Kerala Plus One Computer Application Chapter Wise Previous Questions Chapter 7 Control Statements
Question 1.
a) …………… statement takes the program control out of the loop even though the test expression is true. (March – 2015)
b) Consider the following code fragment. How many times will the character ‘ * ’ be printed on the screen?
for (1=0; i< 10; i =i+2);
{
cout<<“*”;
}
Answer:
a) break or goto .
b) Only one time because of semicolon(;) in the end of the for(i=0;i<10;i=i+2);
Question 2.
Which selection statement tests the value of a variable or an expression against a list of integers or character constants? (Say – 2015)
a) For
b) If
c) Switch
d) Conditional expression
Answer:
c) switch
Question 3.
How many times the following loop will execute? (March – 2016)
int m = 2
do
{
cout<<“Welcome”;
m++ ;
} while (m>10);
Answer:
Only one time
Question 4.
……………. search method is an example for ‘divide and conquer method’. (Say – 2016)
Answer:
d) goto
Question 5.
a) Name the type or loop which can be used to ensure that the body of the loop will surely be executed at least once. (March – 2017)
b) Consider the code given below and predict the output;
for (int i=1; i<=9; i=i+2)
{
if (i==5) continue;
cout<<i<<“ “;
}
Answer:
a) do-while loop(Exit controlled loop)
b) 1 3 7 9. It bypasses one iteration of the loop when i = 5.
Question 6.
a) ……………… is an entry control loop.
b) Explain the memory allocation for the following declaration statement.
Answer:
a) while or for loop
b) To store On integer 4 bytes is used in Geany Editor
int A[10] [10]; → It needs 10*10*4 = 400 bytes7
Question 7.
Differentiate between break and continue statements in C++. (Say – 2016)
Answer:
break statement : It is used to skip over a part of the code i.e. we can premature exit from a loop such as while, do-while, for or switch.
Syntax:
while (expression)
{
if (condition)
break;
}
continue statement : It bypasses one iteration of the loop. That is it skips one iteration and continue the loop with next iteration value.
Syntax:
while (expression)
{
if (condition)
continue;
}Question 8.
Write C++ program forgetting the following output. (Say – 2016)
1
1 2
1 2 3
1 2 3 4
OR
Consider the following C++ program and answer the following questions.
#include<iostreaiti.h>
int main( )
{
int a, p=1;
for(a=1;a<=5;a+=2)
P = P*a;
cout<<p;
}a) Predict the output of the above code.
b) Rewrite the above program using while loop.
Answer:
#include<iostream>
using namespace std;
int main( )
{
int i,j;
for(i=1;i<=4;i++)
{
for(i=1;j<=i;j++)
cout<<j<<“\t”;
cout<<“\n”;
}
}OR
a) The output is 15.
b)
#include<iostream>
using namespace std;
int main( )
{
int a=1, p=1;
while(a<=5)
{
p=p*a;
a+=2;
}
cout<<p;
}Question 9.
Write a program to do the following: (March – 2015)
a) Inputs the values for variables ‘n’ and ‘m’.
b) Prints the numbers between ‘1 ’ and ‘n’ which are exactly divisible by ‘m’.
c) Checks whether the numbers divisible by ‘m’ are odd or even.
OR
Write a program using nested loop that inputs a number ‘n’ which generates an output as follows. Hint: if the value of ‘n’ is 5, the output will be as
25
25 16
25 16 9
25 16 9 4
25 16 9 4 1
Answer:
b)
#include<iostream.h>
#include<conio.h>
void main( )
{
clrscr( );
int i,n,m;
cout<<“Enter values for n and m”;
cin>>n>>m;
for(i=1;i<=n;i++)
if(i%m==0)
cout<<i<<“,”;
getch( );
}c)
#include<iostneam.h>
#include<conio.h>
voicj main( )
{
clrscr( );
int i,n,m;
cout<<“Enter values for n and m”;
cin>>n>>m;
for(i=1;i<=n;i++)
if(i%m==0)
{
cout<<i<<“\t";
if(i%2==0)
{
cout<<“even”<<endl;
else
cout<<“odd”<<endl;
}
getch( );
}OR
#include<iostream.h>
#include<conio.h>
#include<string.h>//forstrlen( )
main( )
{
clrscr( );
int n,i,j;
cout<<“enter a value for n:“;
cin>>n;
for(i=n;i>0;i--)
{
for(j=n;j>=i;j--)
cout<<j*j<<“\t”;
cout<<endl;
}
getch( );
}Question 10.
Write a C++ program to display the Fibonacci series.
Answer:
#include<iostream>
using namespace std;
int main( )
{
int n,fib1=0,Fib2=1,fib3;
cout<<"Enter the limit";
cin>>n;
cout<<"The fibonacci series is";
if(n==1)
cout<<fib1<<",";
else if(n==2)
cout<<fib1<<","<<fib2<<",";
else if (n>2)
{
cout<<fib1<<","<<fib2<<",";
fib3=fib1 +fib2;
while(fib3<=n)
{
cout<<fib3<<",";
fib1=fib2;
fib2=fib3;
fib3=fib1+fib2;
}
}
else
cout<<"lnvalid";
}Question 11.
Write a C++ program to accept an integer number and check whether it is an Armstrong number or not. (March – 2016)
(Hint: Sum of the cubes of the digits of an Armstrong number is equal to that number itself)
OR
Write a C++ program to accept an integer number and print its reverse
(Hint: If 234 is given, the output must be 432).
Answer:
a)
#include<iostream>
using namespace std;
int main( )
{
int n,m,rem,cube=0;
cout<<"Enter a number";
cin>>n;
m=n;
while(n)
{
rem=n%10;
cube=cube+rem*rem*rem;
n=n/10;
}
if(cube==m)
cout<<"The number "<<m<<" is Armstrong";
else
cout<<"The number "<<m<<" is not Armstrong";
}OR
(b)
# include<iostream>
void main ( )
{
int n, rem, rev = 0;
cout<<“Enter a number:”;
cin>>n;
while (n)
{
rem = n%10;
rev = rev*10+rem;
n = n/10;
}
cout<< “The reverse is” << rev;
}Question 12.
Write a program to check whether the given number is palindrome or not. (March – 2017)
OR
Write a program to print the leap years between 2000 and 3000.
(A century year is leap year only if it is divided by 400 and a noncentury year is leap year only if it is divided by 4).
#include<iostream>
using namespace std;
int main( )
{
int n,m,rem,rev=0;
cout<<“Enter a number”;
cin>>n;
m=n;
while(n)
{
rem=n%10;
rev? rev*10+rem; ;
n=n/10;
}
if(rev==m)
cout<<“The number “<<m<<“ is palindrome”;
else
cout<<“The number “<<m<<“ is, not palindrome”;
}OR
#include<iostream>
using namespace std;
int main( )
{
int year;
for(year=2000;year<=3000;year++)
{
if(year%100==0)
{
if(year%400==0)
cout<<year<<endl;
}
else
{
if(year%4==0)
cout<<year<<endl;
}
}
}