Cisco Placement Papers 2026 with Solutions — Aptitude, Technical & Coding

Last Updated: March 2026 | Difficulty: Moderate to High | Questions: 50+ with Solutions

Cisco Recruitment Process 2026 — Complete Overview

Cisco Systems is a global technology leader specializing in networking hardware, software, and telecommunications equipment. For the 2026 batch, Cisco is actively hiring fresh graduates for roles like Software Engineer, Network Engineer, Systems Engineer, Technical Consulting Engineer, and Data Scientist.

Cisco is known for its rigorous technical selection process that emphasizes networking fundamentals, programming skills, problem-solving abilities, and innovation mindset. The company offers excellent career growth with cutting-edge technology exposure and global opportunities.

Why Join Cisco?

Pioneer in networking and cybersecurity technologies
Strong focus on innovation and emerging technologies (IoT, AI/ML, 5G)
Comprehensive learning and certification programs (CCNA, CCNP)
Collaborative culture with emphasis on work-life balance

Cisco Eligibility Criteria 2026

Criteria	Requirements
Education	BE/B.Tech, M.Tech, MCA (CS, IT, ECE, EEE preferred)
Batch	2024, 2025, 2026 passing out
CGPA/Percentage	Minimum 70% or 7.0 CGPA throughout academics
Backlogs	No active backlogs at time of joining
Gap Year	Maximum 1 year gap allowed with valid reason

Cisco CTC for Freshers 2026

Role	CTC Package	Location
Software Engineer I	₹14 - 18 LPA	Bangalore, Hyderabad, Pune
Network Engineer	₹12 - 15 LPA	Bangalore, Chennai, Pune
Systems Engineer	₹13 - 16 LPA	Bangalore, Hyderabad
Technical Consulting Engineer	₹15 - 19 LPA	Bangalore, Hyderabad
Data Scientist	₹16 - 20 LPA	Bangalore, San Jose (US)

Note: Cisco offers competitive packages with RSUs (Restricted Stock Units), joining bonus, and comprehensive benefits.

Cisco Online Test Pattern 2026

Section	Questions	Time (Minutes)	Difficulty
Aptitude (Quant + Logical)	20	25	Moderate
Verbal Ability	10	15	Moderate
Technical (Networking + CS)	20	25	High
Coding	2-3	60	High
Total	52-53	125	Moderate-High

Key Points:

Negative Marking: No negative marking in most tests
Sectional Cut-offs: Yes, especially for Technical section
Coding Platform: HackerRank or Cisco proprietary platform
Language Options: C, C++, Java, Python for coding

Section 1: Quantitative Aptitude (15 Questions)

Q1. A network packet travels through 5 routers. The delay at each router is 2ms, 3ms, 1ms, 4ms, and 2ms. What is the average delay?

Options: a) 2.2ms
b) 2.4ms
c) 2.6ms
d) 2.8ms

Solution:

Total delay = 2 + 3 + 1 + 4 + 2 = 12ms
Average = 12/5 = 2.4ms

Q2. In a subnet with mask 255.255.255.192, how many usable host addresses are available?

Options: a) 30
b) 62
c) 126
d) 254

Solution:

255.255.255.192 = /26 (26 network bits)
Host bits = 32 - 26 = 6
Total addresses = 2⁶ = 64
Usable hosts = 64 - 2 = 62 (subtract network and broadcast addresses)

Q3. A Cisco router processes 1000 packets/second. How many packets does it process in 2.5 hours?

Options: a) 9,000,000
b) 9,500,000
c) 10,000,000
d) 10,500,000

Solution:

Packets per second = 1000
Seconds in 2.5 hours = 2.5 × 60 × 60 = 9000
Total packets = 1000 × 9000 = 9,000,000

Q4. If a binary tree has 15 leaf nodes, what is the minimum height of the tree?

Options: a) 3
b) 4
c) 5
d) 6

Solution:

Minimum height with n leaves in a binary tree = ⌈log₂(n)⌉
⌈log₂(15)⌉ = ⌈3.91⌉ = 4

Q5. A data stream of 1 GB is transmitted at 100 Mbps. How long does transmission take?

Options: a) 80 seconds
b) 82.5 seconds
c) 85 seconds
d) 87.5 seconds

Solution:

1 GB = 8 Gb = 8000 Mb
Speed = 100 Mbps
Time = 8000/100 = 80 seconds

Q6. Find the probability of getting exactly 2 heads in 3 coin tosses.

