PAPERS · ISRO

ISRO Scientist/Engineer Computer Science 2023 question paper

All 95 questions of ISRO's written test for Scientist/Engineer 'SC' (Computer Science), held on 7 January 2024, with ISRO's answer key. ISRO excluded Q.4, Q.5, Q.12, Q.23, Q.28 and Q.46 from evaluation.

Answers from ISRO's revised and final answer key of 16 February 2024.

The question paper (PDF, ISRO)ISRO’s answer key (PDF)

questions · 100 marks: 95questions · 100 marks
Part A: Computer Science: 80Part A: Computer Science
Part B: aptitude: 15Part B: aptitude

How the paper is marked

Part A (Q.1 to Q.80): 1 mark for a correct answer, minus 1/3 for a wrong answer.
Part B (Q.81 to Q.95): 1 or 2 marks as printed with each question, 20 marks in all, no negative marking.

Source: ISRO's recruitment advertisement ISRO:ICRB:02(EMC):2023 of 25 May 2023, item 16(b).

ISRO CS 2023 answer key

ISRO's answer key for ISRO Scientist/Engineer Computer Science 2023
Question	Part	Marks	Answer
Q.1	A	1	B
Q.2	A	1	B
Q.3	A	1	A or B
Q.4	A	1	Excluded
Q.5	A	1	Excluded
Q.6	A	1	A
Q.7	A	1	A
Q.8	A	1	C
Q.9	A	1	C
Q.10	A	1	D
Q.11	A	1	B
Q.12	A	1	Excluded
Q.13	A	1	D
Q.14	A	1	C
Q.15	A	1	A
Q.16	A	1	C
Q.17	A	1	A
Q.18	A	1	B
Q.19	A	1	B
Q.20	A	1	B
Q.21	A	1	B
Q.22	A	1	C
Q.23	A	1	Excluded
Q.24	A	1	C
Q.25	A	1	B
Q.26	A	1	D
Q.27	A	1	A
Q.28	A	1	Excluded
Q.29	A	1	A
Q.30	A	1	A
Q.31	A	1	B
Q.32	A	1	C
Q.33	A	1	D
Q.34	A	1	A
Q.35	A	1	A
Q.36	A	1	A
Q.37	A	1	C
Q.38	A	1	D
Q.39	A	1	D
Q.40	A	1	D
Q.41	A	1	C
Q.42	A	1	D
Q.43	A	1	C
Q.44	A	1	A
Q.45	A	1	D
Q.46	A	1	Excluded
Q.47	A	1	C or D
Q.48	A	1	A
Q.49	A	1	C
Q.50	A	1	C
Q.51	A	1	D
Q.52	A	1	B
Q.53	A	1	C
Q.54	A	1	A
Q.55	A	1	A
Q.56	A	1	B
Q.57	A	1	A
Q.58	A	1	D
Q.59	A	1	B
Q.60	A	1	B
Q.61	A	1	C
Q.62	A	1	D
Q.63	A	1	A or B
Q.64	A	1	C
Q.65	A	1	B
Q.66	A	1	A
Q.67	A	1	B
Q.68	A	1	C
Q.69	A	1	A
Q.70	A	1	B
Q.71	A	1	A
Q.72	A	1	A
Q.73	A	1	A
Q.74	A	1	D
Q.75	A	1	C
Q.76	A	1	C
Q.77	A	1	C
Q.78	A	1	B
Q.79	A	1	B or C
Q.80	A	1	A
Q.81	B	1	D
Q.82	B	1	D
Q.83	B	2	B
Q.84	B	1	A
Q.85	B	1	A
Q.86	B	2	A
Q.87	B	1	A
Q.88	B	1	C
Q.89	B	2	C
Q.90	B	1	C
Q.91	B	1	D
Q.92	B	2	D
Q.93	B	1	A
Q.94	B	1	C
Q.95	B	2	A

ISRO excluded Q.4, Q.5, Q.12, Q.23, Q.28 and Q.46 from evaluation.

