C) 3125

function(5,5) will return 5 x function(5,4)
function(5,4) will return 5 x 5 x function(5,3)
function(5,3) will return 5 x 5 x 5 x function(5,2)
function(5,2) will return 5 x 5 x 5 x 5 function(5,1)
function(5,0) will return 5 x 5 x 5 x 5 x 5 = 3125

D) 0

There are two variables a and b declared. Value initialized for a is 15 and b is 7. Taking mod operation for a by 12(a%12) and the answer is 3 will stored in a. The next mod operation for b is 7 mod (7%4). The answer is 3 will be stored in b. The next line takes the updated value of a and mods it by 1(3%1). Then the answer becomes 0 will be stored in a. The next line takes the updated value of b mod by 1 (3%1) then the answer is 0. Finally adding all the updated values of a and b (0+0 ) and the output of Pseudocode is 0.

C) 0

There are three variables a, b and c declared. Value initialized for a is 2 and b is 2. When we do a bitwise exclusive OR of c i.e (2^2), the answer is 0. Finally, print the value of c.

B) 8

In this question, the value of x is initialized as 0 and y as 1 in the beginning. Later the value 8 is assigned to the variable z and the value of z is assigned to the variable y and the value of y is assigned to the variable x. Finally, the value of x is updated as 8.

A) 64

Explanation:

Here, the left shift operation pushes the values towards the left once; when 1 is left-shifted the value that we will be obtaining will always be 2 to the power something. When one is converted in the beginning and not shifted the value will be 2^0 which is 1. The next iteration will be pushed one place towards the left, therefore the value now will be 2^1 which is 2; this will go on happening until the value stored is greater than 45. The value which is greater than 45 in 2 powers is 64. Now the loop terminates and the last value stored in the variable is 64 and the same will be printed.

C) 15 27 39

c and a is initialized, and d as well; line 5 we are re-initializing c in every iteration of goto; the goto loop terminates only when the value of c is greater than 40 the last value stored in c which breaks the if the condition is 51 and all the numbers 15 27 and 39 will be printed according to the algebraic expression in line number 5.

C) 9 7 5

the first reverse recursion would print 9 and returns to a previous function call, next it prints 7 and returns to the very first function call finally it prints a 5 and completes the execution.

B) 3

All the conditions are true when checked with the condition so the print statement will be executed 3 times.

A) 2

The while loop will only terminate when the value of b becomes zero, this will happen only after the 4th iteration when the last value of the b will be zero and the value of a is 2

D) 64

When the x is reinitialized in line 3 the value stored will be 3. Therefore 8 has to be left-shifted thrice. The shifted value is the 3rd next power of 2 that is 2^6 and the value is 64.

D) 1

AND gate works when both the conditions are true so the first (x and y) condition will be taken as true as both the inputs are true. OR gate works when any of the conditions are true, where it has already got one of its input as true so without checking the other input it will directly assign the value as 1.

D) 7 18 14

The intial values of a=5, b=4, c=3
a = 4 +3 = 7
c = 7 – 4 = 3
c = 3 + 7 = 10
c = 4 + 10 = 14
b = 4 + 14 = 18
Print 7, 18, 14

D) 6 4 16

The loop runs for 5 times; after the 5th iteration the value of a=6; b=4; c=10. So the final answers are 6, 4, 16

D) 6 5 4

Step 1:
It will print i+3, here i value is 3. So i+3 is 6. On the next line, i will be decremented by 1. Then checking the conditions in do-while() i!=0. Here updated i value is 2 (2!=0),so condition is true. The loop continues.
Step 2:
It will print i+3, here updated i value is 2. So i+3 is 5. On the next line i will be decremented by 1. Then checking the conditions in do-while() i!=0. Here updated i value is 1 (1!=0),so condition gets true. The loop continues
Step 3:
It will print i+3, here updated i value is 1. So i+3 is 4. On the next line i will be decremented by 1. Then checking the condition in do while() i!=0. Here updated i value is 0 (0!=0),so condition gets false. Thus the loop gets terminated!

B) 4

There are two variables a and str1. Value initialized for str1 is “goose”. On the next line, we are finding the length of str1 that is 5. Finally, printing the output of a bitwise exclusive OR operator with 1. And the answer is 4

A) 13

There are three variables a, b and c declared. Value initialized for a is 8, b is 51 and c is 2. When we do a bitwise exclusive OR of (8^2), the answer is 10. Again 10 bitwise exclusive OR of a i.e (10 ^ 8) is 2, which will be stored in variable c. Then taking modulo operation for b by 4 (b%4) the answer is 3. Finally adding all the updated values of a,b, and c (8+2+3 ) and the output of Pseudocode is 13.

D) 0

There are two variables a and b declared. Value initialized for a is 15 and b is 7. Taking mod operation for a by 12(a%12) and the answer is 3 will stored in a. The next mod operation for b is 7 mod (7%4). The answer is 3 will be stored in b. The next line takes the updated value of a and mods it by 1(3%1). Then the answer becomes 0 will be stored in a. Next line takes the updated value of b mod by 1 (3%1) then the answer is 0. Finally adding all the updated values of a and b (0+0 ) and the output of Pseudocode is 0.

B) 13

There are three variables a, b and c declared. Value initialized for a is 2, b is 5 and c is 2. Checking the condition using if, b >a and a>c and c>b here if conditions get false. Now else part will be executed. b value will be incremented by 1 and stored in a. Finally adding all the updated values of a, b and c (6+5+2 ) and the output of Pseudocode is 13.

D) 0

There are two variables a and b declared. Value initialized for a is 15 and b is 7. Taking mod operation for a by 12(a%12) and the answer is 3 will stored in a. The next mod operation for b is 7 mod (7%4). The answer is 3 will be stored in b. The next line takes the updated value of a and mods it by 1(3%1). Then the answer becomes 0 will be stored in a. Next line takes the updated value of b mod by 1 (3%1) then the answer is 0. Finally adding all the updated values of a and b (0+0 ) and the output of Pseudocode is 0.

A) 3 5 7 9 5