Options: a) 1/8
b) 3/8
c) 1/4
d) 1/2

Solution:

Total outcomes = 2³ = 8
Favorable outcomes (2 heads): HHT, HTH, THH = 3
Probability = 3/8
Formula: C(3,2) × (1/2)² × (1/2)¹ = 3 × 1/8 = 3/8

Q7. A server handles requests following Poisson distribution with mean 4 requests/second. What is the variance?

Options: a) 2
b) 4
c) 8
d) 16

Solution:

For Poisson distribution: Mean = Variance = λ
Variance = 4

Q8. In a binary search tree with 1000 nodes, what is the maximum number of comparisons needed to find an element?

Options: a) 10
b) 100
c) 500
d) 1000

Solution:

Maximum comparisons = height of tree
Balanced BST: ⌈log₂(1001)⌉ = 10
Worst case (skewed): 1000
But for balanced BST used in practice: ~10

Q9. A hash table has 100 slots. Using chaining, what's the load factor if 50 elements are inserted?

Options: a) 0.25
b) 0.5
c) 1.0
d) 2.0

Solution:

Load factor α = n/m = number of elements / number of slots
α = 50/100 = 0.5

Q10. If round-trip time (RTT) is 100ms and packet size is 1KB, what is the bandwidth-delay product?

Options: a) 100 KB
b) 1000 KB
c) 10 KB
d) 1 KB

Solution:

This question needs bandwidth which is missing. Assuming 1 MBps bandwidth:
Bandwidth-delay product = Bandwidth × RTT
= 1 MBps × 0.1s = 100 KB
Answer depends on given bandwidth; typical answer: 100 KB

Q11. A queue implementation using array has front=10, rear=25 in an array of size 30. How many elements are in the queue?

Options: a) 15
b) 16
c) 25
d) 35

Solution:

Number of elements = rear - front + 1
= 25 - 10 + 1 = 16

Q12. What is the subnet mask for a /20 network?

Options: a) 255.255.0.0
b) 255.255.240.0
c) 255.255.255.0
d) 255.240.0.0

Solution:

/20 means 20 bits for network
255.255.240.0 = 11111111.11111111.11110000.00000000
First 20 bits are 1s
Answer: 255.255.240.0

Q13. A CRC generator polynomial is x³ + x + 1. What is the degree of the polynomial?

Options: a) 1
b) 2
c) 3
d) 4

Solution:

Degree is the highest power of x
x³ + x + 1 has highest power 3
Degree = 3

Q14. In a CSMA/CD network, maximum propagation delay is 25μs. What is the minimum frame size if data rate is 10 Mbps?

Options: a) 250 bits
b) 500 bits
c) 1000 bits
d) 2000 bits

Solution:

Minimum frame transmission time = 2 × propagation delay
= 2 × 25μs = 50μs
Frame size = Data rate × Time = 10 Mbps × 50μs = 10 × 10⁶ × 50 × 10⁻⁶ = 500 bits

Q15. A router has 4 interfaces. In how many ways can IP addresses from different subnets be assigned?

Options: a) 4
b) 8
c) 16
d) 24

Solution:

If 4 different subnets for 4 interfaces: 4! = 24 ways
This is the number of permutations

Section 2: Logical Reasoning (15 Questions)

Q1. If all switches are routers, some routers are gateways, and no gateway is a firewall, then:

Options: a) Some switches are gateways
b) No router is a firewall
c) Some routers are not firewalls
d) All of the above

Solution:

Some routers are gateways, no gateway is firewall
Therefore some routers (those that are gateways) are not firewalls
Some routers are not firewalls follows

Q2. Complete the series: 1, 1, 2, 3, 5, 8, 13, ?

Options: a) 18
b) 20
c) 21
d) 23

Solution:

Fibonacci series: each number is sum of previous two
8 + 13 = 21

Q3. If TCP is coded as 2016, how is UDP coded?

Options: a) 2116
b) 2115
c) 2015
d) 2114

Solution:

T(20), C(3) → 2016? Pattern unclear
Alternative: Position × Position: T(20), C(3) → 20, 3 → 2016? No
U(21), D(4), P(16) → 2116 (positional concatenation)

Q4. In a network topology, A connects to B and C. B connects to D. C connects to E. D and E connect to F. How many paths from A to F?