Part A: Computer Science Q.1 to Q.80

Q.11 markOfficial answer: B
Tap to open full size
Data Structures and Algorithms › Graph algorithms
A minimum spanning tree of a connected graph with $n$ vertices always has $n - 1$ edges. Kruskal's algorithm builds one by taking the lightest edges that do not form a cycle.
Full solution
Q.21 markOfficial answer: B
Tap to open full size
Data Structures and Algorithms › Graph algorithms
Match each algorithm to its defining problem: Floyd-Warshall solves all pair shortest paths, Kruskal's builds a minimum spanning tree, and Dijkstra and Bellman-Ford are single-source methods. Fix the two unambiguous matches first and use them to eliminate options.
Full solution
Q.31 markOfficial answer: A or B
Tap to open full size
Earlier, ISRO's key gave A.
Data Structures and Algorithms › Trees & heaps
Compute subtree heights and the best path through each node in a single post-order traversal, so each node is handled once in constant time.
Full solution
Q.41 markExcluded from evaluation by ISRO
Tap to open full size
Earlier, ISRO's key gave B.
Q.51 markExcluded from evaluation by ISRO
Tap to open full size
Earlier, ISRO's key gave B.
Q.61 markOfficial answer: A
Tap to open full size
Probability and Statistics › Counting & axioms
A simple graph on $n$ labelled vertices is fixed by choosing, for each of the $(2 n)$ vertex pairs, whether the edge is present, giving $2^{(2 n)}$ graphs.
Full solution
Q.71 markOfficial answer: A
Tap to open full size
Compiler Design › Lexical analysis
Read an NFA as a regular expression piece by piece. An alternation is two parallel branches joined by $ε$ -edges. An $ε$ -edge from the join back to an earlier final state is a Kleene star. The edge before that state is a prefix that appears only once.
Full solution
Q.81 markOfficial answer: C
Tap to open full size
Theory of Computation › CFG & PDA
Count parse trees by choosing the root production, which fixes which operator is applied last. Then count the trees of the two sub-strings. With no precedence in the grammar, each operator can be the root.
Full solution
Q.91 markOfficial answer: C
Tap to open full size
Operating Systems › Memory management
Next Fit continues searching from the point of the last allocation, so consecutive requests are placed in adjacent memory. Data allocated together sits together, which is spatial locality.
Full solution
Q.101 markOfficial answer: D
Tap to open full size
Compiler Design › Lexical analysis
Each compiler phase takes the output of the one before it. Tokens come before the parse tree, the parse tree before semantic checks, and the checked tree before intermediate code. Optimization works on that intermediate code, so it comes after it.
Full solution
Q.111 markOfficial answer: B
Tap to open full size
Compiler Design › Parsing
Read LL(k) as: first L = Left-to-right scan of the input, second L = Leftmost derivation, k = number of lookahead symbols. LR parsers instead give a Rightmost derivation in reverse.
Full solution
Q.121 markExcluded from evaluation by ISRO
Tap to open full size
Earlier, ISRO's key gave C.
Q.131 markOfficial answer: D
Tap to open full size
Computer Networks › Data link & MAC
CSMA/CD requires $T_{tx} = L / B \geq 2 T_{p}$ , so the rate that fits the minimum frame to the round trip exactly (and is the highest allowed) is $B = L / (2 T_{p})$ .
Full solution
Q.141 markOfficial answer: C
Tap to open full size
Computer Networks › Layering & switching
Every layer adds its own header, so the overhead is (number of layers x header size). Wasted bandwidth is that overhead divided by the total bytes transmitted (payload plus headers).
Full solution
Q.151 markOfficial answer: A
Tap to open full size
Computer Networks › Data link & MAC
Remember the peak rate of each 802.11 amendment: b = 11 Mbps, a/g = 54 Mbps, ac = about 3.5 Gbps in the usual textbook figure, ad = about 7 Gbps. Fix the three firm ones first and the last pair follows.
Full solution
Q.161 markOfficial answer: C
Tap to open full size
Computer Networks › Data link & MAC
Parity and CRC only detect errors. Reed-Solomon and LDPC carry enough structured redundancy to correct them.
Full solution
Q.171 markOfficial answer: A
Tap to open full size
Data Structures and Algorithms › Linear structures
Apply precedence first, then place every operator after its two operands: the two products are converted separately and the + goes last.
Full solution
Q.181 markOfficial answer: B
Tap to open full size
