Numerical Computation & Estimation PYQs for GATE DA

GATE DA Numerical Computation & Estimation: 6 chapters, 12 previous year questions (80% of Quantitative Aptitude), 683 practice questions and one solved question from each chapter.

About Numerical Computation & Estimation Previous Year Questions (PYQs)

12 previous year questions from Numerical Computation & Estimation in GATE DA, grouped by chapter with the exam year, answer key and step-by-step solution for each.

Numerical Computation & Estimation Weightage in GATE DA

Numerical Computation & Estimation accounts for 12 of 15 Quantitative Aptitude previous year questions in our bank (80%), about 4 per paper across 3 papers.

Numerical Computation & Estimation Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Number Systems, Digit Problems and Divisibility	Digit Constraints and Number Formation	2	17%	103
Exponents, Logarithms and Infinite Series	Exponent and Logarithm Laws, Infinite Series Summation	3	25%	167
Percentages, Ratios and Financial Computation	Percentage Change in Geometry, Investment Returns and Weighted Percentages	2	17%	114
Descriptive Statistics and Data Normalization	Mean, Median and Mode Constraints, Standardization and Z-Score Normalization	2	17%	127
Combinatorial Counting and Bijections	Digit-Based Counting with Divisibility Constraints, Bijections and Self-Inverse Mappings	2	17%	112
Geometry and Mensuration	Circle Geometry and Chord-Angle Relations	1	8%	60

One Solved Question from Each Numerical Computation & Estimation 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 = $\frac{1}{2} \times 102 \times 10 = 502 cm^{2}$ .

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Numerical Computation & Estimation PYQs for GATE DA

About Numerical Computation & Estimation Previous Year Questions (PYQs)

Numerical Computation & Estimation Weightage in GATE DA

Numerical Computation & Estimation Chapter Matrix

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One Solved Question from Each Numerical Computation & Estimation Chapter

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