Class 10 Computer Science Chapter 5 SEBA : Nested loops in C – Exercise Answers

Class 10 Computer Science Chapter 5 SEBA : Nested loops in C – Exercise Answers

SEBA Class 10 Computer Science Chapter 5 – Nested Loops in C: In this chapter, students learn about nested loops in C programming and how one loop can be placed inside another loop to solve different programming problems. This page provides exercise answers, solutions and explanations for Chapter 5 according to the SEBA Class 10 Computer Science syllabus.

These Class 10 Computer Science Nested Loops in C exercise answers are explained in a simple and student-friendly way to help learners understand the concepts, write C programs correctly and prepare for examinations.

SEBA Class 10 Computer Science Chapter 5: Nested Loops in C : Exercise Questions and Answers :

1. Write C programs to display the following patterns using nested loop construct.

a. Print the following pattern:
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
Answer:
#include<stdio.h>
int main( )
{
            int k, a;
            for (k=1; k<=5; k++)
            {
                        for(a=1; a<=3; a++)
                        printf(“%d”, a);
                        printf(“\n”);
            }
return 0;
}
a. Print the following pattern:
1 2 1
1 2 1
1 2 1
1 2 1
1 2 1
Answer:
#include<stdio.h>
int main( )
{
            int k, a, b;
            for (k=1; k<=5; k++)
            {
            for(a=1; a<=2; a++)
                        printf(“%d”, a);
            for(b=1; b<=1; b++)
                        printf(“%d”, b);
                        printf(“\n”);
            }
return 0;
}
c. Print the following pattern:
4 3 2 1
4 3 2 1
4 3 2 1
4 3 2 1
4 3 2 1
Answer:
#include<stdio.h>
int main( )
{
            int k, a;
            for (k=1; k<=5; k++)
            {
                        for(a=4; a>=1; a- -)
                        printf(“%d”, a);
                        printf(“\n”);
            }
return 0;
}
d. Print the following pattern:
        2
    2  3  4
2  3  4  5  6
Answer:
#include<stdio.h>
int main( )
{
            int k, a, b;
            for (k=1; k<=3; k++)
            {
                        for(a=1; a<=(3-k); a++)
                        printf(” “);
                        for(b=2; b<=(2*k); b++)
                        printf(“%d”, b);
                        printf(“\n”);
            }
return 0;
}
e. Print the following pattern:
        1
    1  2  1
1  2  3  2  1
Answer:
#include<stdio.h>
int main( )
{
            int k, a, b, c;
            for (k=1; k<=3; k++)
            {
                        for(a=1; a<=(3-k); a++)
                                    printf(” “);
                        for(b=1; b<=k; b++)
                                    printf(“%d”, b);
                        for(c= (k-1) ; c>=1 ; c- -)
                                    printf(“%d”, c);
                        printf(“\n”);
            }
return 0;
}
f. Print the following pattern:
               *
         *    *    *
    *   *   *    *   *
*   *   *   *    *   *   *
    *   *   *    *   *
          *   *    *
              *
Answer:
#include<stdio.h>
int main( )
{
int k, a, b, c;
          for(k=1 ; k<=4 ; k++)
          {
             for(a=1 ; a<=(4-k) ; a++)
                printf(” “);
             for(b=1 ; b<=(k*2-1); b++)
                printf(“*“);
              printf(“\n”);
          }
          for(k=1 ; k<=3 ; k++)
          {
            for(a=1 ; a<=k ; a++)
                printf(” “);
            for(b=5 ; b>=(2*k-1); b- -)
                printf(“*“);
             printf(“\n”);
          }
return 0;
}
g. Print the following pattern:
        X                      X
        X  X                  X  X
        X  X  X             X  X  X
        X  X  X  X  X     X  X  X  X
        X  X  X  X  X  X  X  X  X  X  X
 
Answer:
#include<stdio.h>
int main( )
{
     int k, a, b ;
     for(k=1 ; k<=5 ; k++)
     {
          for(a=1 ; a<=k ; a++)
              printf(“x“);
          for(b=1 ; b<=(5-k); b++)
              printf(” “);
          for(a=1 ; a<=k ; a++)
              printf(“x“);
       printf(“\n”);
     }
return 0;
}
2. Modify the solution of question no. 1 to accept the number of lines as the input. The program should make the display pattern accordingly (Hint write separate programs).
Answer:
Please find the modified program below of Q. 1 (a) :-
#include<stdio.h>
int main( )
{
          int k, a, N;
          printf(“Enter the value of N:”);
          scanf(“%d” , &N);
         for (k=1; k<=N; k++)
          {
            for(a=1; a<=3; a++)
               printf(“%d”, a);
             printf(“\n”);
            }
 return 0;
}
4. What is nested loop? Why do we use nested loops in our programs?
Answer:
Whenever we write a loop inside another loop, it is called a nested loop. A nested loop is divided into two parts: outer loop and inner loop.
We use nested loops for the following reasons:-
i) To reduce the length of the program.
ii) To save time and effort.
iii) To make the program easier and simpler than using separate loops.
5. Do we need to use same type of loops as outer and inner loops? Justify your answer with some code segments.
Answer:
No, we don not need to use same type of loops as outer and inner loops.
Some code segments are given below to justify:
#include<stdio.h>
int main( )
{
    int k, i ;
    for (k=1; k<=5; k++)
     {
          i=0 ;
          while(i<=3)
           {
             printf(“I read in class 10 \n”) ;
            i++ ;
            }
        printf(“\n”) ;
      }
return 0 ;
}

On the above code, we used a for loop as the outer loop, and a while loop as the inner loop.
Output of the code:
I read in class 10
I read in class 10
I read in class 10
6. Do we need to use same type of loops as outer and inner loops? Justify your answer with some code segments.
Answer:
Yes, we can put third loop inside the inner loop of a nested loop construct.
A program given below to justify:
#include<stdio.h>
int main( )
{
          int a, b, c ;
          for(a=1 ; a<=3; a++)
          {
              for(b=1 ; b<=3 ; b++)
              {
                for(c=1 ; c<=1 ; c++)
                printf(” X “) ;
              }
              printf(“\n”) ;
          }
return 0 ;
}

Output of the program:
X X X
X X X
X X X

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