by Krishna Rai | Feb 19, 2022 | Python Programming, Technical
In this article, we will be discussing some of the InfyTQ questions asked in the InfyTQ sample test released officially by Infosys. These InfyTQ sample questions will give you a fair idea about the difficulty level of the Infytq exam.
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Check out InfyTQ Final Round Python Coding Questions here
InfyTQ Questions | Official Sample Paper Python Qualifying Round
Q1.
Consider the Python code below.
Assume that the necessary imports have been done.
class Shape (metaclass-ABCMeta) :def _init_(self) :print ( “I am in init”)
def draw_shape(self) :passdef set_color(self) :passclass Circle(Shape) :def draw_shape(self) :
print(“Draw Circle”)
Which of the following statement(s) is/are TRUE?
i) Class Circle cannot be instantiated as it does not implement the set_color() method
ii) The above code will not compile because class Shape has two abstract methods
A. Both i) and ii) are True
B. Neither of i) or ii) is True
C. Only i) is True
D. Only ii) is True
Answer: C
Explanation:
The abstract base class may have more than one abstract method.
Here class Shape has more than one abstract method. It is totally possible for a base class to have more than one abstract method but all these methods have to be implemented by the child class or a TypeError would occur.
Statement 1 says that the child class Circle cannot be instantiated which is true, as it does not implement all the abstract methods of the base class.
Statement 2 says the code will not compile because of the base class having more than one abstract method which is false.
Q.2
Consider the below code snippet.
Identify the most efficient test data set for testing the below code using the ‘Logic Coverage’ technique.
if previous_year_percentage>=75 and previous_year_percentage<=85:
scholarship=5000elif previous_year_percentage>85 and previous_year_percentage<=95:
scholarship=8000elif previous_year_percentage>95:
scholarship=1000else:
scholarship=0
A. 79, 87, 91, 99
B. 78, 80, 92, 99
C. 74, 77, 90, 100
D. 74, 75, 76, 84, 85, 86, 94, 95, 96, 97
Answer: C
Explanation:
Logic corresponds to the internal structure of the code and this testing is adopted for safety-critical applications such as software used in the aviation industry. This Test verifies the subset of the total number of truth assignments to the expressions.
Basically, you would be testing all the conditional statements for their respective True and False inputs. Here option C and option D provide logic coverage but the efficient one among them would be option C.
Q3.
Consider the following Python code snippet.
class ClassA :
_param1=200def _init_(self) :
self._param1 = 100def method1(self) :
@staticmethod
def method2( ) :
obj1=ClassA( )
obj1.method1( )
Note: Line numbers are for reference only.
Fill in the blanks at Line1, Line2, and Line3 to get the output as 302.
Choose TWO CORRECT options
A: Line1:ClassA._param1=(ClassA._param1+1)+self._param1 Line2:print(self._param 1+1) Line3:self.method2()
B: Line1:ClassA._param1=(ClassA._param1+2)+ClassA._param1 Line2:print(ClassA._param1) Line3:ClassA.method2()
C: Line1:ClassA._param1=(ClassA._param1+1)+self._param1 Line2:print(ClassA._param1+1) Line3:ClassA.method2()
D: Line1:ClassA._param1=(ClassA._param1+2)+self._param1 Line2:print(ClassA._param 1) Line3:method2()
E: Line1:ClassA._param1=ClassA._param 1 +(self._param1+2) Line2:print(ClassA._param 1) Line3:ClassA.method2()
Answer: You are supposed to select C and E
Explanation:
A:
There will be an error because of a self._param1 call outside the class
B:
This option calls for ClassA.method2()
and method2 is defined as → print(ClassA._param1)
but the method2 is a static method and the values for ClassA.param1 has already been initialized in method1→ (ClassA._param1+2)+ClassA._param1
which is equivalent to → (200+2)+200
That ends up giving us 402 as the output.
C:
This option calls for ClassA.method2()
and method2 is defined as → print(ClassA._param1+1)
but the method2 is a static method and the values for ClassA.param1 has already been initialized in method1→ (ClassA._param1+1)+self._param1
which is equivalent to → (200+1)+100
That ends up giving us 302 as the output.
D:
This option calls for method2()
and method2 is not defined outside the classes so basically it would not return anything
E:
This option again would be a call for the method2→ ClassA.method2()
and method2 is defined as → print(ClassA._param1)
but the method2 is a static method and the values for ClassA.param1 has already been initialized in method1→ ClassA._param1+(self._param1+2)
which is equivalent to → 200+(100+2)
That ends up giving us 302 as the output.
