GATE 2026 | Electrical Engineering | Solved Question Paper | Exam Date : 14.02.2026

Q.1. The shopkeeper sells lemons. In this sentence, the word “lemons” is the:

(a) object
(b) subject
(c) predicate
(d) verb

Ans. (a) Object

Explanation: A sentence mainly consists of three parts: subject, verb, and object. The subject is the doer of the action, which is “The shopkeeper.” The verb shows the action being performed, which is “sells.” The object is the person or thing that receives the action of the verb. Since “lemons” are being sold, they receive the action of the verb. Therefore, “lemons” is the object of the sentence.

Q.2. The figure below is supposed to show three non-overlapping shapes—one oval and two triangles. Which one of the following figures P, Q, R, or S fits the missing portion indicated by ‘?’ and completes the oval and the two triangles?

(a) P
(b) Q
(c) R
(d) S

Ans. (a) P

Explanation:
To solve this figure completion problem, we carefully observe the shapes that must be completed:

The Oval: The given figure already shows part of an oval, but the right curved portion is missing. Therefore, the required piece must contain a smooth curved edge that completes the oval.
The Two Triangles: Parts of two triangles are visible in the figure. The missing piece must include straight diagonal edges that connect correctly to form two distinct triangles without overlapping.
Why Option P is Correct: Figure P has the correct combination of one curved edge (to complete the oval) and straight edges (to complete both triangles). Its shape aligns perfectly with the existing outlines and completes all three shapes neatly.

Hence, option (a) P is the correct answer.

Q.3. At how many points will the curves $ y = x^2 $ and $ y = -x^2 – 2x – 5 $ intersect in the real $ (x, y) $ plane?

(a) 0

(b) 1

(d) 3

Ans. (a) 0

Explanation:

Two curves intersect at points where the same value of $x$ gives the same value of $y$.

First curve:
For $ y = x^2 $, the value of $y$ is always greater than or equal to zero for all real values of $x$.
Hence, this curve lies on or above the $x$-axis.

Second curve:
\[
y = -x^2 – 2x – 5 = -\left[(x+1)^2 + 4\right]
\]
Since $ (x+1)^2 \ge 0 $, the expression inside the bracket is always at least $4$.
Therefore, $ y \le -4 $ for all real values of $x$, and this curve lies entirely below the $x$-axis.

Because one curve is always non-negative and the other is always negative, they never intersect. Hence, the number of points of intersection is 0.

Q.4. If Anish had scored a hundred runs in today’s match, he would have been made the captain of his team. He would have then become the youngest captain in the history of his team. Unfortunately, he got out without scoring any runs. Hence, there will be no change in the captaincy for now.

Based on the paragraph above, which one of the following statements is true?

(a) Anish made a hundred runs but was denied captaincy.

(b) Anish was the captain of his team before today’s match.

(d) Anish is the youngest player in his team.

Ans. (c)

Explanation: The passage states that if Anish had scored a hundred runs, he would have become the youngest captain in the team’s history. Since he failed to score the runs, there was no change in captaincy. This means the present captain continues in the role. Because Anish would have been the youngest captain, it logically follows that the current captain must be older than Anish.

Q.5. Which one of the following figures P, Q, R, or S, correctly shows the $45^\circ$ clockwise-rotated version of figure (I)?

(a) P

(b) Q

(d) S

Ans. (b) Q

Explanation:

To answer this question, we must visualize the rotation of figure (I) by $45^\circ$ in the clockwise direction.

Clockwise rotation: The figure is turned to the right, similar to rotating a clock hand.
The $45^\circ$ angle: A $90^\circ$ rotation would change horizontal lines into vertical ones. Since $45^\circ$ is exactly half of $90^\circ$, all horizontal and vertical lines in the original figure become diagonal.
Comparison of options: On carefully comparing the arrangement of black blocks after rotation, only figure Q preserves the original pattern while correctly aligning it at a $45^\circ$ clockwise tilt.

Q.6. Match the words in Column I with their synonyms in Column II.

