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Quiz – 1

Question 1

Time: 00:00:00

What will be the output of the following pseudocode for i=140?

integer fun(int i)

if((i%2)!=0)

return i;

else

return fun(fun(i=1));

End function fun()

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Explanation:
In the first step i=140 is passed through the function fun(), and if the condition becomes false when it is checked so else will get executed.

When else gets executed, function fun() is called again and i=1 is passed and now condition in IF gets true and 'i' whose value is 1 gets returned.

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Question 2

Time: 00:00:00

#include<stdio.h>

int main()

{

int x=2,y=0,z=3;

x>y ?( printf("%d", z)):( return z);

}

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Explanation:
The operator ?: works as follows expr1 is evaluated first if it is true expr2 is evaluated otherwise expr3 is evaluated. hence in expressions the return statement can not be used in C-language. The return statement is used for returning from a function , you can't use inside ternary operator

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Question 3

Time: 00:00:00

What will be the output of below code?

Integer a, b, c, d, e

Set a=50 , b=3, c=3 e=0

while(c>0)

d=a mod b

e= e + d + a

c= c - 1

End while

Print e

165

160

170

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#include<iostream>



using namespace std;



int main()



{

    int a, b, c, d, e;

    a=50 , b=3, c=3, e=0;

    while(c>0)

    {

        d=a % b;

        e= e + d + a;

        c= c - 1;

    }

    cout<<e;

    return 0;

}

While loop will run 3 times and every time d will become 2, e will update 52 in first step, 108 in second step and 160 in final step which will be the answer

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Question 4

Time: 00:00:00

for (i = 1; i <= 6; i++)

for (j = i; j < 6; j++)

Print blank space

for (k = 1; k < (i * 2); k++)

Print *

End for

Line break

End for

* *** ***** ****** *******

* ** *** ***** ******* *********

* *** ***** ******* ********* ***********

None

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Refer this code for the given pseudo code for better understanding, this pseudo code will print a pyramid of * with six rows each row having an odd number of * from 1 to 11 i.e the first row will have 1st star and the 6th row will have 11 starts

#include<stdio.h>

int main ()

{

int i, j, k;

for (i = 1; i <= 6; i++)

{

for (j = i; j < 6; j++)

{

printf (" ");

}

for (k = 1; k < (i * 2); k++)

{

printf ("*");

}

printf ("\n");

}}

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Question 5

Time: 00:00:00

What will be the output of the following algorithm?

Start

Declare a=0, I and b

for I =0 to 4

Increment a by 1

if I = 3 then

print hello

get out of the loop

End if

End for

print a

Hello4

Hello

Hello3

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Explanation:
Inside the loop with a value of I from 0 to 4, it will increment the value of 4 times that will be equal to 1, 2, 3, and 4.
but first inside the for loop hello will be printed when the value of I = 3
and at last, a will be printed i.e 4so the output will be hello4

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Question 6

Time: 00:00:00

What will be the output of below code?

#include<stdio.h>

int

f (int n)

{

  static int a = 0;

  if (n <= 0)

    {

      return 1;

    }

  if (n > 3)

    {

      a = n;

      return f (n - 2) + 2;

    }

  return f (n - 1) + a;

}



int main ()

{

  printf ("Result: %d", f (5));

  return 0;

}

Error

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Explanation:
There are no errors in the given code, the output produced by the code is 18

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Question 7

Time: 00:00:00

What will be the output of below code?

integer a

if((a mod 10) IS EQUAL TO 0)

a=a*2

else if((a mod 5 ) IS EQUAL TO 0)

a=a/5

else

a=a-1

end if

If we pass input as : a=25, a=16

5, 15

15, 5

25, 5

Compiler Error

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#include<iostream>

using namespace std;

int main()

{

int a =16;

if((a % 10)== 0)

a=a*2;

else if((a % 5 )== 0)

a=a/5;

else

a=a-1;

cout<<a;

return 0;

}

The above code checks if the value of a is completely divisible by 10 or 5 or not if the value is completely divisible by 10 then the value of a is changed to a * 2 and same as given in the pseudocode. The above code is the c++ representation of the pseudocode

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Question 8

Time: 00:00:00

integer a,b;

Set a=2; b=50;

while(b>0)

a = b%2 + a;

if( a MOD 2 Is Equal To 0)

Printf a

else

Print b-1

b = b/5

a=a+1

End while

2 9 4

4 9 2

8 18 4

Error

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Explanation:
Step 1:
Condition true for the while loop
a = 2
Condition true for if
Value of an i.e 2 will get printed
Now b=10
and a=3

Step 2:
Condition true for the while loop
a = 3
Condition false for if, else will get printed
i.e. 9 will get printed
Now b=2
and a=4

Step 3:
Condition true for the while loop
a = 4
Condition true for if
Value of an i.e 4 will get printed
Now b=0
and a=5

Step 4:
Condition false for while. Exit

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Question 9

Time: 00:00:00

Integer array1[10] = {2, 3, 56, 34}

Integer k, a, j, n

Set a = 3, n = 4

for(each k from 0 to n-1)

Set array1[n] = array1[0]

for(each j from 0 to n-1)

Set array1[j] = array1[j+1]

End for

End for

for(each k from 0 to n-1)

Print array1[k]

End for

2 3 56 34

12 13 56 34

3 4 12 13

None of the mentioned

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#include<iostream>

using namespace std;

int main()

{

int array1[10] = {2, 3, 56, 34};

int k, a, j, n;

a = 3, n = 4;

for(k=0 ;k<=n-1;k++)

{

array1[n] = array1[0];

for(j=0;j<=n-1;j++)

array1[j] = array1[j+1];

}

for(k=0;k<=n-1;k++)

cout<<array1[k]<<" ";

return 0;

}

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Question 10

Time: 00:00:00

What will be the output of the following pseudocode?

Integer x,y,z

Set x=3

Set y=90

while(y is greater than 0);

y=y/3

x=x+6

c=x+y

while(c is greater than 30):

if(c mod 3 is equals to 0):

Write x

else:

Write y

c=c/5

Write c

9 18 15

6 33 9

9 33 6

None of the mentioned

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This pseudocode performs some simple mathematical operations until the value of y is greater than 0 and c is greater than 30 and prints the value of x, y, and z.
below is the Python code for this pseudo code:
x=3

x=3

y=90

while(y>0):

y=y//3

x=x+6

c=x+y

while(c>30):

if(c % 3 == 0):

print(x)

else:

print (y)

c=c//5

print(c)

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["0","40","60","80","100"]

["Need more practice!","Keep trying!","Not bad!","Good work!","Perfect!"]

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