Solution of C Programming by E Balaguruswamy

CHAPTER-1

Problem exercise no 1.1&1.2:

Coding of the programme:

#include<stdio.h> #include<conio.h> void main() { clrscr(); printf("---\n");

printf("I First line :A.Z.M.Shakilur Rahman I\nI Second line :12/a ,Ali sonar lane I\nI Third line:Bogra,5800 I\n");

printf("---"); getch();

}

Output:

Problem exercise no. 1.3:

Coding of the programme:

#include<stdio.h> #include<conio.h> void main() {clrscr(); printf("*\n* *\n* * *\n* * * * "); getch(); } Output: * * * * * * * * * *

Problem exercise no :1.4

Coding of the problem:

#include<stdio.h> #include<conio.h> void main() {clrscr(); printf("a>>--->b"); getch(); }

Output

: a>>--->b

Problem exercise no:1.5

Coding of the problem:

#include<conio.h> #define pi 3.14159 void main() { float r,A; clrscr();

printf("\n\tENTER THE RADIUS OF A CIRCLE="); scanf("%f",&r); A=pi*r*r; printf("\n\n\tArea=%f sqr unit",A); getch(); }

Output:

ENTER THE RADIUS OF A CIRCLE=2 Area=12.566360 sqr unit

Problem exercise no:1.6

CODING:

#include<stdio.h> #include<conio.h> void main() { int b,c; clrscr(); for(b=1;b<=10;b++) { c=5*b; printf("\n\t%d*%d=%d\n",5,b,c); getch(); } } Output :

Problem exercise no:1.7

Coding of the programme:

#include<stdio.h> #include<conio.h> void add(); void sub(); void main() { clrscr(); add(); sub(); getch(); } void add() { printf("\n\t%d+%d=%d",20,10,30); }

void sub() { printf("\n\t%d-%d=%d",20,10,10); }

Output :

20+10=30 20-10=10

Problem exercise no:1.8

Coding:

#include<stdio.h> #include<conio.h> void main() { int a,b,c,x; clrscr();

printf("Enter values of a,b&c\n"); scanf("%d%d%d",&a,&b,&c); x=a/(b-c); printf("result=%d",x); getch(); }

Output:

a)

Enter values of a,b&c 250

85 25 result=4

b)NO OUTPUT

Problem exercise no:1.9 (b)

Coding :

#include<stdio.h> #include<conio.h> void main() { float a,F,C; clrscr();

printf("ENTER TEMPERATURE IN FARENHITE\n"); scanf("%f",&F);

a=5*(F-32); C=a/9;

printf("\nIn celsius scale=%f",C); getch();

}

Output :

ENTER TEMPERATURE IN FARENHITE 10

In Celsius scale=-12.222222

Problem exercise no:1.9 (a)

Coding :

#include<stdio.h> #include<conio.h> void main() { float a,F,C; clrscr();

printf("ENTER TEMPERATURE IN CELSIUS\n"); scanf("%f",&C);

a=(9*C)/5; F=a+32;

printf("\nIn farenhite scale=%f",F); getch();

}

Output:

ENTER TEMPERATURE IN CELSIUS 10

In frenhite scale=50.00000

Problem exercise no: 1.10

Coding of the problem:

#include<stdio.h> #include<conio.h> #include<math.h> void main() { clrscr(); float a,b,c,S,A;

printf("\n\tENTER THE THREE SIDES OF A TRIANGLE="); scanf("%f%f%f",&a,&b,&c);

S=(a+b+c)/2;

A=sqrt(S*(S-a)*(S-b)*(S-c));

printf("\n\tArea of the triangle=%f",A); getch();

}

Sample output:

ENTER THE THREE SIDES OF A TRIANGLE=10 12

14

Area of the triangle=58.787754

Problem exercise no:1.11

Coding:

#include<stdio.h> #include<conio.h> #include<math.h> void main() { float D,x1,x2,y1,y2; clrscr();

printf("ENTER CO-ORDINATES x1,x2,y1,y2=\n"); scanf("%f%f%f%f",&x1,&x2,&y1,&y2); D=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); printf("Result=%f",D); getch(); }

Output :

ENTER CO-ORDINATES x1,x2,y1,y2= 2

4 8 5

Result=3.605551

Problem exercise no:1.12

Coding:

#include<stdio.h> #include<conio.h> #include<math.h> #define pi 3.14159 void main() { float r,x1,x2,y1,y2,A; clrscr(); x1=0; x2=0; y1=4; y2=5; r=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); A=pi*r*r; printf("Result=%f",A); getch(); }