So you can choose C and E.
Q4.
What is the output of the code given below?
def fun(input_list,index):try:
output_list = [ 0 ]*len(input_list)
output_list [index] = input_list[index]/int(input_list[index+1])
return output_list
except ValueError :
print("Invalid value")
except ZeroDivisionError :
print("Division by zero")
finally :
print("End of function")
try :
list1 = [2,4,'6',2,8]
list2 = fun(list1, 4)
except TypeError :
print ("Invalid type")
except IndexError :
print ("Invalid index")
finally :
print ("End of program")
A:
Invalid value
End of function
End of program
B:
Division by zero
End of function
End of program
C:
End of function
Invalid index
End of program
D:
End of function
Invalid type
End of program
Answer: C
Explanation: The syntax of the try-catch block would basically be:
try:
# Some Code….
except:
# optional block
# Handling of exception (if required)
finally:
# Some code …..(always executed)
So here, the list is passed to the function and the execution of the fun method begins, but there would be an index error because of output_list [index] = input_list[index]/int(input_list[index+1]), this would be caught in the second try-catch block. Since there is nothing to be caught in the first try-catch block it would execute and print the statement in the finally block of the first try-catch block which is End of function.
Next up, the index error will be handled and the respective statement Invalid index will be printed.
In the end, the finally block of the second try-catch block will be executed to print End of program.
So the order of the output would be:
End of function
Invalid index
End of program
Q5.
Consider the Python code below
class ClassA:def first_method(self):
print("Johnny Johnny..., ")
def second_method (self):
print ("Yes Papa. ")
def third_method (self):
print("Eating Sugar..., ")
class ClassB (ClassA):def second_method (self):
super () .first_method ()
super ().second_method ()
super() .third_method ()
print("No Papa. ")
def third_method (self):
print("Telling Lies..., ")
class ClassC (ClassB):def first_method (self):
print("Open your mouth..., Ha. Ha. Ha.")
def second_method (self):
print ("No Papa. ")
def third_method (self):
super().second_method()
super() .third_method()
self. second_method()
obj_A=ClassA()
obj_B=ClassB()
obj_C=ClassC()
Which among the below options if written at #Line 1, prints the rhyme correctly?
Choose TWO CORRECT options.
The expected output for your reference:
Johnny Johnny…,
Yes Papa.
Eating Sugar…,
No Papa.
Telling Lies…,
No Papa.
Open your mouth…,
Ha. Ha. Ha.
A:
obj_A.third_method()
obj_B.second_method()
obj_C.first_method()
obj_A.second_method()
B:
obj_C.third_method()
obj_B.second_method()
obj_A.first_method()
obj_C.first_method()
C:
obj_C.third_method()
obj_C.first_method()
D:
obj_B.second_method()
obj_B.third_method()
obj_C.second_method()
obj_C.first_method()
Answer: C
Explanation:
The super() builtin returns a proxy object (temporary object of the superclass) that allows us to access methods of the base class. In Python, super() has two major use cases. Allows us to avoid using the base class name explicitly and working with Multiple Inheritance.
In this particular case, the base class methods would be called whenever there is a super keyword used. The correct order of the execution would be when the instance is created related to ClassC and the method in the base class is used via C by the super keyword.
Q6.
Number 14 needs to be searched using BINARY SEARCH in the following sorted list of numbers:
1, 3, 7, 9, 14, 19, 45
How many comparisons will be required to conclude that the number 14 is found at 5th position?
Note: We have used integer division for finding the middle element and the index starts with 0 (zero)
A. 2
B. 3
C. 4
D. 1
Answer: B
Explanation:
1st iteration: low= 0
high= 6
mid= (0+6)//2 = 3
a[3]<key, which means 9<14
2nd iteration: low = mid+1 = 4
high = 6
mid = (4+6)//2 = 5
a[5]>key, which means 19>14
3rd iteration: low = 4
high = mid – 1 = 4
mid = (4+4)//2 = 4
a[4]==key, which means 14==14.
Q7.
What will be the output of the following Python code snippet?
def func (var1, var2, var3):if(var1<=var2):
if(var3>-var2):
if(var1+var2>-var3):
print(-1)
else:
print(-2)
else:
if(var1+var3>var2):
print (-3)
else:
print (-4)
func(156,2100,9500)
A. -2
B. -3
C. -4
D. -1
Answer: D
Explanation:
The first condition is if(var1<=var2) → 156 <= 2100 #True
It will get inside to the next condition if(var3>-var2) → 9500 > (-2100) #True
It will get inside the next condition if(var1+var2>-var3) → (156+2100) > (-9500) #True
Final the last statement print will be executed which prints -1.
Q8.
Consider the below inputs:
input_linked_list (Head to Tail): 1 -> 2 -> 5 -> 3
input_stack (Top to Bottom): 4, 2, 5, 10
def generate (input_linked_list , input_stack):
temp= input_linked_list.get_head ( )
element=0while(temp.get_next ( ) is not None):
temp.set_data (temp.get_data ( )+temp.get_next ( ). get_data ( )+element)
if temp.get_data ( ) %2 !=0:
temp.set_data(temp.get_data ( ) +input_stack.pop ( ) )
element=temp.get_data ( )
else:
input_stack.push (element )
element=temp.get_next ( ).get_data ( )
temp=temp.get_next ( )
temp.set_data(temp.get_data ( )+input_stack.pop ( ) )
What will be the content of Input_linked_list from head to tail and input_stack from top to bottom after the execution of the function generate?
Assumption: Stack and LinkedList classes, with the necessary methods, are available
A:
input_linked_list (Head to Tail): 7 -> 14 -> 20 -> 5
input_stack (Top of Bottom): 5, 10
B:
input_linked_list (Head to Tail): 5 -> 7 -> 10 -> 5
input_stack (Top of Bottom): 2, 5, 10
C:
input_linked_list (Head to Tail): 7 -> 14 -> 20 -> 3
input_stack (Top of Bottom): 5, 10
D:
input_linked_list (Head to Tail): 7 -> 14 -> 20 -> 5
input_stack (Top of Bottom): 10
Answer: B
Explanation:
All the nodes are checked for the values to be a multiple of 2 and when they are not a value of the stack is popped and added to them and the previous values are overwritten.
Q9.
Consider the following Python code. Choose the correct option with respect to the number of static, instance and local variables in ClassOne:
Class ClassOne :
__var_one=1001
def__init__(self,var_two) :
self.__var_two=var_two
self.__var_five=5def method_one(self) :
var_four=50self.__var_five=ClassOne.__var_one+self.__var_two+var_four
A. 0 static variable, 3 instance variables, 2 local variables
B. 1 static variable, 2 instance variables, 2 local variables
C. 1 static variable, 3 instance variables, 1 local variables
D. 2 static variable, 2 instance variables, 1 local variables
Answer: C
Explanation:
static variables: __var_one
instance variables: var_two, __var_five, var_four
local variables: var_four
Q 10.
Consider the following code snippet written in Python:
def fun(num) :if num<1 :
return 0elif num%2 == 0 :
return fun (num-1)
else :
return num+fun(num-2)
print (fun (8) )
A. 8
B. 12
C. 20
D. 16
Answer: D
Q 11.
What will be the output of below Python code?
class Vehicle ( ):def __init__(self, color):
self.color=color
class FourWheeler(Vehicle):def __init__(self,no_of_gears,color):
super().__init__(color)
self.no_of_gears=no_of_gears
class TwoWheeler(Vehicle):def __init__(self,engine_cc):
super().__init__("Red")
self.engine_cc=engine_cc
self.color="Blue"
maruti=FourWheeler(4, "Red")
print (maruti.color,maruti.no_of_gears)
activa=TwoWheeler(80)
print(activa.color, activa.enigne_cc)
A:
Error as object of TwoWheeler class cannot be created since required number of arguments have not been passed
B:
Red 4
Red 80
C:
Error: Cannot modify the value of attribute color inside TwoWheeler class
D:
Red 4
Blue 80
Answer: C
Explanation: We will not be able to modify the value of attribute colour inside TwoWheeler class, it would indirectly throw an error as ‘TwoWheeler’ object has no attribute ‘enigne_cc’
Q 12.
