Python Coding Challenge - Question with Answer (01171125)

 

Explanation:

Assign a value to num

num = 5

We set num = 5.

We will use this value for checking divisibility.

Create the list a

a = [5, 10, 11, 13]

This list contains four numbers:

5, 10, 11, 13

Initialize sum variable

s = 0

We create s to store the total of selected numbers.

Starting value = 0

Start loop — go through each value in list

for v in a:

The loop takes each value v from the list:

First: v = 5

Then: v = 10

Then: v = 11

Then: v = 13

Check divisibility using IF + continue

    if v % num == 0:

        continue

We check:

If v is divisible by num (5), skip it.

Check each:

v v % 5 Divisible? Action

5 0             Yes         Skip

10 0             Yes         Skip

11 1              No         Add

13 3              No         Add

continue means: skip the rest of the loop and go to the next number.

Add the non-skipped values

    s += v

Only numbers NOT divisible by 5 get added:

s = 11 + 13 = 24

Print the final answer

print(s)

So the output is:

24

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Python Coding Challenge - Question with Answer (01161125)

 Explanation:

List Initialization

a = [1, 4, 7, 10]

You create a list a that contains four numbers:

1, 4, 7, 10

Variable Initialization

s = 0

You create a variable s to store the sum.

Initially, sum = 0.

Loop Through Each Value

for v in a:

This loop will pick each value from list a, one by one.

So v becomes:

1 → 4 → 7 → 10

Check Condition

if v % 3 == 1:

You check if the number leaves remainder 1 when divided by 3.

Let’s check each:

v v % 3 Remainder Condition v%3==1?

1 1             Yes             ✔ Add

4 1             Yes             ✔ Add

7 1             Yes             ✔ Add

10 1             Yes             ✔ Add

All numbers satisfy the condition.

Add to Sum

s += v

Add all numbers that passed the condition.

s = 1 + 4 + 7 + 10 = 22

Print the Final Sum

print(s)

This prints:

22

Final Output: 22

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Python Coding Challenge - Question with Answer (01151125)

Explanation:

Line 1: Initialization

count = 0

A variable count is created and initialized to 0.

This variable will act as the loop counter.

Line 2: Start of While Loop

while count < 5:

The loop will continue as long as count is less than 5.

The condition is checked before each iteration.

Line 3: Increment Counter

count += 1

In each iteration, count is increased by 1.

Equivalent to count = count + 1.

So count will take values: 1, 2, 3, 4, 5 during the iterations.

Line 4: Continue Condition

if count == 4:

    continue

If count equals 4, the continue statement executes.

continue means: skip the rest of this iteration and move to the next iteration.

Here, print(count) will be skipped when count == 4.

Line 5: Print Statement

print(count, end=" ")

If count is not 4, it will be printed.

end=" " ensures that the output is printed on the same line separated by a space.

Execution Flow

Iteration count value if count==4? Action Output so far

1             1                     No         print 1         1

2             2                     No         print 2         1 2

3             3                     No         print 3 1 2 3

4             4                     Yes continue (skip) 1 2 3

5             5                     No print 5 1 2 3 5

Final Output

1 2 3 5

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Python Coding Challenge - Question with Answer (01141125)

 

Explanation:

1. Class Declaration

class Counter:

This line defines a class named Counter.

2. Class Variable (Shared Across All Calls)

    x = 1

This is a class variable, not tied to any object.

Its value is shared every time the method is called.

Initial value: 1

3. Method Without self

    def nxt():

        Counter.x *= 2

        return Counter.x

Explanation:

def nxt(): → This method does not use self because we are not creating objects.

Counter.x *= 2 → Every time the method is called, the value of x is doubled.

return Counter.x → The updated value is returned.

4. Loop That Calls the Method Several Times

for _ in range(4):

This loop runs 4 times.

5. Printing the Result Each Time

    print(Counter.nxt(), end=" ")

Each loop iteration calls Counter.nxt()

The returned value is printed

end=" " keeps everything on one line with spaces

Final Output

2 4 8 16

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Python Coding Challenge - Question with Answer (01131125)

 

Explanation:

Create a List

nums = [2, 4, 6]

A list named nums is created with three integer elements: 2, 4, and 6.

This will be used to calculate the average of its elements later.

Initialize Loop Control Variables

i = 0

s = 0

i → Acts as a loop counter, starting from 0.

s → A sum accumulator initialized to 0.

It will store the running total of the numbers in the list.

Start the While Loop

while i < len(nums):

The loop runs as long as i is less than the length of the list nums.

Since len(nums) is 3, this means the loop will run while i = 0, 1, 2.

This ensures every element in the list is processed once.

Add the Current Element to the Sum

    s += nums[i]

At each loop iteration:

The element at index i is accessed: nums[i].

It is added to the sum s.

Example of how s changes:

When i=0: s = 0 + 2 = 2

When i=1: s = 2 + 4 = 6

When i=2: s = 6 + 6 = 12

Increment the Loop Counter

    i += 1

After processing one element, i increases by 1.

This moves to the next element of the list in the next iteration.

Print the Final Result

print(s // len(nums))

Once the loop ends, the total sum s is 12.

The expression s // len(nums) performs integer division:

12 // 3 = 4

Hence, it prints the average (integer form) of the list elements.

Final Output

4

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Python Coding Challenge - Question with Answer (01121125)

 Explanation:

1. Defining the Class

class Test:

This line defines a class named Test.

A class in Python is a blueprint for creating objects (instances).

2. Defining a Method Inside the Class

def square(self, n):

    return n * n

This defines a method named square inside the class Test.

self refers to the object that will call this method.

n is a parameter (the number whose square we will calculate).

return n * n computes the square of n and returns it.

3. Creating an Object of the Class

t = Test()

Here, an object t is created from the class Test.

Now, t can access all the methods of the class, including square().

4. For Loop Iteration

for i in range(1, 4):

The range(1, 4) generates the sequence 1, 2, 3.

So the loop will run 3 times with i = 1, then 2, then 3.

5. Calling the Method Inside the Loop

print(t.square(i), end=" ")

On each loop iteration, the method square() is called with the current value of i.

The method returns the square of i.

The print(..., end=" ") prints each result on the same line separated by spaces.

Let’s see what happens in each iteration:

Iteration Value of i t.square(i) Printed Output

1 1 1×1 = 1 1

2 2 2×2 = 4 1 4

3 3 3×3 = 9 1 4 9

Final Output

1 4 9 

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Python Coding Challenge - Question with Answer (01111125)

 

Explanation:

1. Class Definition

class Test:

    x = 5

Explanation:

Test is a class with a class variable x initialized to 5.

Class variables are shared across all instances unless overridden in an instance.

2. Creating Instances

t1 = Test()

t2 = Test()

Explanation:

t1 and t2 are instances of the Test class.

Initially, neither instance has its own x, so they both refer to the class variable Test.x, which is 5.

3. Overriding Instance Variable

t1.x = 10

Explanation:

Here, we assign 10 to t1.x.

Python now creates an instance variable x inside t1 that shadows the class variable x.

t2.x still refers to the class variable Test.x.

Test.x remains unchanged (5).

4. Printing Values

print(t1.x, t2.x, Test.x)

Step-by-step Output:

t1.x → 10 (instance variable of t1)

t2.x → 5 (still refers to class variable Test.x)

Test.x → 5 (class variable remains unchanged)

Final Output:

10 5 5

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Python Coding Challenge - Question with Answer (01101125)

 

Explanation:

 Creating a list of pairs

pairs = [(1,2),(3,4),(5,6)]

pairs is a list containing three tuples.

Each tuple has two numbers: (x, y).

Initializing sum

s = 0

A variable s is created to store the running total.

Starts at 0.

 Loop starts

for x, y in pairs:

The loop iterates over each tuple in pairs.

On each iteration:

x receives the first value.

y receives the second value.

Iterations:

(1, 2)

(3, 4)

(5, 6)

Modify x inside loop

x += y

Adds y to x.

This does not change the original tuple (tuples are immutable).

This only changes the local variable x.

Step-by-step:

x = 1 + 2 = 3

x = 3 + 4 = 7

x = 5 + 6 = 11

Add new x to sum

s += x

Adds the updated value of x into s.

Running total:

s = 0 + 3 = 3

s = 3 + 7 = 10

s = 10 + 11 = 21

Line 6 — Final output

print(s)

Displays the total sum.

Final Output:

21

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Python Coding Challenge - Question with Answer (01091125)

Explanation:

1. List Initialization

nums = [5, 2, 8, 1]

A list named nums is created.

It contains four elements: 5, 2, 8, 1.

2. Initialize Result Variable

r = 0

A variable r is created to store the running total.

It is initially set to 0.

3. Start of the Loop

for i in range(len(nums)):

len(nums) = 4, so range(4) gives: 0, 1, 2, 3.

This loop runs once for each index in the list.

i represents the current index.

4. Update the Result

    r += nums[i] - i

For each index i:

Take the value at that index → nums[i]

Subtract the index → nums[i] - i

Add that result to r

5. Step-By-Step Calculation

i = 0

nums[0] = 5

5 − 0 = 5

r = 0 + 5 = 5

i = 1

nums[1] = 2

2 − 1 = 1

r = 5 + 1 = 6

i = 2

nums[2] = 8

8 − 2 = 6

r = 6 + 6 = 12

i = 3

nums[3] = 1

1 − 3 = −2

r = 12 − 2 = 10

6. Print Final Output

print(r)

Prints the final result.

Output:

10

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Python Coding Challenge - Question with Answer (01081125)

 

Step-by-Step Explanation

1. Dictionary

   d = {"a":2, "b":4, "c":6}

   This dictionary has keys (a, b, c) and values (2, 4, 6).

2. Initialize a variable

   x = 1

   We start x with 1 because we are going to multiply values.

3. Loop through the values of dictionary

   for v in d.values(): 
       x *= v

   • On each loop, v takes one value from the dictionary.

   • x *= v means x = x * v.

   Let's see how x changes:

   • First value v = 2: x = 1 * 2 = 2

   • Next value v = 4: x = 2 * 4 = 8

   • Next value v = 6: x = 8 * 6 = 48

4. Print Result

   print(x)

   Output will be:

   48

---

Final Output

48

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Python Coding Challenge - Question with Answer (01071125)

 

Step-by-Step Explanation

Step	Value of i	Condition i > 4	Action Taken	Value of x
1	1	No	x = x + 1	1
2	3	No	x = x + 3	4
3	5	Yes (5>4)	continue → Skip adding	4
4	6	Yes (6>4)	continue → Skip adding	4

---

What is happening?

• The loop goes through each number in the list a.

• If a number is greater than 4, the continue statement skips adding it.

• Only numbers less than or equal to 4 are added to x.

So:

x = 1 + 3 = 4

---

✅ Final Output

4

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Python Coding Challenge - Question with Answer (01061125)

 

Step-by-Step Execution

Loop Step	i Value	Expression (i + value)	New value Calculation	Updated value
Start	—	—	—	2
1st loop	1	1 + 2 = 3	2 + 3	5
2nd loop	2	2 + 5 = 7	5 + 7	12
3rd loop	3	3 + 12 = 15	12 + 15	27

---

✅ Final Output

27

---

Why this happens?

Each loop uses the updated value to calculate the next addition.
So the value grows faster (this is an example of cumulative feedback in loops).

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Python Coding Challenge - Question with Answer (01051125)

 

Step 1:

count = 0
We start with a variable count and set it to 0.

Step 2:

for i in range(1, 5):

• range(1,5) means numbers from 1 to 4
  (remember, Python stops 1 number before 5)

So the loop runs with:

i = 1
i = 2
i = 3
i = 4

Step 3:

count += i means:

count = count + i

So step-by-step:

i	count (before)	count = count + i	count (after)
1	0	0 + 1	1
2	1	1 + 2	3
3	3	3 + 3	6
4	6	6 + 4	10

Step 4:

print(count) prints 10

---

✅ Final Output:

10

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Python Coding Challenge - Question with Answer (01041125)

 

Step-by-step explanation:

1. range(3) → gives numbers 0, 1, 2.

2. The loop runs three times — once for each i.

3. Inside the loop:

   • The first statement is continue.

   • continue immediately skips the rest of the loop for that iteration.

   • So print(i) is never reached or executed.

 Output:

(no output)

💡 Key point:

When Python hits continue, it jumps straight to the next iteration of the loop — skipping all remaining code below it for that cycle.

So even though the loop runs 3 times, print(i) never runs at all.

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Python Coding Challenge - Question with Answer (01031125)

 

Step-by-step explanation:

1. Dictionary d → {1:10, 2:20, 3:30}

   • Keys → 1, 2, 3

   • Values → 10, 20, 30

2. .items()

   • Returns key-value pairs as tuples:
     (1, 10), (2, 20), (3, 30)

3. Loop:

   • First iteration → k=1, v=10 → print(1+10) → 11

   • Second iteration → k=2, v=20 → print(2+20) → 22

   • Third iteration → k=3, v=30 → print(3+30) → 33

4. end=' '

   • Keeps output on one line separated by spaces.

✅ Output:

11 22 33

 Concept Used:
Looping through a dictionary using .items() gives both key and value, allowing arithmetic or logic to be performed on them together.

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Python Coding Challenge - Question with Answer (01011125)

 

 Step-by-step explanation:

1. arr = [5, 10, 15]
   → A list (array) with three elements.

2. range(0, len(arr), 2)
   → This generates a sequence of numbers starting from 0 up to len(arr)-1 (which is 2),
   → but it skips every 2nd number.
   
   So the output of range(0, len(arr), 2) is:
   👉 [0, 2]

3. Loop runs like this:

   • When i = 0 → arr[i] = arr[0] = 5

   • When i = 2 → arr[i] = arr[2] = 15

4. print(arr[i], end=' ')
   → Prints the selected elements on the same line (because of end=' ').

✅ Output:

5 15

👉 The code prints elements at even indices (0 and 2) from the list.

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Python Coding Challenge - Question with Answer (01311025)

 

Step 1: The list data

data is a list of lists:

[
  [1, 2],
  [3, 4]
]

So, it has two elements:

• First element → [1, 2]

• Second element → [3, 4]

---

🔸 Step 2: The for loop

for x, y in data:

This line unpacks each inner list into two variables x and y.

• On first iteration → x = 1, y = 2

• On second iteration → x = 3, y = 4

---

🔸 Step 3: Inside the loop

print(x + y, end=' ')

• First iteration → x + y = 1 + 2 = 3

• Second iteration → x + y = 3 + 4 = 7

end=' ' keeps the output on the same line.

---

✅ Final Output:

3 7

---

💡 Quick Tip:

This is called sequence unpacking — it works when each element in the list (like [1, 2]) contains exactly two items to assign to x and y.

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Python Coding Challenge - Question with Answer (01301025)

 

Step-by-Step Execution

1. range(3) → means the loop will normally run for i = 0, 1, 2.

---

2. First iteration:

   i = 0

   • print(i) → prints 0

   • if i == 1: → condition False, so it continues.

---

3. Second iteration:

   i = 1

   • print(i) → prints 1

   • if i == 1: → condition True

   • Executes break → this stops the loop immediately.

---

4. Because the loop is terminated using break,
   the else block is skipped.

---

Output

0
1

---

Key Concept

In Python,
for...else and while...else blocks work like this:

• The else part runs only if the loop completes normally (no break).

• If a break statement is used, the else block will not execute.

---

Try Changing It

If you remove the break, like this:

for i in range(3):
    print(i)
else: 
    print('Finished')

Then output will be:

0
1
2
Finished

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Python Coding Challenge - Question with Answer (01291025)

 

Explanation:

Import the NumPy Library

import numpy as np

This line imports the NumPy library and gives it a short alias name np.

NumPy provides functions for creating and manipulating arrays efficiently.

Create a NumPy Array

a = np.arange(1,10).reshape(3,3)

np.arange(1,10) → creates a 1D array from 1 to 9 (end is exclusive of 10).

Result: [1, 2, 3, 4, 5, 6, 7, 8, 9]

.reshape(3,3) → converts it into a 3x3 matrix (2D array).

So now:

a =

[[1 2 3]

 [4 5 6]

 [7 8 9]]

Initialize a Variable

total = 0

total will store the running sum of some elements from the array.

Starts at 0.

Loop Through Each Row

for i in range(3):

range(3) → generates numbers 0, 1, 2

These are the row indices of the array a.

So the loop runs 3 times:

1st iteration → i = 0

2nd iteration → i = 1

3rd iteration → i = 2

Access and Add Last Column Element

total += a[i, 2]

a[i, 2] means:

Row = i

Column = 2 (the third column, since index starts at 0)

Now let’s see each iteration:

i a[i,2] total (after addition)

0 3 0 + 3 = 3

1 6 3 + 6 = 9

2 9 9 + 9 = 18

Print the Result

print(total)

Output:

18

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