Computer Organization and Architecture › I/O, interrupts & DMA
In a vectored interrupt the device sends the vector, which tells the processor where its service routine is, so no polling is needed.
Full solution
Q.191 markOfficial answer: B
Tap to open full size
Computer Organization and Architecture › Pipelining
Execution time is $N \times CPI \times t_{clock}$ . A super-scalar design issues several instructions per cycle, which lowers CPI below 1 and shortens the run for the same instruction count.
Full solution
Q.201 markOfficial answer: B
Tap to open full size
Computer Organization and Architecture › Pipelining
Instructions flow through a pipeline in program order, so the buffer holding them between stages must be first-in, first-out (an instruction prefetch queue).
Full solution
Q.211 markOfficial answer: B
Tap to open full size
Computer Organization and Architecture › ALU & control unit
An operation performed on data held in registers is a micro-operation. Each machine instruction is carried out as a sequence of such micro-operations.
Full solution
Q.221 markOfficial answer: C
Tap to open full size
Computer Networks › Data link & MAC
Parity and XOR checks are blind to certain even-numbered error patterns, while a CRC with a suitably chosen generator polynomial detects every double-bit error within a frame shorter than the generator's period.
Full solution
Q.231 markExcluded from evaluation by ISRO
Tap to open full size
Earlier, ISRO's key gave A.
Q.241 markOfficial answer: C
Tap to open full size
Data Structures and Algorithms › Complexity
Compare growth rates by dividing out common factors or by taking logarithms: $lo g n$ is smaller than any power $n^{ϵ}$ , a function $n^{l o g n} = 2^{(l o g n)^{2}}$ is super-polynomial but still below $2^{n}$ because $(lo g n)^{2} = o (n)$ .
Full solution
Q.251 markOfficial answer: B
Tap to open full size
Data Structures and Algorithms › Sorting & hashing
Radix sort makes one pass per digit, and each pass handles every item against the $k$ possible digit values, so the count is $d \times n \times k$ with $d = 3$ , $n = 6$ , $k = 10$ .
Full solution
Q.261 markOfficial answer: D
Tap to open full size
Data Structures and Algorithms › Trees & heaps
If a node has a left subtree, its inorder predecessor is the rightmost (largest) node of that left subtree: go left once, then right as far as possible.
Full solution
Q.271 markOfficial answer: A
Tap to open full size
Programming in C › Pointers & arrays
Adding $n$ to a pointer advances it by $n$ elements, so the address changes by $n \times sizeof$ (element type) bytes.
Full solution
Q.281 markExcluded from evaluation by ISRO
Tap to open full size
Earlier, ISRO's key gave D.
Q.291 markOfficial answer: A
Tap to open full size
Data Structures and Algorithms › Sorting & hashing
Heap sort builds the heap in $O (n)$ and then performs $n - 1$ extractions, each a sift-down of at most $lo g_{2} n$ levels. The total is $O (n lo g n)$ for every input, and it needs no extra array.
Full solution
Q.301 markOfficial answer: A
Tap to open full size
Database Management and Warehousing › SQL
Each new row is built from the current maxima, which always belong to the latest row. So the $Y$ values follow $Y_{k} = 2 Y_{k - 1} + 1$ , giving $Y_{k} = 2^{k + 1} - 1$ , while $X_{k} = k + 1$ .
Full solution
Q.311 markOfficial answer: B
Tap to open full size
Database Management and Warehousing › SQL
A comma-separated list of tables in FROM with no condition is the Cartesian product, which is what CROSS JOIN means.
Full solution
Q.321 markOfficial answer: C
Tap to open full size
Database Management and Warehousing › SQL
When the condition for a DELETE depends on another table, delete from the target table and filter it with IN (SELECT ...) on the related table.
Full solution
Q.331 markOfficial answer: D
Tap to open full size
Q.341 markOfficial answer: A
Tap to open full size
Database Management and Warehousing › Normal forms
Each normal form is tied to a kind of dependency. Functional dependencies lead to 2NF, 3NF and BCNF. Multivalued dependencies lead to 4NF. Join dependencies lead to 5NF, also called Project-Join Normal Form.
Full solution
Q.351 markOfficial answer: A
Tap to open full size
Digital Electronics › Combinational
A multiplexer picks one of many inputs. A 1-to- $N$ demultiplexer needs $lo g_{2} N$ select lines. An $N$ -to- $lo g_{2} N$ encoder needs only $lo g_{2} N$ OR gates, one per output bit.
Full solution
Q.361 markOfficial answer: A
