Accenture Pseudo Code Quiz-1

1. In the first iteration, it takes 7 (rightmost digit), multiplies it by 8^0=1, and adds it to s, making s=7.


2. In the second iteration, it takes 2 (now the rightmost digit), multiplies it by 8^1=8, and adds it to s, making s=23.


3. In the third iteration, it takes 1 (now the rightmost and only digit), multiplies it by 8^2=64, and adds it to s, making s=87.

Loop will end up giving output as 2( if we considered f as an int)  as value of f is initialized to 1 and then following BODMAS rule the next step "s = s +(i / f)" will be followed and i is incremented.

First iteration of while loop: count <= limit. 1<=6. So, we will enter while loop.  count=2, Now we will print n3. So value of n3 is 1. 1 will be printed. n3=n1+n2. So, n3=1,n1=1,n2=1

Second iteration: Count=3, Print n3. So 1 will be printed , n3=2, n1=1, n2=2

Third iteration: Count=4, Print n3. So 2 will be printed , n3=3, n1=2,  n2=3

Fourth iteration: Count=5 , Print 3, n3=5, n1=3, n2=5

Fifth iteration: Count=6, Print 5, n3=8, n1=5, n2=8

Sixth iteration: Count=7, Print 8, n3=13, n1=8, n2=13

Hence output will be 112358

First iteration: Digit: 0, Num_remainder: 263
second iteration: Digit: 3, Num_remainder: 26
Third iteration: Digit: 6, Num_remainder: 2
Fourth iteration: Digit: 2, We will enter  if. Even_counter: 1, Num_remainder: 0

Exit while loop

Even counter=1. So, 1 will be printed

First iteration of while loop:

t=t+a=0+56=56

b=b-1=876-1=875

Similarly we will do further iterations.

At the end 49056 will be printed as output.

Error in line 7.

The code is to sort the given array in ascending order.  But in line 7 , we are checking if second value is greater than first value. If true then we need to swap. But if second  value is greater than first , that means the two numbers are in ascending order, and we need to do anything. The correct code for line 7 will be: If a[i] > a[j] then   

If the above condition holds true than we need to swap the elements.

a=9

b=7

c=2

b = b mod c =7 mod 2=1

a = a mod c= 9 mod 2=1

a+b=1+1=2

So, 2 will be returned by the function

a = 30, b = 60, C = 90

sum=180

((sum EQUALS 180) and (a NOT EQUALS 0) and (b NOT EQUALS 0) and (c NOT EQUALS 0))

So, (true) and (true) and (true) and (true)

So, output will be "success"

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 check
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
check the condition in do while() i!=0. Here updated i value is 0 (0!=0), so the condition gets false.
Thus the loop gets terminated!

the loop will execute for 25 time.

First iteration : b lie between a to z. Then ch=c;

Second iteration : c lie between a to z. Then ch=d;

similarly it will run till ch=z