Options: a) 2
b) 3
c) 4
d) 5

Solution:

Paths: A-B-D-F and A-C-E-F
2 paths

Q5. Find the odd one: IPv4, IPv6, TCP, UDP

Options: a) IPv4
b) IPv6
c) TCP
d) UDP

Solution:

IPv4 and IPv6 are Network Layer protocols
TCP and UDP are Transport Layer protocols
IPv4 (or IPv6) is at different layer

Q6. If ping = 192.168.1.1 succeeds but ping = google.com fails, the problem is likely with:

Options: a) Physical layer
b) Data link layer
c) DNS
d) Transport layer

Solution:

IP ping works → physical, data link, network layers OK
Domain name ping fails → DNS resolution problem

Q7. A packet takes 3 hops to reach destination. Each hop has 10% packet loss probability. What's the probability of successful delivery?

Options: a) 70%
b) 72.9%
c) 90%
d) 95%

Solution:

Success probability per hop = 90% = 0.9
Success for 3 hops = 0.9³ = 0.729 = 72.9%

Q8. Statement: All protocols have headers. TCP is a protocol. Conclusion: TCP has a header.

Options: a) True
b) False
c) Cannot say
d) Irrelevant

Solution:

All protocols → have headers
TCP is a protocol → TCP has header
Valid syllogism, conclusion is true

Q9. What comes next: 2, 6, 12, 20, 30, ?

Options: a) 40
b) 42
c) 44
d) 48

Solution:

Pattern: n(n+1)
1×2=2, 2×3=6, 3×4=12, 4×5=20, 5×6=30, 6×7=42

Q10. In a binary tree, each node has 0 or 2 children. If there are 15 leaf nodes, how many internal nodes?

Options: a) 13
b) 14
c) 15
d) 16

Solution:

For full binary tree: L = I + 1 (where L=leaves, I=internal nodes)
15 = I + 1
I = 14

Q11. If A AND B = C, and A OR B = D, which is always true?

Options: a) C ≤ D
b) C > D
c) C = D
d) C ≥ D

Solution:

AND gives minimum, OR gives maximum
C = min(A,B), D = max(A,B)
Therefore C ≤ D always

Q12. A network cable can be cut into pieces of 3m, 4m, or 5m without waste. What is the minimum cable length?

Options: a) 12m
b) 30m
c) 60m
d) 120m

Solution:

Cable length must be divisible by 3, 4, and 5
LCM(3, 4, 5) = 60m

Q13. Packet switching vs Circuit switching analogy:

Options: a) Restaurant reservation vs walk-in
b) Walk-in vs restaurant reservation
c) Highway vs local road
d) Train vs bus

Solution:

Packet switching = Walk-in (share resources, no dedicated path)
Circuit switching = Restaurant reservation (dedicated resources)

Q14. Complete: HTTP : Port 80 :: HTTPS : ?

Options: a) 21
b) 443
c) 25
d) 8080

Solution:

HTTP uses port 80
HTTPS uses port 443

Q15. A routing table has 3 entries. What's the maximum number of routing decisions needed to find a match?

Options: a) 1
b) 2
c) 3
d) 4

Solution:

Worst case: check all 3 entries
Maximum decisions = 3

Section 3: Verbal Ability (10 Questions)

Q1. Choose the synonym of "ROBUST"

Options: a) Weak
b) Fragile
c) Strong
d) Delicate

Q2. Choose the antonym of "EPHEMERAL"

Options: a) Brief
b) Transient
c) Permanent
d) Fleeting

Q3. Fill in the blank: The new security protocol provides ___ protection against cyber attacks.

Options: a) Weak
b) Robust
c) Fragile
d) Temporary

Q4. Spot the error: "Neither the server nor the clients was responding to requests."

Options: a) Neither the server
b) nor the clients
c) was responding
d) to requests

Solution:

"Clients" is plural, so use "were responding"

Q5. Rearrange: (P) is crucial (Q) network security (R) for modern (S) businesses

Options: a) PQRS
b) QPRS
c) PRQS
d) QPRS

Solution:

"Network security is crucial for modern businesses"
Order: Q (Network security) + P (is crucial) + R (for modern) + S (businesses)

Q6. Choose the correctly spelled word:

Options: a) Occured
b) Occurred
c) Ocurred
d) Occureed

Q7. Choose the meaning of idiom: "Back to the drawing board"

Options: a) Start over
b) Complete successfully
c) Take a break
d) Celebrate victory

Q8. Active to Passive: "The engineer configured the router."