Tap to open full size
Digital Electronics › Sequential
A serial shift register delays data by one clock period per stage, so total delay = (number of bits) × (clock period).
Full solution
Q.371 markOfficial answer: C
Tap to open full size
Digital Electronics › Combinational
A full adder adds three bits (two operands plus carry-in) and gives a two-bit result: sum and carry-out. A half adder has only two inputs because it has no carry-in.
Full solution
Q.381 markOfficial answer: D
Tap to open full size
Digital Electronics › Boolean logic
A NAND gate with its inputs tied together is an inverter. A NAND of two inverted signals is an OR gate, by De Morgan's theorem.
Full solution
Q.391 markOfficial answer: D
Tap to open full size
Earlier, ISRO's key gave C.
Digital Electronics › Boolean logic
XOR outputs LOW when its two inputs are equal and HIGH when they differ. XNOR is the opposite: HIGH when the inputs are equal.
Full solution
Q.401 markOfficial answer: D
Tap to open full size
Digital Electronics › Boolean logic
Joining both inputs of a NAND or NOR gate gives a single-input gate. Because $A \cdot A = A$ and $A + A = A$ , the output is $\overline{A}$ , an inverter.
Full solution
Q.411 markOfficial answer: C
Tap to open full size
Linear Algebra › Eigenvalues
If one row of a square matrix is a scalar multiple of another, the rows are linearly dependent, the matrix is singular and its determinant is $0$ . Look for such a relation before expanding.
Full solution
Q.421 markOfficial answer: D
Tap to open full size
Discrete Mathematics › Graph theory
Handshaking lemma gives degree sum $2 e$ ; the radius is the smallest eccentricity, so some vertex always attains it; girth is the number of edges in the shortest cycle; equal degrees mean a regular graph, not a multigraph.
Full solution
Q.431 markOfficial answer: C
Tap to open full size
Discrete Mathematics › Graph theory
Euler's formula for a connected planar graph, $V - E + F = 2$ , gives the missing count when two of the three are known. The outer (unbounded) region counts as a face.
Full solution
Q.441 markOfficial answer: A
Tap to open full size
Probability and Statistics › Counting & axioms
With equally likely outcomes, probability is favourable count over total count. "Not white" can be counted directly (10 of 15) or found as 1 minus the probability of white.
Full solution
Q.451 markOfficial answer: D
Tap to open full size
Data Structures and Algorithms › Complexity
Count the entries of the result ( $m \times p$ ) and multiply by the length of the inner sum for each entry ( $n$ ): three nested loops, each running once over its own dimension.
Full solution
Q.461 markExcluded from evaluation by ISRO
Tap to open full size
Earlier, ISRO's key gave C.
Q.471 markOfficial answer: C or D
Tap to open full size
Earlier, ISRO's key gave C.
Discrete Mathematics › Mathematical logic
Every variable doubles the number of possible assignments, so $n$ variables give $2^{n}$ rows.
Full solution
Q.481 markOfficial answer: A
Tap to open full size
Operating Systems › Processes & threads
Linux creates every thread with clone(), and each thread is a separate kernel task that the scheduler sees. That makes the threading model one-to-one.
Full solution
Q.491 markOfficial answer: C
Tap to open full size
Operating Systems › Synchronisation
A lock holder temporarily inherits the priority of the highest-priority task waiting for the lock, so a medium-priority task cannot preempt it and delay the waiter indefinitely.
Full solution
Q.501 markOfficial answer: C
Tap to open full size
Operating Systems › Memory management
A segment number is an index into the process's segment table. The compiler decides what the segments are (code, data, stack, and so on). The loader builds the segment table when the program is loaded, so it is the loader that numbers the segments and fills in each one's base and limit.
The question and its official answer
Q.511 markOfficial answer: D
Tap to open full size
Computer Organization and Architecture › I/O, interrupts & DMA
Hardware interrupts are asynchronous and come from external devices. Traps, exceptions and system calls are synchronous and are raised by the running program (software).
Full solution
Q.521 markOfficial answer: B
Tap to open full size
Operating Systems › CPU scheduling
Long term scheduler: admits jobs and so controls the degree of multiprogramming. Medium term scheduler: swaps processes in and out. Short term scheduler: runs most often, so it must be the fastest.
