Quantitative Aptitude PYQs for GATE DA

GATE DA Quantitative Aptitude: 2 units and 7 chapters, weightage from 15 previous year questions across 3 papers, a study order by exam weight and 843 practice questions.

About Quantitative Aptitude Previous Year Questions (PYQs)

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

GATE DA Quantitative Aptitude Unit-wise Weightage from Past Papers

We counted every GATE DA Quantitative Aptitude previous year question in our bank (15 questions from 3 papers) and grouped them by unit.

Unit	Chapters	PYQs	Share of section	Avg per paper
Numerical Computation & Estimation	6	12	80%	4
Data Interpretation	1	3	20%	1

Suggested Quantitative Aptitude Study Order for GATE DA

Numerical Computation & Estimation: 80% of past Quantitative Aptitude questions, about 4 per paper.
Data Interpretation: 20% of past Quantitative Aptitude questions, about 1 per paper.

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

Units in GATE DA Quantitative Aptitude

All Quantitative Aptitude chapters

One Solved Question from Each Quantitative Aptitude Chapter

Question 1 · Number Systems, Digit Problems and Divisibility · 2026 MCQ

P, Q, R, S, X,

and

Y

are distinct single-digit whole numbers taking values from 0 to 9.

P Q

is a two-digit number with

Q

being in the units place and

P

in the tens place. Similarly,

R S

is a two-digit number.
It is known that

P Q

and

R S

are consecutive numbers and

(P Q)^{2} + (R S)^{2} = X Y P,

with

X Y P

being a three-digit number.
The value of

Y

is __________

A.
4
B.
5
C.
6
D.
7

Correct Answer:

Step-by-Step Solution

Insight: Let $P Q = n$ and $R S = n + 1$ (consecutive). Then $n^{2} + (n + 1)^{2} = 2 n^{2} + 2 n + 1 = X Y P$ . The units digit of this expression must equal $P$ , and $P$ is the tens digit of $n$ . This double role of $P$ is the key constraint.

Exam route:

Step 1: $2 n^{2} + 2 n + 1 \equiv P (mod 10)$ , where $n = 10 P + Q$ .

Step 2: Try $P = 1$ (since $n \leq 21$ for $X Y P$ to be 3-digit). $2 n^{2} + 2 n \equiv 0 (mod 10) ⟹ n (n + 1) \equiv 0 (mod 5) ⟹ n \equiv 0$ or $4 (mod 5)$ .

Step 3: $n \in {10, 14, 15, 19}$ . Check each:

$n = 10$ : $221 ⟹ X = 2, Y = 2$ (not distinct).
$n = 14$ : $421 ⟹ X = 4, Y = 2, P = 1, Q = 4$ . But $X = Q = 4$ (not distinct).
$n = 15$ : $481 ⟹ X = 4, Y = 8, P = 1, Q = 5$ . $R S = 16 ⟹ R = 1 = P$ (not distinct).
$n = 19$ : $761 ⟹ X = 7, Y = 6, P = 1, Q = 9$ . $R S = 20 ⟹ R = 2, S = 0$ . Digits ${1, 9, 2, 0, 7, 6}$ all distinct. ✓

Step 4: $Y = 6$ .

Learning route:

The constraint that $P$ is both the tens digit of $P Q$ and the units digit of $X Y P$ is the pivot. We express $X Y P = 100 X + 10 Y + P$ and $P Q = 10 P + Q$ . The equation $n^{2} + (n + 1)^{2} = 100 X + 10 Y + P$ forces $2 n^{2} + 2 n + 1 \equiv P (mod 10)$ . Since $n = 10 P + Q$ , we test small $P$ (as $n \leq 21$ for 3-digit result). Only $n = 19$ yields all distinct digits.

Trap: Ignoring that $P$ must be the tens digit of $P Q$ leads to $n = 20 ⟹ 841 ⟹ Y = 4$ , but $P = 2 \neq = 1$ .

Verification: $1 9^{2} + 2 0^{2} = 361 + 400 = 761$ . $X = 7, Y = 6, P = 1$ . $P Q = 19, R S = 20$ . All digits ${1, 9, 2, 0, 7, 6}$ distinct. ✓

Question 2 · Exponents, Logarithms and Infinite Series · 2026 MCQ

Consider two distinct positive real numbers

m, n

, with

m > n

.
Let

x = n^{l o g_{10} (m)}

and

y = m^{l o g_{10} (n)}

. The relation between

x

and

y

is _______.

A.
$x > y$
B.
$x < y$
C.
$x = y$
D.
$x = lo g_{10} (y)$

Correct Answer:

Step-by-Step Solution

Insight: Taking the logarithm of both expressions reveals that their log values are identical due to the commutative property of multiplication.

Exam route: Apply $lo g_{10}$ to both $x$ and $y$ . Use the power rule to bring the exponents down. Compare the results.

Learning route:

Given $x = n^{l o g_{10} m}$ and $y = m^{l o g_{10} n}$ .

Take $lo g_{10}$ of both sides for $x$ :

$lo g_{10} x = lo g_{10} (n^{l o g_{10} m}) = (lo g_{10} m) \times (lo g_{10} n)$ .

Take $lo g_{10}$ of both sides for $y$ :

$lo g_{10} y = lo g_{10} (m^{l o g_{10} n}) = (lo g_{10} n) \times (lo g_{10} m)$ .

Since multiplication is commutative, $(lo g_{10} m) \times (lo g_{10} n) = (lo g_{10} n) \times (lo g_{10} m)$ .

Therefore, $lo g_{10} x = lo g_{10} y$ .

Since the logarithm function is one-to-one, this implies $x = y$ .

The relation is $x = y$ , matching option C.

Question 3 · Percentages, Ratios and Financial Computation · 2025 MCQ

A rectangle has a length $L$ and a width $W$ , where $L > W$ . If the width, $W$ , is increased by $10%$ , which one of the following statements is correct for all values of $L$ and $W$ ?

A.
Perimeter increases by $10%$ .
B.
Length of the diagonals increases by $10%$ .
C.
Area increases by $10%$ .
D.
The rectangle becomes a square.

Correct Answer:

Step-by-Step Solution

Insight: Area is the product of length and width; a percentage change in only one dimension translates directly to the exact same percentage change in the area, regardless of the other dimension.

Exam route: Multiply the original area by the width multiplier (1.1) to get the new area. The factor is 1.1, meaning a 10% increase.

Learning route:

Identify the original dimensions: Length = $L$ , Width = $W$ . Original Area = $L \times W$ .
Identify the change: Width is increased by 10%, so the new width is $1.1 W$ . Length remains $L$ .
Calculate the new area: $New Area = L \times 1.1 W = 1.1 L W$ .
Calculate the percentage change in area:

$\frac{1.1 L W - L W}{L W} \times 100% = 0.1 \times 100% = 10%$

Verify other options to be absolutely sure:

Perimeter: Old = $2 L + 2 W$ , New = $2 L + 2.2 W$ . Increase is $0.2 W$ , which is not 10% of $2 L + 2 W$ .
Diagonal: Old = $L^{2} + W^{2}$ , New = $L^{2} + 1.21 W^{2}$ . The increase is not a flat 10%.
Square: $L > W$ , but $1.1 W$ is not guaranteed to equal $L$ .

The correct statement is that the area increases by 10%.

Question 4 · Descriptive Statistics and Data Normalization · 2026 MCQ

Five integers are picked from 0 to 20, with possible repetitions, such that their mean is 12, median is 18, and they have a single mode of 20.
Ignoring permutations, the number of ways to pick these five integers is _____

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: This is a statistical constraint satisfaction question, recognisable because it asks to reconstruct a specific data set given its mean, median, and mode.

Step 1: Use the mean to find the sum. For 5 integers with a mean of 12, the sum is $5 \times 12 = 60$ .

Step 2: Use the median to fix the middle value. Sorting the 5 integers as $x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$ , the median is $x_{3}$ . Thus, $x_{3} = 18$ .

Step 3: Use the mode to fix the highest values. The mode is 20, meaning 20 must appear more frequently than any other number. Since $x_{3} = 18$ , the only way to include 20 is if $x_{4} = 20$ and $x_{5} = 20$ .

Step 4: Find the remaining sum. We have $x_{1} + x_{2} + 18 + 20 + 20 = 60$ , which simplifies to $x_{1} + x_{2} = 2$ .

Step 5: Find integer pairs for $x_{1}$ and $x_{2}$ . Since the integers are from 0 to 20 and $x_{1} \leq x_{2} \leq 18$ , the possible pairs summing to 2 are $(0, 2)$ and $(1, 1)$ .

Step 6: Apply the "single mode" constraint. If the pair is $(1, 1)$ , the set is ${1, 1, 18, 20, 20}$ , which has two modes (1 and 20). This violates the single mode condition. If the pair is $(0, 2)$ , the set is ${0, 2, 18, 20, 20}$ , which has a single mode of 20.

Answer: There is exactly 1 valid way to pick these integers.

Question 5 · Combinatorial Counting and Bijections · 2026 NAT

The number of bijections $f (\cdot)$ from the set $S = {1, 2, 3, 4}$ to itself such that $f (f (n)) = n$ , for all $n \in S$ , is __________ . (Answer in integer)

Correct Answer:

10.00

Step-by-Step Solution

Insight: The condition $f (f (n)) = n$ defines an involution, meaning the permutation consists entirely of 1-cycles (fixed points) and 2-cycles (swaps).

Exam route: Use the involution recurrence $I (n) = I (n - 1) + (n - 1) I (n - 2)$ . With $I (1) = 1$ and $I (2) = 2$ , we get $I (3) = 2 + 2 (1) = 4$ , and $I (4) = 4 + 3 (2) = 10$ .

Learning route:

We classify the bijections by their cycle structure for $n = 4$ :

Zero swaps (4 fixed points): There is exactly $(0 4) = 1$ way.
One swap (2 fixed points): Choose 2 elements to swap out of 4. This is $(2 4) = 6$ ways.
Two swaps (0 fixed points): Choose 2 elements for the first swap ( $(2 4) = 6$ ), and the remaining 2 form the second swap ( $(2 2) = 1$ ). Since the two swaps are indistinguishable, we divide by $2!$ . This gives $\frac{6 \times 1}{2} = 3$ ways.

Total involutions = $1 + 6 + 3 = 10$ .

Common Trap: Forgetting to divide by $2!$ in the two-swap case leads to $6 \times 1 = 6$ ways, incorrectly totaling $1 + 6 + 6 = 13$ .

Verification: The recurrence $I (4) = I (3) + 3 I (2) = 4 + 3 (2) = 10$ perfectly matches the manual enumeration.

Question 6 · Geometry and Mensuration · 2026 MCQ

In the given figure,

P, Q,

and

R

are three points on a circle of radius 10 cm with

O

as its center,

\overline{P Q} = \overline{R Q}

, and

∠ P QR = 4 5^{\circ}

. The figure is representative.

The area of the shaded region

P QR O

is ______________

c m^{2}

A.
50
B.
$252$
C.
$502$
D.
100

Correct Answer:

Step-by-Step Solution

Insight: The quadrilateral PQRO is a kite with perpendicular diagonals, making its area simply half the product of the diagonals.

Exam route:

By the Central Angle Theorem, $∠ P O R = 2 \times ∠ P QR = 2 \times 4 5^{\circ} = 9 0^{\circ}$ .
$△ P O R$ is a right isosceles triangle with legs $O P = O R = 10$ . Thus, hypotenuse $P R = 1 0^{2} + 1 0^{2} = 102$ .
Since $P Q = R Q$ and $O P = O R$ , the line $O Q$ is the perpendicular bisector of $P R$ .
The area of quadrilateral $P QR O$ is $\frac{1}{2} \times diagonal_{1} \times diagonal_{2} = \frac{1}{2} \times P R \times O Q$ .
$O Q$ is a radius, so $O Q = 10$ .
Area = $\frac{1}{2} \times 102 \times 10 = 502$ .

Learning route:

Identify the given: Circle with center $O$ , radius $r = 10$ . Points $P, Q, R$ on circle. $P Q = R Q$ , $∠ P QR = 4 5^{\circ}$ .
By the Central Angle Theorem, the angle subtended by arc $P R$ at the center is $∠ P O R = 2 \times ∠ P QR = 9 0^{\circ}$ .
In $△ P O R$ , $O P = O R = 10$ (radii) and $∠ P O R = 9 0^{\circ}$ . By Pythagoras, $P R = 1 0^{2} + 1 0^{2} = 102$ .
The quadrilateral $P QR O$ is a kite because adjacent sides are equal ( $O P = O R = 10$ , and $P Q = R Q$ is given).
A key property of a kite is that its diagonals are perpendicular. Thus, $O Q ⊥ P R$ .
The area of any quadrilateral with perpendicular diagonals is $\frac{1}{2} \times d_{1} \times d_{2}$ .
Here, $d_{1} = P R = 102$ , and $d_{2} = O Q = 10$ (since $Q$ is on the circle, $O Q$ is a radius).
Area = $\frac{1}{2} \times 102 \times 10 = 502 cm^{2}$ .

Question 7 · Data Interpretation: Tables, Charts and Graphs · 2025 MCQ

4 \times 4

digital image has pixel intensities

(U)

as shown in the figure. The number of pixels with

U \leq 4

is:

A.
3
B.
8
C.
11
D.
9

Correct Answer:

Step-by-Step Solution

Key idea: This is a systematic counting question on a matrix, recognisable because it provides a grid of numerical values and asks for the count of cells satisfying a specific inequality condition.

Step 1: Identify the condition. We need to count cells where the pixel intensity U \leq 4.

Step 2: Examine each row systematically.

Row 1: 0, 1, 0, 2 (All 4 values are \leq 4) -> Count = 4
Row 2: 4, 7, 3, 3 (4, 3, 3 are \leq 4) -> Count = 3
Row 3: 5, 5, 4, 4 (4, 4 are \leq 4) -> Count = 2
Row 4: 6, 7, 3, 2 (3, 2 are \leq 4) -> Count = 2

Step 3: Sum the counts. Total = 4 + 3 + 2 + 2 = 11.

Answer: C

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Quantitative Aptitude PYQs for GATE DA

About Quantitative Aptitude Previous Year Questions (PYQs)

GATE DA Quantitative Aptitude Unit-wise Weightage from Past Papers

Suggested Quantitative Aptitude Study Order for GATE DA

Units in GATE DA 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

Step-by-Step Solution