Consider the following Python function for a vehicle management system. Objective of the code is to record details of vehicles-
def vehicle_record(reg_no,mileage,veh_type,purpose,kickstart,no_of_gears,miles_run) :
record=””
if(veh_type = = “TwoWheeler”) :
record=reg_no+”:TwoWheeler:”+ str(mileage)+”:”+kickstart
elif(veh_type = = “FourWheeler”) :
record=reg_no+”:FourWheeler:”+str(mileage)+”:”+no_of_gears
if(purpose = = “Commercial”):
record=record+”:Commercial FourWheeler:”+miles_run
return record
record=vehicle_record(“KA09Z7211”,15,”FourWheeler”,”Commercial”,”No”,”4”,”89000”)
What will be the optional class structure if this was to be re-written in Object-oriented programming?
A. 4 independent classes: Vehicle, TwoWheeler, FourWheeler, CommercialFourWheeler
B. 3 classes with inheritance: TwoWheeler and FourWheeler inherit from Vehicle
C. 4 classes with inheritance: TwoWheeler, FourWheeler and CommercialFourWheeler which inherit from Vehicle
D. 4 classes with inheritance: TwoWheeler and FourWheeler inherit from Vehicle; CommercialFourWheeler inherit from FourWheeler
Answer: D
Explanation: The main objective of object-oriented programming is code reusability and in the above case it can be done by segregation the parameters into parent-child class pairs. The optimum design would be as given below:
So in this segregation the common features of vehicle class can be used by all its child classes and likewise the common features of four wheeler class can be used by the commercial class.
This arrangement is suggested in option D
Q 13.
Consider the code given below.
def display_cat_details(color, name1, name2=none, name3=None) : pass
display_cat_details(“Brown”, “Fluffy”,”Snow”)
Which of the following function signatures when replaced in Line 1 would continue to execute successfully?
Choose two correct answers
A. def display_cat_details(color, *names,name1)
B. def display_cat_details(color, name_list)
C. def display_cat_details(color, *names)
D. def display_cat_details(color, name1,*names)
Answer: Option C and D are supposed to be selected
Explanation:
A. Is again a invalid syntax as display_cat_details() takes 2 positional arguments but 3 were given
B. Same reason as A
C. This would take in all the 3 respective parameters and execute successfully
D. This also would take all the 3 parameters even though there are only 2 parameters given in the function definition, because of *names
Q 14.
Consider the below code:
my_queue1 = Queue(3)
my_queue2 = Queue(3)
for index in range(0, 3):
my_queue.enqueue(index *3)
for index in range(0, 3):
if(index == 2):
break
my_queue2.enqueue(my_queue1.dequeue ( ) )
my_queue2.enqueue(12)
Assumption: Queue class, with the necessary methods, is available
What is the status of my_queue1 and my_queue2 after the execution of above code?
A:
my_queue1(front->rear):0, 3, 6
my_queue2(front->rear):0, 3, 12
B:
my_queue1(front->rear):6
my_queue2(front->rear):0, 3,
C:
my_queue1(front->rear):6
my_queue2(front->rear):0, 3, 12
D:
my_queue1(front->rear):Queue is empty.
my_queue2(front->rear):0, 3, 12
Answer: C
Explanation: Here the queue1 is being filled as 0, 3, 6 (front to rear). When this is dequeued the order would be 0 first, then 3 and finally 6. So the 1st dequeue would give us 0, so now queue2 becomes 0 (front to rear). Next dequeue of queue1 would give us 3, which would be the 2nd enqueue for queue2. So now queue2 would be 0, 3 (front to rear). The dequeue of queue1 would stop here as the index reaches 2. Final enqueue of queue2 would happen with a 12. So finally queue2 becomes 0, 3, 12 (front to rear).
Q 15.
The following numbers are to be stored in a hash table(arriving in the order shown) using the hash function.
h(k)=k%4
4, 7, 16, 9, 17, 22
Identify for which numbers collision will NOT occur after mapping the numbers with the given hash function.
A. 4, 7, 16 and 9
B. 7, 9, 17 and 22
C. 7 and 22
D. 4, 16, 9 and 17
Answer: C
Explanation:
A. There would be a collision between 4 and 16 which would both give the value 0 for 4%4 and 16%4.
B. There would be a collision between 9 and 7 which would both give the value 1.
C. There would be no collision between 7 and 22.
D. All four numbers would end up in a collision.
Q 16.
Alister is a Python developer who has written the below code.
He desires the output to be 123 Volvo, but, he is unable to proceed due to an error
class Car:def ___init__(self,car_id,car_name) :
self.__car_id=car_id self.car_name=car_name
carl1=Car(123, “Volvo”)
print(car1.car_id,car1.car_name)
Choose two options from below which he should implement to get the desired output?
