
C.
$X (i) = X (i - 1) I (i); Y (i) = Y (i - 1) I (i); i > 0$
D.
$X (i) = X (i - 1) + I (i); Y (i) = Y (i - 1) I (i - 1); i > 0$

Correct Answer:

Step-by-Step Solution

Insight: Test the given recurrence relations step-by-step for $i = 1$ and $i = 2$ using the table values to eliminate incorrect options.

Exam route: For $i = 1$ , $X (1) = X (0) + I (1) = 20 + (- 4) = 16$ . $Y (1) = Y (0) \times I (1) = 20 \times (- 4) = - 80$ . This matches Option A. Verify for $i = 2$ : $X (2) = 16 + 10 = 26$ , $Y (2) = - 80 \times 10 = - 800$ . Matches perfectly.

Learning route:

Observe the table: We have sequences for Input $I$ , Output $X$ , and Output $Y$ over iterations $i = 0, 1, 2, 3$ .
Test Option A for $i = 1$ :

$X (1) = 20 + (- 4) = 16$ . (Matches table)

$Y (1) = 20 \times (- 4) = - 80$ . (Matches table)

Test Option A for $i = 2$ :

$X (2) = 16 + 10 = 26$ . (Matches table)

$Y (2) = - 80 \times 10 = - 800$ . (Matches table)

Test Option A for $i = 3$ :

$X (3) = 26 + 15 = 41$ . (Matches table)

$Y (3) = - 800 \times 15 = - 12000$ . (Matches table)

Conclusion: Option A is the only recurrence relation that consistently reproduces the table's values.

Question 6 · Charts and Graphical Data Interpretation · 2024_Set2 MCQ

The pie charts depict the shares of various power generation technologies in the total electricity generation of a country for the years 2007 and 2023.

The renewable sources of electricity generation consist of Hydro, Solar and Wind. Assuming that the total electricity generated remains the same from 2007 to 2023, what is the percentage increase in the share of the renewable sources of electricity generation over this period?

A.
25%
B.
50%
C.
77.5%
D.
62.5%

Correct Answer:

Step-by-Step Solution

Insight: Since total electricity generated remains constant across both years, percentage shares can be used directly in the percentage change formula without converting to absolute values.

Exam route: Sum the renewable percentages for each year, then apply (new − old)/old × 100.

Learning route:

Step 1: Identify renewable sources from the question statement: Hydro, Solar, and Wind.

Step 2: Sum renewable share for 2007 from the pie chart: Hydro 30% + Solar 5% + Wind 5% = 40%.

Step 3: Sum renewable share for 2023 from the pie chart: Hydro 35% + Solar 20% + Wind 10% = 65%.

Step 4: Since total generation is constant, the percentage increase in share equals the percentage increase in actual generation. The old value is the anchor denominator.

Step 5: Apply the percentage change formula:

$Percentage Change = (\frac{65 - 40}{40}) \times 100 = \frac{25}{40} \times 100 = 62.5%$

Verification: If total generation = 100 units, renewables went from 40 to 65 units. $(65 - 40) /40 = 0.625 = 62.5%$ . ✓

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Quantitative Aptitude PYQs for GATE CS

About Quantitative Aptitude Previous Year Questions (PYQs)

GATE CS Quantitative Aptitude Unit-wise Weightage from Past Papers

Suggested Quantitative Aptitude Study Order for GATE CS

Units in GATE CS Quantitative Aptitude

All Quantitative Aptitude chapters

One Solved Question from Each Quantitative Aptitude Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution