Mayank Mayur CS Practical file 2024-25 Class-11th-C

actical-file-2024-25-class-11th-c

February 17, 2025

[2]: # Q1-WAP to calculate the area and cicumference of a circle if radius is given
# Ans:-

r=7
�=3.14
a=�*r*r
c=2*�*r
print("The area of the circle is",a)
print('The circumfernce of the circle is',c)

The area of the circle is 153.86

The circumfernce of the circle is 43.96

[1]: #Q2-WAP to make a simple calculator that reads two numbers and an operator. On␣
↪selection of an operator, perform operation and print the result

#Ans-

a=int(input("enter 1st number: "))

b=int(input("enter 2nd number: "))
op=input("enter (+.-,/,//,*,**,%)")

if op=='+':
print("Result is",a+b)
elif op=='-':
print("Result is",a-b)
elif op=='/':
print("Result is",a/b)
elif op=='//':
print("Result is",a//b)
elif op=='*':
print("Result is",a*b)
elif op=='**':
print("Result is",a**b)
elif op=='%':
print("Result is",a%b)
else:
print("invalid Chacracter please try again")

1
 enter 1st number: 2
enter 2nd number: 10
enter (+.-,/,//,*,**,%)**
Result is 1024

[12]: #Q3 You are driving a little too fast and the police officer stops you and␣
↪issues a ticket. Write code to compute the result, encoded as an integer␣

↪value: 0=no ticket, 1=small ticket, 2=big ticket.

#If speed is 60 or less, the result is 0. If speed is between 61 and 80␣

↪inclusive, the result is 1. If speed is 81 or more, the result is 2. Unless␣

↪it is your birthday, on that day, your speed can be 5 higher in all cases.␣

↪WAP to model above scenario.

#Ans-

bday=input("Is your birthday today!! enter y/n : ")

speed=int(input("enter the speed of your vehicle you good sir : "))

if bday=="y":
if speed<65:
Ticket=0
elif (speed>65 and speed<85):
Ticket=1
elif speed>85:
Ticket=2

else :
if speed<60:
Ticket=0
elif speed>60 and speed<80:
Ticket=1
elif speed>80:
Ticket=2

if Ticket==0:
print("The speed is =", speed ,'and bday is =', bday ,", So No ticket for␣
↪you")

elif Ticket==1:
print("The speed is =", speed ,'and bday is =', bday ,", Small ticket")
elif Ticket==2:
print("The speed is =", speed ,'and bday is =', bday ,", Big ticket")

Is your birthday today!! enter y/n : y

enter the speed of your vehicle you good sir : 56
The speed is = 56 and bday is = y , So No ticket for you

2
[1]: #Q6 Write a program which will find all such numbers which are divisible by 7␣
↪but are not multiple of 5, between 2000 and 3200 (both included).

#The numbers obtained should be printed in a comma-separated sequence on a␣

↪single line.

#ans:-

for x in range (2000,3201):

if (x%7==0 and x%5!=0):
print(x,end=", ")

2002, 2009, 2016, 2023, 2037, 2044, 2051, 2058, 2072, 2079, 2086, 2093, 2107,
2114, 2121, 2128, 2142, 2149, 2156, 2163, 2177, 2184, 2191, 2198, 2212, 2219,
2226, 2233, 2247, 2254, 2261, 2268, 2282, 2289, 2296, 2303, 2317, 2324, 2331,
2338, 2352, 2359, 2366, 2373, 2387, 2394, 2401, 2408, 2422, 2429, 2436, 2443,
2457, 2464, 2471, 2478, 2492, 2499, 2506, 2513, 2527, 2534, 2541, 2548, 2562,
2569, 2576, 2583, 2597, 2604, 2611, 2618, 2632, 2639, 2646, 2653, 2667, 2674,
2681, 2688, 2702, 2709, 2716, 2723, 2737, 2744, 2751, 2758, 2772, 2779, 2786,
2793, 2807, 2814, 2821, 2828, 2842, 2849, 2856, 2863, 2877, 2884, 2891, 2898,
2912, 2919, 2926, 2933, 2947, 2954, 2961, 2968, 2982, 2989, 2996, 3003, 3017,
3024, 3031, 3038, 3052, 3059, 3066, 3073, 3087, 3094, 3101, 3108, 3122, 3129,
3136, 3143, 3157, 3164, 3171, 3178, 3192, 3199,

[2]: #Q7 Develop a program to classify students amongst various categories as per␣
↪their age entered. Read age of N students and count no of students falling␣

↪in each category

#Ans-

n=int(input("Enter the number of People = "))

gpa=gpb=gpc=gpd=gpe=0
for x in range(1,n+1):
age=int(input("Enter the age of the person = "))
if age>=12 and age<15:
gpa+=1
elif age>=15 and age<17:
gpb+=1
elif age>=17 and age<19:
gpc+=1
elif age<12:
gpd+=1
else:
gpe+=1
print("People in Group A: 12 yrs and above but less than 15 yrs =",gpa)
print("People in Group B: 15 yrs and above but less than 17 yrs =",gpb)
print("People in Group C: 17 yrs and above but less than 19 yrs =",gpc)
print("People in Group D: Less than 12 yrs =",gpd)
print("People in Group E: Above 19 yrs =",gpe)

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 Enter the number of People = 5
Enter the age of the person = 10
Enter the age of the person = 13
Enter the age of the person = 78
Enter the age of the person = 16
Enter the age of the person = 18
People in Group A: 12 yrs and above but less than 15 yrs = 1
People in Group B: 15 yrs and above but less than 17 yrs = 1
People in Group C: 17 yrs and above but less than 19 yrs = 1
People in Group D: Less than 12 yrs = 1
People in Group E: Above 19 yrs = 1

[1]: #q4 Write a program to print following patterns upto N lines

#i)
n=int(input("enter a number = "))
for i in range(1,n+1):
for x in range(1,i+1):
print("*",end=" ")
print()

enter a number = 5
*
* *
* * *
* * * *
* * * * *

[2]: #q4 Write a program to print following patterns upto N lines

#ii]

n=int(input("enter a number = "))

for i in range(1,n+1):
print(" "*(n-i),end="")
print("*"*i,end="")
print()

enter a number = 6
*
**
***
****
*****
******

[2]: #Q8.Write a program to find if a number entered is a palindrome or not

n=int(input("enter a number"))
rev=0

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 x=n
while(n>0):
digit=n%10
rev=rev*10+digit
n=n//10
if rev==x:
print(x,"is a palindrome")
else:
print(x,"is not a plaindrome")

enter a number1234
1234 is not a plaindrome
9. According to a study, the approximate level of Intelligence of a person can be calculated using
the following formula: i=2(y+0.5x) Write program, which will produce a table of values of i, y and
x where y varies from 1 to 6 and for each value of y , x varies from 5.5 to 12.5 in steps of 0.5.
[1]: for y in range(1,7):
for x in range(5,13):
x+=0.5
print("value of the y = ",y)
print("value of the x = ",x)
i=2*(y+(0.5*x))
print("value of the i = ",i)
print()

value of the y = 1
value of the x = 5.5
value of the i = 7.5

value of the y = 1
value of the x = 6.5
value of the i = 8.5

value of the y = 1
value of the x = 7.5
value of the i = 9.5

value of the y = 1
value of the x = 8.5
value of the i = 10.5

value of the y = 1
value of the x = 9.5
value of the i = 11.5

value of the y = 1
value of the x = 10.5

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value of the i = 12.5

value of the y = 1
value of the x = 11.5
value of the i = 13.5

value of the y = 1
value of the x = 12.5
value of the i = 14.5

value of the y = 2
value of the x = 5.5
value of the i = 9.5

value of the y = 2
value of the x = 6.5
value of the i = 10.5

value of the y = 2
value of the x = 7.5
value of the i = 11.5

value of the y = 2
value of the x = 8.5
value of the i = 12.5

value of the y = 2
value of the x = 9.5
value of the i = 13.5

value of the y = 2
value of the x = 10.5
value of the i = 14.5

value of the y = 2
value of the x = 11.5
value of the i = 15.5

value of the y = 2
value of the x = 12.5
value of the i = 16.5

value of the y = 3
value of the x = 5.5
value of the i = 11.5

value of the y = 3
value of the x = 6.5

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value of the i = 12.5

value of the y = 3
value of the x = 7.5
value of the i = 13.5

value of the y = 3
value of the x = 8.5
value of the i = 14.5

value of the y = 3
value of the x = 9.5
value of the i = 15.5

value of the y = 3
value of the x = 10.5
value of the i = 16.5

value of the y = 3
value of the x = 11.5
value of the i = 17.5

value of the y = 3
value of the x = 12.5
value of the i = 18.5

value of the y = 4
value of the x = 5.5
value of the i = 13.5

value of the y = 4
value of the x = 6.5
value of the i = 14.5

value of the y = 4
value of the x = 7.5
value of the i = 15.5

value of the y = 4
value of the x = 8.5
value of the i = 16.5

value of the y = 4
value of the x = 9.5
value of the i = 17.5

value of the y = 4
value of the x = 10.5

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value of the i = 18.5

value of the y = 4
value of the x = 11.5
value of the i = 19.5

value of the y = 4
value of the x = 12.5
value of the i = 20.5

value of the y = 5
value of the x = 5.5
value of the i = 15.5

value of the y = 5
value of the x = 6.5
value of the i = 16.5

value of the y = 5
value of the x = 7.5
value of the i = 17.5

value of the y = 5
value of the x = 8.5
value of the i = 18.5

value of the y = 5
value of the x = 9.5
value of the i = 19.5

value of the y = 5
value of the x = 10.5
value of the i = 20.5

value of the y = 5
value of the x = 11.5
value of the i = 21.5

value of the y = 5
value of the x = 12.5
value of the i = 22.5

value of the y = 6
value of the x = 5.5
value of the i = 17.5

value of the y = 6
value of the x = 6.5

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 value of the i = 18.5

value of the y = 6
value of the x = 7.5
value of the i = 19.5

value of the y = 6
value of the x = 8.5
value of the i = 20.5

value of the y = 6
value of the x = 9.5
value of the i = 21.5

value of the y = 6
value of the x = 10.5
value of the i = 22.5

value of the y = 6
value of the x = 11.5
value of the i = 23.5

value of the y = 6
value of the x = 12.5
value of the i = 24.5

10. WAP to print the sum of n terms of the following series - A) x+ x2/2 + x3/3 + x4/4 +....+xn/n

[2]: x=int(input("Enter the value of x"))

n=int(input("Enter the value of n"))
result=0
for i in range(1,n+1):
y=(x**i)/i
result+=y
print("The result is",result)

Enter the value of x1

Enter the value of n2
The result is 1.5
B) 1/ (sqrt(2) + 2 / sqrt(3) + 3/ sqrt(4) +.......n/sqrt(n+1)

[5]: n=int(input("enter the value of n"))

result=0
for i in range(1,n+1):
y=(i/((i+1)**1/2))
result+=y
print("The result is",result)

9
 enter the value of n3
The result is 3.833333333333333
11. Read two strings and check if string1 is prefix, postfix or nothing from the string2.
[4]: s=input("Enter the main string - ")
sub=input("enter the substring - ")
news=""
for i in range(len(sub)):
x=s[i]
news+=x
if news==sub:
print(sub,"is the prefix of",s)
else:
print(sub,"is not the prefix of",s)

Enter the main string - evergreen

enter the substring - ever
ever is the prefix of evergreen
12. Write a program to read a string and print i)Frequency of all characters ii) Word of highest
length iii) The string in title case iv) Read full name as a string and print only the initial letters
except last name Eg- Enter name: Mohan Das Karam Chand Gandhi, O/p M D K C Gandhi.

[5]: #I)Frequency
s=input("enter the string - ")
print("Frequency of the letters")
news=""
for i in range(len(s)):
if s[i] not in news:
news+=s[i]
print(s[i],"comes",s.count(s[i]),"times")
print()
print()

#II)Highest length
print("Highest length")
l=s.split()
print(l)
high=l[0]
y=0
for x in l:
if len(high)<len(x):
high=x
print(high)
print()
print()

#III)Title case

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print("Title case")
print(s.title())
print()
print()

#IV)M D K C Gandhi
print("M D K C Gandhi")
g=s.title()
o=g.split()
print(o)
name=""
for i in range(len(o)-1):
name+= o[i][0].upper()+" "
print(name+o[-1])

enter the string - mohan das karam chand gandhi

Frequency of the letters
m comes 2 times
o comes 1 times
h comes 3 times
a comes 6 times
n comes 3 times
comes 4 times
d comes 3 times
s comes 1 times
k comes 1 times
r comes 1 times
c comes 1 times
g comes 1 times
i comes 1 times

Highest length
['mohan', 'das', 'karam', 'chand', 'gandhi']
gandhi

Title case
Mohan Das Karam Chand Gandhi

M D K C Gandhi
['Mohan', 'Das', 'Karam', 'Chand', 'Gandhi']
M D K C Gandhi
13. In Cryptography, a Caesar cipher is a very simple encryption techniques in which each letter
in the plain text is replaced by a letter some fixed number of positions down the alphabet list in a

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 cyclic manner. For example, with a shift of 3, A would be replaced by D, B would become E, and
so on. The method is named after Julius Caesar, who used it to communicate with his generals.
ROT-13 (”rotate by 13 places”). Create menu driven program to encrypt and decrypt which takes
in a string and rotation (rotate by n places) and encrypts the string then decrypts the string. The
program should be menu driven. Also if rotation is not specified then the encryption should take
place by 13.
[9]: s=input("Enter a string: ")
rotno=input('Do you want to give a specific rotation number y/n: ')

if rotno=='y':
n=int(input("enter the rotation number: "))
elif rotno=='n':
n=13
choice=int(input("Choose to 1.Encrypt or 2.Decrypt: "))
newcode=''

if choice==1:
for i in s:
x=ord(i)+n
if((x>90 and x<97) or (x>122)):
x=x-26
newcode+=chr(x)
print()
print(s,"is encryted to",newcode)
elif choice==2:
for i in range(len(s)):
x=ord(s[i])-n
if((x<65) or (x<97 and x>90)):
x=x+26
newcode+=chr(x)
print()
print(s,"is decryted to",newcode)

Enter a string: ZOOM

Do you want to give a specific rotation number y/n: y
enter the rotation number: 3
Choose to 1.Encrypt or 2.Decrypt: 1

ZOOM is encryted to CRRP

5.WAP that reads n digit number. After reading the number, compute and display the sum of the
odd positioned digits, multiply all even positioned digits and add these two numbers.
[4]: n=int(input("enter a number = "))
s=str(n)
sumodd=0
prodeven=1

12
 for i in range(len(s)):
if i%2==0:
prodeven*=int(s[i])
else:
sumodd+=int(s[i])
result=prodeven+sumodd
print("The result is",result)

enter a number = 1234

The result is 9
14. Write a program which takes 2 digits, X, Y as input and generates a 2-dimensional array. The
element value in thei-th row and j-th column of the array should be i*j.
[5]: r=int(input("enter rows:"))
c=int(input("enter cols:"))
l=[]
for i in range(r):
k=[]
for j in range(c):
k.append(i*j)
l.append(k)
print(l)

for i in range(r):
for j in range(c):
print(l[i][j],end=' ')
print()

enter rows:3
enter cols:5
[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 6, 8]]
0 0 0 0 0
0 1 2 3 4
0 2 4 6 8
16. Write a program to find the second largest number in a given list.
[1]: n=int(input("enter the number of elements:"))
l=[int(input("Enter a number:")) for i in range(n)]
l.sort()
print(l)
print(f'the second largest number is {l[-2]}')

enter the number of elements:4

Enter a number:3
Enter a number:43
Enter a number:12
Enter a number:65

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 [3, 12, 43, 65]
the second largest number is 43
15. Write a program to generate a random 6 digits secure OTP. The digits should not repeat.
[ ]: import random

n=random.randint(0,9)
s=""
while (len(s)!=6):
if str(n) not in s:
s+=str(n)
print(f"Random OTP is {s}")

17. Write a program that takes a list of words and creates a dictionary with word as the
key and number of occurrences of the word as the value. For example if list is given as
[’the’,’of’,’an’,’is’,’an’,’the’] Then the dictionary should be d=’the’:2,’of’:1,’an’:2,’is’:1

[4]: import collections

s=input('Enter words with spaces: ')
l=s.split()
d=dict(collections.Counter(l))
print(d)

Enter words with spaces: the of an is an the

{'the': 2, 'of': 1, 'an': 2, 'is': 1}
19. Write a program to create a dictionary of names of flowers as keys and the colours in which
they exist as values in tuples, for example D=’rose’:(’red’,’black’,’pink’),’lily’:(’white’,’violet’) . Now
write a code to print a) The flower which exists in maximum number of colors b) The colour in
which most of the flowers exist
[14]: #A
d={'rose':('red','black','pink','white'),'lily':('white','violet'),'tulips':
↪('pink','yellow','purple','white'),'daisy':('red','orange','yellow')}

max=0
for i in d:
if len(d[i])>=max:
max = len(d[i])
flower = i
else:
max = max
print(f"Flower with the maximum colours is {flower} and it has {max} no. of␣
↪colours.")

#B
v=list(d.values())
d={}
for i in v:
for j in i:

14
 if j not in d:
d[j]=i.count(j)
else:
d[j]+=i.count(j)
max=0
color=""
for i in d:
if d[i]>max:
max=d[i]
color=i

print(color,"is the color which is most flowers exist in")

Flower with the maximum colours is tulips and it has 4 no. of colours.
white is the color which is most flowers exist in
20. There are 36 possible combinations if we throw two dice. A simple pair of loops over range
(6) +1 will enumerate all combinations. The sum of the two dice is more interesting than the
actual combination. Create a dictionary of all combinations, using the sum of numbers on the two
dice as the key. Each value of dictionary will be a tuple with all possible combination. 12:[(6,6)],
11:((6,5),(5,6))……

[13]: l=[]
d={}
for i in range(1,7):
for j in range(1,7):
l.append((i,j))
for x in l:
if sum(x) not in d:
d[sum(x)]=[(x)]
else:
d[sum(x)].append((x))
print(d)

{2: [(1, 1)], 3: [(1, 2), (2, 1)], 4: [(1, 3), (2, 2), (3, 1)], 5: [(1, 4), (2,
3), (3, 2), (4, 1)], 6: [(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)], 7: [(1, 6),
(2, 5), (3, 4), (4, 3), (5, 2), (6, 1)], 8: [(2, 6), (3, 5), (4, 4), (5, 3), (6,
2)], 9: [(3, 6), (4, 5), (5, 4), (6, 3)], 10: [(4, 6), (5, 5), (6, 4)], 11: [(5,
6), (6, 5)], 12: [(6, 6)]}

[ ]:

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