Full solution
Q.531 markOfficial answer: C
Tap to open full size
Operating Systems › Synchronisation
A counting semaphore both counts filled slots and blocks or wakes threads, so wait(full) makes the consumer sleep on an empty buffer and signal(full) from the producer wakes it.
Full solution
Q.541 markOfficial answer: A
Tap to open full size
Operating Systems › Memory management
Memory protection hardware catches any access outside a process's own allowed memory and raises a trap, which hands control to the OS in kernel mode.
Full solution
Q.551 markOfficial answer: A
Tap to open full size
Theory of Computation › Finite automata
Identify the dead state, then follow the only path that avoids it. Here that path is one $a$ from the start, followed by any number of $b$ 's looping on the final state.
Full solution
Q.561 markOfficial answer: B
Tap to open full size
Theory of Computation › Finite automata
The NFA accepts strings whose second-to-last symbol is $b$ , so a DFA must remember the last two symbols, which gives $2^{2} = 4$ distinguishable states (Myhill-Nerode).
Full solution
Q.571 markOfficial answer: A
Tap to open full size
Theory of Computation › CFG & PDA
Split the grammar into its two parts. $T$ generates $b^{k}$ with $k \geq 1$ , and $S \to a S b$ wraps it with $m$ matching pairs of $a$ and $b$ . So the number of $b$ 's exceeds the number of $a$ 's by exactly $k \geq 1$ .
Full solution
Q.581 markOfficial answer: D
Tap to open full size
Theory of Computation › Turing machines
Recursive languages are closed under complement. A language is recursive exactly when both it and its complement are recursively enumerable. Recursively enumerable languages are not closed under complement.
Full solution
Q.591 markOfficial answer: B
Tap to open full size
Theory of Computation › Context-free languages
Context-free languages are closed under union and under intersection with a regular language (product of a PDA and a DFA). The result of that intersection is context-free in general, not regular.
Full solution
Q.601 markOfficial answer: B
Tap to open full size
Theory of Computation › CFG & PDA
Unroll the grammar to get a closed form. The rule $S \to aa S b$ pairs two $a$ 's with one $b$ , and $T \to T b ∣ a$ gives one $a$ followed by any number of $b$ 's. So $L (G) = {a^{2 n + 1} b^{m} : m \geq n}$ , which has an odd number of $a$ 's, all $a$ 's first, and at least $n$ $b$ 's.
Full solution
Q.611 markOfficial answer: C
Tap to open full size
Artificial Intelligence › Logic
The connectives of First Order Logic are $\neg$ , $\land$ , $\lor$ , $\Rightarrow$ and $\Leftrightarrow$ . Any other symbol, such as <>, is not one.
Full solution
Q.621 markOfficial answer: D
Tap to open full size
Artificial Intelligence › Uncertainty
"Hybrid" means the network mixes the two variable types: continuous nodes (often Gaussian) and discrete nodes, linked by conditional distributions such as conditional Gaussians or threshold models.
Full solution
Q.631 markOfficial answer: A or B
Tap to open full size
Earlier, ISRO's key gave A.
Q.641 markOfficial answer: C
Tap to open full size
Q.651 markOfficial answer: B
Tap to open full size
Machine Learning › Clustering
Clustering methods are either partitional (one flat set of $k$ disjoint clusters, as in K-Means) or hierarchical (a nested tree, built top-down by divisive or bottom-up by agglomerative methods).
Full solution
Q.661 markOfficial answer: A
Tap to open full size
Q.671 markOfficial answer: B
Tap to open full size
Q.681 markOfficial answer: C
Tap to open full size
Q.691 markOfficial answer: A
Tap to open full size
Q.701 markOfficial answer: B
Tap to open full size
Q.711 markOfficial answer: A
Tap to open full size
Q.721 markOfficial answer: A
Tap to open full size
Q.731 markOfficial answer: A
Tap to open full size
Q.741 markOfficial answer: D
Tap to open full size
Q.751 markOfficial answer: C
Tap to open full size
Q.761 markOfficial answer: C
Tap to open full size
Q.771 markOfficial answer: C
Tap to open full size
Q.781 markOfficial answer: B
Tap to open full size
Q.791 markOfficial answer: B or C
Tap to open full size
Earlier, ISRO's key gave C.
Q.801 markOfficial answer: A
Tap to open full size
Database Management and Warehousing › ER model
An E-R model captures structure: entities, attributes and relationships. It is a static view. Processes (functional view) and behaviour over time (dynamic view) are modelled with other diagrams.
Full solution