Column I	Column II
(i) Lonely	(p) Verbatim
(ii) Literal	(q) Solitary
(iii) Lousy	(r) Deadly
(iv) Lethal	(s) Terrible

Options:

(a) (i)-(q); (ii)-(p); (iii)-(s); (iv)-(r)
(b) (i)-(q); (ii)-(s); (iii)-(r); (iv)-(p)
(c) (i)-(s); (ii)-(p); (iii)-(q); (iv)-(r)
(d) (i)-(r); (ii)-(s); (iii)-(p); (iv)-(q)

Answer: (a)

Explanation:

Lonely → Solitary: Both describe being alone.
Literal → Verbatim: Both imply exact wording or meaning.
Lousy → Terrible: Both indicate very poor quality.
Lethal → Deadly: Both mean capable of causing death.

Hence, the correct matching is (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r), which corresponds to option (a).

Q.7. In the given figure, $ \overline{PQ} $ is the diameter of a circle with center $ O $. Two points $ R $ and $ S $ are chosen on the circle such that $ \angle ROS = 80^\circ $. When $ \overline{PR} $ and $ \overline{QS} $ are extended, they meet at $ T $. The value of $ \angle RTS $ is __________.

(a) $40^\circ$

(b) $50^\circ$

(d) $80^\circ$

Ans. (b) $50^\circ$

Explanation:

Step 1: Analyze the Central Triangle
In $\triangle ROS$, $OR = OS$ (radii of the same circle). Therefore, $\triangle ROS$ is an isosceles triangle. Since $\angle ROS = 80^\circ$, the base angles are:

$\angle ORS = \angle OSR = \frac{180^\circ – 80^\circ}{2} = 50^\circ$
Step 2: Use the Straight Line Property
Since $PQ$ is a diameter, $POQ$ is a straight line, meaning the sum of angles around point $O$ on one side is $180^\circ$:

$\angle POR + \angle ROS + \angle SOQ = 180^\circ$

Substituting the known value:

$\angle POR + 80^\circ + \angle SOQ = 180^\circ \implies \angle POR + \angle SOQ = 100^\circ$
Step 3: Relate to the Large Triangle $\triangle PTQ$
In $\triangle OPR$ and $\triangle OQS$, we again have isosceles triangles ($OP=OR$ and $OQ=OS$). The base angles (which are also the interior angles of our target triangle $\triangle PTQ$) are:

$\angle RPQ = \frac{180^\circ – \angle POR}{2}$ and $\angle SQP = \frac{180^\circ – \angle SOQ}{2}$

Adding these two base angles together:

$\angle RPQ + \angle SQP = \frac{360^\circ – (\angle POR + \angle SOQ)}{2}$

Substituting the result from Step 2:

$\angle RPQ + \angle SQP = \frac{360^\circ – 100^\circ}{2} = \frac{260^\circ}{2} = 130^\circ$
Step 4: Solve for $\angle RTS$
In the large triangle $\triangle PTQ$, the sum of all interior angles is $180^\circ$:

$\angle RTS + \angle RPQ + \angle SQP = 180^\circ$

$\angle RTS + 130^\circ = 180^\circ$

$\mathbf{\angle RTS = 50^\circ}$

Q.8. Based on the relationship between each polygon and the number inside it, the value of is _______.

(a) 720

(b) 596

(d) 240

Ans. (a) 720

Explanation:

To determine the value of X, observe the pattern between the number of sides of each polygon and the number written inside it. The number inside each polygon is the factorial of the number of its sides, denoted by $ n! $.

Triangle (3 sides):
$ 3! = 3 \times 2 \times 1 = 6 $
Square (4 sides):
$ 4! = 4 \times 3 \times 2 \times 1 = 24 $
Pentagon (5 sides):
$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $
Hexagon (6 sides):
Following the same pattern,

$ X = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 $

Therefore, the value of X is
$720$, which corresponds to option (a).