Options: a) The router was configured by the engineer
b) The router is configured by the engineer
c) The router has been configured by the engineer
d) The router had been configured by the engineer

Q9. Choose the word most appropriate: "The ___ implementation ensured zero downtime."

Options: a) Careless
b) Seamless
c) Hasty
d) Sloppy

Q10. Choose the synonym of "REDUNDANT"

Options: a) Necessary
b) Superfluous
c) Essential
d) Required

Section 4: Technical / CS Fundamentals (10 Questions)

Q1. Which layer of OSI model does IP belong to?

Options: a) Transport
b) Network
c) Data Link
d) Application

Q2. What is the maximum size of an IPv4 header?

Options: a) 20 bytes
b) 40 bytes
c) 60 bytes
d) 80 bytes

Solution:

Base header = 20 bytes
Options can add up to 40 bytes
Maximum = 60 bytes

Q3. Which protocol uses three-way handshake?

Options: a) UDP
b) TCP
c) HTTP
d) IP

Q4. What is the default subnet mask for Class B networks?

Options: a) 255.0.0.0
b) 255.255.0.0
c) 255.255.255.0
d) 255.255.255.255

Q5. Which routing algorithm is used by OSPF?

Options: a) Distance Vector
b) Link State
c) Path Vector
d) Flooding

Q6. What is the purpose of NAT?

Options: a) Encrypt data
b) Translate private to public IP addresses
c) Route packets
d) Filter traffic

Q7. Which command is used to check network connectivity?

Options: a) ipconfig
b) ping
c) traceroute
d) netstat

Q8. What is VLAN used for?

Options: a) Increase bandwidth
b) Create logical network segments
c) Encrypt traffic
d) Route between networks

Q9. Which port does SSH use?

Options: a) 20
b) 22
c) 23
d) 25

Q10. What does CIDR stand for?

Options: a) Classless Inter-Domain Routing
b) Classful Internet Data Routing
c) Common Internal Data Routing
d) Classless Internet Domain Routing

Section 5: Coding Problems (5 Questions)

Q1. Implement a function to check if two IP addresses are in the same subnet.

Solution (Python):

def ip_to_binary(ip):
    """Convert IP address to binary string"""
    octets = list(map(int, ip.split('.')))
    return ''.join(format(octet, '08b') for octet in octets)

def are_in_same_subnet(ip1, ip2, subnet_mask):
    """Check if two IPs are in the same subnet"""
    bin1 = ip_to_binary(ip1)
    bin2 = ip_to_binary(ip2)
    mask = ip_to_binary(subnet_mask)
    
    # Apply mask to both IPs
    network1 = ''.join('1' if m == '1' and b == '1' else '0' for m, b in zip(mask, bin1))
    network2 = ''.join('1' if m == '1' and b == '1' else '0' for m, b in zip(mask, bin2))
    
    return network1 == network2

# Alternative using integer arithmetic
def are_in_same_subnet_v2(ip1, ip2, mask):
    def ip_to_int(ip):
        octets = list(map(int, ip.split('.')))
        return (octets[0] << 24) + (octets[1] << 16) + (octets[2] << 8) + octets[3]
    
    ip1_int = ip_to_int(ip1)
    ip2_int = ip_to_int(ip2)
    mask_int = ip_to_int(mask)
    
    return (ip1_int & mask_int) == (ip2_int & mask_int)

# Test
print(are_in_same_subnet("192.168.1.10", "192.168.1.20", "255.255.255.0"))  # True
print(are_in_same_subnet("192.168.1.10", "192.168.2.10", "255.255.255.0"))  # False
print(are_in_same_subnet_v2("10.0.0.5", "10.0.0.100", "255.255.0.0"))  # True

Time Complexity: O(1)
Space Complexity: O(1)

Q2. Find the first non-repeating character in a string.

Solution (Python):

def first_non_repeating(s):
    from collections import Counter
    
    # Count frequency
    count = Counter(s)
    
    # Find first with count 1
    for char in s:
        if count[char] == 1:
            return char
    
    return None  # No non-repeating character

# One-pass solution using ordered dictionary
def first_non_repeating_optimized(s):
    from collections import OrderedDict
    
    char_order = OrderedDict()
    
    for char in s:
        if char in char_order:
            char_order[char] = False  # Mark as repeating
        else:
            char_order[char] = True   # Mark as non-repeating
    
    for char, is_unique in char_order.items():
        if is_unique:
            return char
    
    return None

# Test
print(first_non_repeating("swiss"))      # 'w'
print(first_non_repeating("network"))    # 'e'
print(first_non_repeating("aabbcc"))     # None
print(first_non_repeating_optimized("cisco"))  # 'c'

Time Complexity: O(n)
Space Complexity: O(1) - assuming fixed character set

Q3. Implement Dijkstra's shortest path algorithm.