Part B: Aptitude Q.81 to Q.95

Part B is the same aptitude test as in the ISRO Scientist/Engineer Electronics 2023 paper, held the same day, with the same question numbers. Its solutions are listed under ISRO EC 2023.

Q.811 markOfficial answer: D
Tap to open full size
General Aptitude › Reasoning
In picture analogies, find the relation in the complete row (material to the tool that applies it) and carry the same relation to the incomplete row.
Full solution
Q.821 markOfficial answer: D
Tap to open full size
Earlier, ISRO's key gave B.
General Aptitude › Spatial
In a figure analogy, list every change in the example pair separately (outline shape, shading direction, small symbol's fill, orientation and corner) and apply each one exactly as drawn, not as you expect it to behave.
The question and its official answer
Q.832 marksOfficial answer: B
Tap to open full size
General Aptitude › Spatial
Work out which item each operator acts on. Both operators here change only the 1st item, so the 2nd and 3rd items pass through untouched.
Full solution
Q.841 markOfficial answer: A
Tap to open full size
Q.851 markOfficial answer: A
Tap to open full size
General Aptitude › Reasoning
To show a survey underestimates a group, find the option that moves members of that group out of the reported count (here, dishonest people who reported being honest). Anything that adds people to the count, or only changes the degree of those already in it, cannot cause an underestimate.
Full solution
Q.862 marksOfficial answer: A
Tap to open full size
General Aptitude › Reasoning
In a circular seating puzzle, fix one person, number the seats, and decide which way a person's right hand points from the direction they face (towards the anticlockwise side when facing the centre). Then place people with the direct conditions and test each possible position for the others against the negative conditions.
Full solution
Q.871 markOfficial answer: A
Tap to open full size
General Aptitude › Reasoning
The three groups of dominoes alternate between counting down and counting up. The last group must count down again, so with $[1∣0]$ at its end the missing dominoes are $[3∣2]$ and $[2∣1]$ .
Full solution
Q.881 markOfficial answer: C
Tap to open full size
General Aptitude › Reasoning
In number-figure puzzles, test the simplest relation first: the sum of the outer numbers compared with the centre. Confirm that one rule fits every complete figure before applying it to the incomplete one.
Full solution
Q.892 marksOfficial answer: C
Tap to open full size
General Aptitude › Spatial
Assign one operation to each symbol using the chains where it appears together with symbols already decoded. Then apply the operations to the query tile in order. Track three things separately: colour, the direction of the diagonal, and the position of the straight line.
The question and its official answer
Q.901 markOfficial answer: C
Tap to open full size
General Aptitude › Reasoning
Combine an "all" statement with a "some" statement. The members picked out by "some" also fall under the "all" group, which gives a new "some" statement. A conclusion that brings in an idea the facts never mention (such as "better" or "simple") is not forced.
Full solution
Q.911 markOfficial answer: D
Tap to open full size
General Aptitude › Reasoning
A course of action must be a sensible, proportionate response to the problem in the statement. An unverified allegation calls for an inquiry, not punishment, and the regulator's directive is not the thing at fault.
Full solution
Q.922 marksOfficial answer: D
Tap to open full size
General Aptitude › Numerical
When each term is a multiple of the previous one plus a growing amount, find the rule from the early terms, then test whether the later terms continue from it. The wrong term is the one that fails while the terms after it still agree with the rule applied to the correct value.
Full solution
Q.931 markOfficial answer: A
Tap to open full size
General Aptitude › Data interpretation
Read each point off the line graph, take the absolute difference for each pair of consecutive days given, and compare. A steep segment of the line marks a large change.
Full solution
Q.941 markOfficial answer: C
Tap to open full size
General Aptitude › Data interpretation
Percentage change is the change divided by the starting value, times 100. Here the starting year is 1983.
Full solution
Q.952 marksOfficial answer: A
Tap to open full size
General Aptitude › Geometry
For a fixed perimeter, the rectangle of largest area is a square: with $l + b$ fixed, $l b \leq (\frac{l + b}{2})^{2}$ .
Full solution

WHERE THE MARKS CAME FROM

Marks by subject in ISRO Scientist/Engineer Computer Science 2023
Subject	Qs	Marks
General Aptitude	14	19
Data Structures and Algorithms	9	9
Operating Systems	7	7
Theory of Computation	7	7
Digital Electronics	6	6
Computer Networks	5	5
Computer Organization and Architecture	5	5
Database Management and Warehousing	5	5
Compiler Design	3	3
Discrete Mathematics	3	3
Artificial Intelligence	2	2
Probability and Statistics	2	2
Linear Algebra	1	1
Machine Learning	1	1
Programming in C	1	1
Other questions	24	24

We sorted the questions into subjects ourselves; the exam paper does not label them.

Every ISRO paper ISRO Scientist/Engineer: who can apply CSE/IT study material DS & AI study material