Q.9. Consider a linear arrangement of seven bulbs, each of which can be in the ON or OFF state. The initial configuration of the bulbs is shown in the figure. In every step, the states of the bulbs are changed based on the following rules:

Any OFF bulb with exactly one ON neighbor at the end of the previous step is turned ON.
Any ON bulb with both neighbors ON at the end of the previous step is turned OFF.
The state of any bulb not meeting the above conditions remains unchanged.

The states of the bulbs at the end of Step 1 and Step 2 are also shown in the figure. The number of bulbs that are ON at the end of Step 8 is _______.

(a) $5$

(b) $4$

(d) $0$

Ans. (b) $4$

Explanation:

Let an ON bulb be represented by $1$ and an OFF bulb by $0$. We apply the given rules step by step.

Initial state:
$0\ 0\ 0\ 1\ 0\ 0\ 0$
(Only the 4^th bulb is ON)
Step 1:
The 3^rd and 5^th bulbs turn ON.Result: $0\ 0\ 1\ 1\ 1\ 0\ 0$ (3 bulbs ON)
Step 2:
The 2^nd and 6^th bulbs turn ON, while the 4^th bulb turns OFF.Result: $0\ 1\ 1\ 0\ 1\ 1\ 0$ (4 bulbs ON)
Step 3:
The end bulbs turn ON.Result: $1\ 1\ 1\ 0\ 1\ 1\ 1$ (6 bulbs ON)
Step 4:
The 2^nd and 6^th bulbs turn OFF.Result: $1\ 0\ 1\ 0\ 1\ 0\ 1$ (4 bulbs ON)
Steps 5 to 8:
In this configuration, no bulb satisfies the conditions for change.
Hence, the pattern becomes stable and remains unchanged.

Therefore, at the end of Step 8, the number of bulbs that are ON is $4$.

Q.10. $P$ and $Q$ are two positive integers such that $ P^2 = Q^2 + 13 $. The product of the numbers $P$ and $Q$ is _________.

(a) $13$

(b) $26$

(d) $42$

Ans. (d) $42$

Explanation:

Starting from the given relation,
$ P^2 = Q^2 + 13 $.

Rearranging the equation:
$ P^2 – Q^2 = 13 $
Using the identity:
$ a^2 – b^2 = (a-b)(a+b) $
Hence,
$ (P – Q)(P + Q) = 13 $
Factor analysis:
Since $13$ is a prime number and $P, Q$ are positive integers,
the only possible factor pair is:$ P – Q = 1 $ and $ P + Q = 13 $
Solving the equations:
Adding the two equations:
$ (P – Q) + (P + Q) = 1 + 13 $
$ 2P = 14 \Rightarrow P = 7 $
Substituting back:
$ Q = 6 $
Final product:
$ P \times Q = 7 \times 6 = 42 $

Therefore, the product of $P$ and $Q$ is
$42$.

Q.11. Consider the infinite-length, discrete-time sequence $ x[n] = 0.9^{|n|} $, where $ n $ is an integer. The region of convergence (ROC) of its Z-transform $ X(z) $ is given by:

(Note: $ z $ is a complex variable)

(a) $ |z| > 0.9 $

(b) $ |z| < 0.9 $

(d) $ \{z : |z| < 0.9\} \cup \{z : |z| > \dfrac{1}{0.9}\} $

Ans. (c)

Explanation:

To determine the Region of Convergence (ROC), we split the sequence
$ x[n] = 0.9^{|n|} $ into two parts based on the value of $ n $.

Right-sided part ($ n \ge 0 $):
Here, $ x[n] = (0.9)^n u(n) $.
This is a right-sided sequence, whose Z-transform converges for
$ |z| > 0.9 $.
Left-sided part ($ n < 0 $):
Here, $ x[n] = (0.9)^{-n} = \left(\dfrac{1}{0.9}\right)^n u(-n-1) $.
This is a left-sided sequence, whose Z-transform converges for
$ |z| < \dfrac{1}{0.9} $.
Overall ROC:
The ROC of the complete sequence is the intersection of the two regions.
Hence, it is the annular region between the two circles.

Therefore, the ROC is
$ 0.9 < |z| < \dfrac{1}{0.9} $.

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