Question 1: Write a program to input any number and check whether number is even or odd.
Solution:
Logic --> here we take anyone random number
from user and test whether it is even or odd number using condition. To test
even number, it’s quite easy because even number are those that is divisible by
2 and those not will be even. Test given number by modular division by 2, if
true, it is even number otherwise it is odd.
Program -->
#include<stdio.h>
#include<conio.h>
void min()
{int a;
clrscr();
printf("input any number");
scanf("%d",&a);
if (a%2==0)
printf("Number is even");
else
printf("Number is odd");
getch();
}
Output: Input any number....10
Number is even
Question 2: Write a program to find the roots of the following equation: ax^2+bx+c=0
Solution:
Logic --> Here we take all values of a,b and c
which is required for us however x is not required because of unknown variable.
We can find root by formula as:
x= (-b±√(b^2-4ac))/2a
However,
we take + and - value separately. Root has both negative and positive value.
Step
1: First we find value inside root i.e (b^2-4ac) and save in new location.
Step
2: Then we check whether it is less than zero or not. If true root will have
imaginary value.
Step
3: Otherwise calculate two roots separately and finally display required
result.
Program
-->
#include<stdio.h>
#include<conio.h>
#include<math.h>
void
main()
{int
a,b,c,r1,r2,d;
clrscr();
printf("Input
value of a,b and c");
scanf("%d%d%d",&a,&b,&c);
d=
((b*b)-(4*a*c));
if
(d<0)
printf("roots
are imaginary");
else
{r1=
(-b+sqrt (d))/ (2*a);
r2=
(-b-sqrt (d))/ (2*a);
printf("%d%d",r1,r2);
getch();
}
Output:
Input valur of a,b and c....5 10 3
-0.367,
-1.632
Note:
we use "math.h" header file for defining sqrt function in step 13 and
14.
Question 3: Write a program to find divisions obtained by students in class. The marks obtained by students in 5 different subjects are input through the keyboard. The students gets the division as per following rules ...
1. Percentage above or equal to 80-Distinction
2. Percentage above or equal to 60and less than 80-First
3. Percentage above or equal to 45 and less than 60-Second
4. Percentage above or equal to 35 and less than 45- Third
5. Percentage less than35-Fail
6. Result is fail, if a student fails in any one subject.
Solution:
Logic --> here, we take marks obtained by
student in five different subjects and calculate their percentage and finally
conclude their division. For this, we have to find first total marks obtained
by student in five different subject i.e. total marks=
marks1+marks2+marks3+marks4+marks5. Then we have to find percentage or average i.e.
percentage=total marks/5. Finally we check for division from provided criteria
in above question.
Step 1: Input marks obtained by student in
five different subject i.e. marks1, marks2, marks3, marks4, marks5.
Step 2: Calculate total marks as marks1+marks2+marks3+marks4+marks5.
Step 3: Calculate percentage. (Percentage=total
marks/5)
Step 4: Check first whether student is fail in
anyone subject otherwise he/she got any one division which can be calculated by
condition.
Program -->
#include<stdio.h>
#include<conio.h>
void main()
{int m1,m2,m3,m4,m5,Total,Per;
clrscr();
printf ("input marks obtained by student in
five different subject");
scanf("%d%d%d%d%d",&m1,&m2,&m3,&m4,&m5);
Total=m1+m2+m3+m4+m5;
printf("\n Total marks=%d",Total);
if (m1>35&&m2>35&&m3>35&&m4>35&&m5>35)
{Per=total/5;
if (Per>80)
printf("Distinction");
else if(Per>60&&Per<80)
printf("First Division");
else if(Per>45&&Per<60)
printf("Second Division");
else if(Per>35&&Per<45)
printf("Third
Division");
else
printf("Fail");
printf("\nPercentage=%d",per);
}
else
printf("Fail");
getch();
}
Output:
Input marks obtained by student in five different subject...45 50 60 70 80
Total
marks=305
First
Division
Percentage=61
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