Note: Line numbers are for reference only.
A. Add a method to the Car class as follows:def get_car_id(self): return car_id
B. Add a method to the Car class as follows:def get_car_id(self): return self._ car_id
C. Change Line 4 to : print(car1.get_car_id(),car1.car_name)
D. Change Line 4 to : print(car1.___car_id,car1.car_name)
Answer: Option B and C supposed to be selected
Explanation: Here we are asked to print 123 Volvo which is being sent as an argument to the function present inside the class Car. But there are 2 problems here.
- The car_id is a private variable inside class Car which means that you would have to define a function inside class Car to access the variable.
- The print statement has car1.car_id which is again not possible as the simple object created for a class cannot access the private variables inside the class
The solutions for these are as follows:
- Defining a function to access the private variable, which is given in option B→ Add a method to the Car class as follows:def get_car_id(self): return self._ car_id
- Use the print statement to access the private variable using the function in class Car which is given in option C→ Change Line 4 to : print(car1.get_car_id(),car1.car_name)
Hence, Option B and C supposed to be selected
Q 17.
Given the following linked_list:
linked_list(Head to Tail): 1->4->6->7->9
What would be the state of the linked list after the following Python function is executed when the head node of the above-given linked_list and num=5 are passed as input to the function?
def func(head, num) :
temp = head
if(head == None) :
returnelse :
index = 1while(index < num) :
val = temp.get_data ( )
temp = temp.get_next ( )
head.set_data ( val )
index +=1
temp1 = head.get_data ( )
temp.set_data(temp 1)
Assumption: LinkedList class, with the necessary methods, is available.
A. Linkedlist data(Head to Tail): 7 4 6 7 9
B. Linkedlist data(Head to Tail): 1 4 6 7 7
C. Linkedlist data(Head to Tail): 7 4 6 7 7
D. Linkedlist data(Head to Tail): 1 4 6 7 9
Answer: B
Explanation:
The iteration stops when the index value is greater than or equal to 5. Hence the last value 9 will be overwritten with 7.
Q 18.
How many ‘*’(stars)will be printed after executing the below Python code?
num1=5
num2=4while (num2 >= 1) :
print (“*”)
for index in range (1, num1+1) :
print (“*”)
num2 -= 1print (“*”)
A. 6
B. 5
C. Infinite loop
D. 7
Answer: D
Explanation: In the beginning of while loop 1 star will be printed. After the control enters the for loop star will be printed 5 times and 1 last star will be printed when the for loop breaks. So altogether 7 stars will be printed.
Q 19.
Consider the below code:
def vending_machine(insert_money,item_id) :if item_id == 101 and insert_money :
insert money= True
item_id = 101
print(vending_machine(insert_money.item_id))
Identify the statements to be replaced in Line 1 and Line 2 respectively such that the output is as below:
Disperse the product
Function executed successfully
A:
Line1 : return “Disperse the product”
Line2 : return “Function executed successfully”
B:
Line1 : print (“Disperse the product”)
Line2 : return “Function executed successfully”
C:
Line1 : return “Disperse the product”
Line2 : print (“Function executed successfully”)
D:
Line1 : print (“Disperse the product”)
Line2 : print (“Function executed successfully”)
Answer: B
Explanation:
A. Would just give Disperse the product as the output. As item ID would match.
B. Would first print Disperse the product and then return Function executed successfully
C. Would just give Disperse the product as the output. As item ID would match.
D. Would first print Disperse the product and then print Function executed successfully and the return none to the function call and the same would be printed to the output screen.
Q 20.
John is visiting a zoo. After roaming for sometime, he finds that he has lost his way. John wants to reach the entry/exit gate. He remembers ‘Vans Ice-Cream’, a landmark which he saw while strolling in the zoo. John finds that there are 3 ways to reach Vans Ice-Cream from his current location and from Vans Ice-Cream there are 4 ways to reach the entry/exit gate.
Considering the above scenario, identify the most suitable data structure that can represent all possible ways to reach the entry/exit gate from John’s current location?
A. Graph
B. Tree
C. Stack
D. Queue
Answer: A
Explanation: Graphs are awesome data structures that you use every day through Google Search, Google Maps, GPS, and social media. They are used to represent elements that share connections. The elements in the graph are called Nodes and the connections between them are called Edges. This is a similar instance.
Did you know that Infosys will be conducting their coding contest, i.e HackWithInfy for the 2022 batch?