Solution (Python):

import heapq

def dijkstra(graph, start):
    """
    graph: dict of {node: [(neighbor, weight), ...]}
    Returns: dict of shortest distances from start to each node
    """
    distances = {node: float('infinity') for node in graph}
    distances[start] = 0
    
    # Priority queue: (distance, node)
    pq = [(0, start)]
    
    while pq:
        current_dist, current_node = heapq.heappop(pq)
        
        # Skip if we've already found a better path
        if current_dist > distances[current_node]:
            continue
        
        # Check neighbors
        for neighbor, weight in graph[current_node]:
            distance = current_dist + weight
            
            # If found shorter path, update
            if distance < distances[neighbor]:
                distances[neighbor] = distance
                heapq.heappush(pq, (distance, neighbor))
    
    return distances

# Test with network topology
network = {
    'A': [('B', 4), ('C', 2)],
    'B': [('A', 4), ('C', 1), ('D', 5)],
    'C': [('A', 2), ('B', 1), ('D', 8), ('E', 10)],
    'D': [('B', 5), ('C', 8), ('E', 2), ('F', 6)],
    'E': [('C', 10), ('D', 2), ('F', 3)],
    'F': [('D', 6), ('E', 3)]
}

print(dijkstra(network, 'A'))
# {'A': 0, 'B': 3, 'C': 2, 'D': 8, 'E': 10, 'F': 11}

Time Complexity: O((V + E) log V) with heap
Space Complexity: O(V)

Q4. Design a Least Recently Used (LRU) Cache.

Solution (Python):

from collections import OrderedDict

class LRUCache:
    def __init__(self, capacity):
        self.capacity = capacity
        self.cache = OrderedDict()
    
    def get(self, key):
        if key not in self.cache:
            return -1
        
        # Move to end (most recently used)
        self.cache.move_to_end(key)
        return self.cache[key]
    
    def put(self, key, value):
        if key in self.cache:
            # Update and move to end
            self.cache.move_to_end(key)
        
        self.cache[key] = value
        
        # Remove least recently used if over capacity
        if len(self.cache) > self.capacity:
            self.cache.popitem(last=False)
    
    def display(self):
        print(list(self.cache.items()))

# Manual implementation without OrderedDict
class LRUCacheManual:
    def __init__(self, capacity):
        self.capacity = capacity
        self.cache = {}
        self.usage = []  # Most recent at end
    
    def get(self, key):
        if key not in self.cache:
            return -1
        
        # Update usage
        self.usage.remove(key)
        self.usage.append(key)
        return self.cache[key]
    
    def put(self, key, value):
        if key in self.cache:
            self.usage.remove(key)
        elif len(self.cache) >= self.capacity:
            # Remove LRU
            lru = self.usage.pop(0)
            del self.cache[lru]
        
        self.cache[key] = value
        self.usage.append(key)

# Test
cache = LRUCache(3)
cache.put(1, 'A')
cache.put(2, 'B')
cache.put(3, 'C')
cache.display()  # [(1, 'A'), (2, 'B'), (3, 'C')]

cache.get(1)     # Access key 1
cache.put(4, 'D')  # Evicts key 2 (LRU)
cache.display()  # [(3, 'C'), (1, 'A'), (4, 'D')]

Time Complexity: O(1) for get and put
Space Complexity: O(capacity)

Q5. Detect a cycle in a directed graph.

Solution (Python):

def has_cycle_directed(graph):
    """
    Detect cycle in directed graph using DFS
    graph: dict of {node: [neighbors]}
    """
    WHITE, GRAY, BLACK = 0, 1, 2
    color = {node: WHITE for node in graph}
    
    def dfs(node):
        color[node] = GRAY  # Currently being processed
        
        for neighbor in graph.get(node, []):
            if color[neighbor] == GRAY:
                # Back edge found - cycle exists
                return True
            if color[neighbor] == WHITE and dfs(neighbor):
                return True
        
        color[node] = BLACK  # Processing complete
        return False
    
    for node in graph:
        if color[node] == WHITE:
            if dfs(node):
                return True
    
    return False

# Alternative: Kahn's Algorithm (Topological Sort)
def has_cycle_kahn(graph):
    """Detect cycle using in-degree method"""
    from collections import deque
    
    # Calculate in-degrees
    in_degree = {node: 0 for node in graph}
    for node in graph:
        for neighbor in graph[node]:
            in_degree[neighbor] = in_degree.get(neighbor, 0) + 1
    
    # Start with nodes having in-degree 0
    queue = deque([node for node in graph if in_degree[node] == 0])
    processed = 0
    
    while queue:
        node = queue.popleft()
        processed += 1
        
        for neighbor in graph.get(node, []):
            in_degree[neighbor] -= 1
            if in_degree[neighbor] == 0:
                queue.append(neighbor)
    
    # If not all nodes processed, cycle exists
    return processed != len(graph)

# Test
dag = {
    'A': ['B', 'C'],
    'B': ['D'],
    'C': ['D'],
    'D': ['E'],
    'E': []
}

graph_with_cycle = {
    'A': ['B'],
    'B': ['C'],
    'C': ['A']  # Cycle: A->B->C->A
}

print(has_cycle_directed(dag))           # False
print(has_cycle_directed(graph_with_cycle))  # True
print(has_cycle_kahn(dag))               # False
print(has_cycle_kahn(graph_with_cycle))  # True

Time Complexity: O(V + E)
Space Complexity: O(V)

Cisco Interview Tips 2026

1. Master Networking Fundamentals

Cisco interviews heavily focus on networking concepts. Be thorough with OSI model, TCP/IP, routing protocols (OSPF, BGP, EIGRP), VLANs, ACLs, NAT, and subnetting. CCNA-level knowledge is expected.

2. Understand Cisco's Product Portfolio

Research Cisco's key products: Catalyst switches, ISR/ASR routers, Nexus data center switches, Meraki cloud-managed devices, and security solutions (Firepower, Umbrella).

3. Practice Packet Tracer/GNS3

Hands-on experience with network simulation tools demonstrates practical skills. Build simple networks, configure routing, and troubleshoot connectivity issues.

4. Prepare for Technical Deep Dives

Be ready to explain concepts in depth: "How does TCP congestion control work?" "Explain the OSPF neighbor establishment process." "How does VLAN trunking work?"

5. Showcase Programming Skills

For software roles, demonstrate Python or network automation skills (Ansible, Netconf/RESTconf). Cisco emphasizes network programmability and automation.

6. Problem-Solving Approach

When faced with network troubleshooting scenarios, structure your approach systematically: Define problem → Gather information → Isolate cause → Test hypothesis → Implement solution.

7. Stay Updated on Industry Trends

Be aware of SDN (Software Defined Networking), SD-WAN, 5G, IoT, and cloud networking. Cisco values candidates who understand industry evolution.

Frequently Asked Questions

What is the Cisco placement paper pattern for 2026?

What is the difficulty level of Cisco technical test?

How to prepare for Cisco placement in 1 month?

Week 1: Master networking fundamentals (OSI model, TCP/IP, subnetting, VLANs)
Week 2: Study routing protocols (OSPF, BGP), switching concepts, and network security basics
Week 3: Practice aptitude, revise CS fundamentals (DSA, OS), and solve previous year papers
Week 4: Focus on coding (arrays, trees, graphs, networking-related problems), take mock tests Daily study of 4-5 hours is recommended with hands-on practice using Packet Tracer.

What is the CTC offered by Cisco for freshers in 2026?

Software Engineer I: ₹14 - 18 LPA
Network Engineer: ₹12 - 15 LPA
Systems Engineer: ₹13 - 16 LPA
Technical Consulting Engineer: ₹15 - 19 LPA
Data Scientist: ₹16 - 20 LPA Packages include base salary, RSUs, joining bonus, and comprehensive benefits.

What networking certifications help in Cisco recruitment?

CCNA (Cisco Certified Network Associate) - Highly recommended
CCNP (Cisco Certified Network Professional) - Advantage for advanced roles
Network+ - Foundation-level certification
AWS/Azure networking certifications - For cloud networking roles
Python network automation courses - Increasingly valuable

All the best for your Cisco placement preparation! Keep learning and stay curious about networking technologies.