Output :

Result=3.14159

Problem exercise no:1.13

Coding:

#include<stdio.h> #include<conio.h> #include<math.h> #define pi 3.14159 void main() { float D,r,x1,x2,y1,y2,A; clrscr(); x1=2; x2=2; y1=5; y2=6; D=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); r=D/2;

A=pi*r*r; printf("Result=%f",A); getch(); }

Output :

Result=0.785398

Problem exercise no:1.14

Coding:

#include<stdio.h> #include<conio.h> void main() { int a,b,c; clrscr(); a=5; b=8; c=18; printf("%dx+%dy=%d",a,b,c); getch(); }

Output :

5x+8y=18

Problem exercise no:1.15

Coding:

#include<stdio.h> #include<conio.h> void main() { float x,y,sum,difference,product,division; clrscr();

printf("ENTER TWO NUMBERS=\n"); scanf("%f%f",&x,&y); sum=x+y; difference=x-y; product=x*y; division=x/y; printf("\n\tSum=%f\tDifference=%f\n\n\tProduct=%f\tDivision= %f",sum,difference,product,division); getch(); }

Output :

ENTER TWO NUMBERS= 10

5

Sum=15.000000 Difference=5.000000 Product=50.000000 Division=2.000000

solution of C programming by

E.Balagurusamy

CHAPTER-2

CHAPTER

2

Constants,

Variables,

and Data Types

REVIEW

QUESTIONS

STATE WHETHER THE FOLLOWING STATEMENTS

ARE TRUE OR FALSE:

(a) Any valid printable ANSII character can be used in an identifier. ( False ) (b) All variables must be given a type when they are declared. ( True )

(c) Declarations can appear anywhere in a program. ( False )

(d) ANSI C treats the variable name and Name to be same. ( False ) (e) The underscore can be used anywhere in an identifier. ( True ) (f) The keyword void is a data type in C. ( True )

(g) Floating point data constants, by default, denote float type values. ( False ) (h) Like variables, constants have a type. ( True )

(i) Character constants are coded using double quotes. (False )

(j) Initialization is the process of assigning a value to a variable at the time of declaration. ( true )

(k) All static variables are automatically initialized to zero. ( True )

Fill in the blanks with appropriate words:

(a)The keyword ………..can be used to create a data type identifier. Answer: int

(b) ……… is the largest value that an unsigned short int type variable can store. Answer: 255

(c) A global variable is also known as ……….variable. Answer: external

(d) A variable can be made constant by declaring it with the qualifier ………. At the time of initialization.

Answer: constant

Question: What are trigraph characters? How are they

useful?

Answer:

Trigraph characters is one kinds of character which consists of three characters ( Two question marks and followed by another ).

Some keyboard does not support some characters. But we can use them by trigraph characters. If a keyboard does not support square brackets, we can still use them in a program using the trigraph ??( and ??) .

Question: Describe the four basic data types. How

could we extend the range of values they represent?

Answer:

The basic four data types are: (1) Char

(2) Int (3) Float (4) Void

We could extend the range of integer by using long beforeinteger.

We can extend the range of float by using double. To extend the precision further we may use long double.

Question: What is an unsigned integer constant? What

is the significant of declaring a constant unsigned?

Answer:

The integer constant which does not take any + or – sign before it is called an unsigned integer constant.

We can take the value double using an unsigned integer constant. For example, a signed integer constant have a value between -32768 to +32767, but an unsigned integer constant takes the value between 0 to 65535.

Question: Describe the characteristics and purpose of

escape sequence characters.

Answer:

C supports some special back slash character constants, that are used in output function. This characters are known as escape sequence characters. For example, the symbol “\n” stands for new line character.

Characteristics :

(1) They acts as a single character.

(2) Each escape sequence character consists of two characters. (3) The first character must be a back slash .

Purpose:

(1) In a program we used it for new line. (2)In a program we used it for horizontal tab.

Question: What is a variable and what is meant by

“value” of a variable?

Answer:

A variable is a data name that may used to store a data value. Like constants that remains unchanged during the execution a program.

The meant of value it is a variable name and it can take any value like character, int, float and double.

Question: How do variables and symbolic names

differ?

Answer:

A variable may be used to store data value. A variable may take different values at different times during execution of a program. Variables has need to declare at the beginning of the body but after the main.

Symbolic names is a unique constants. This constants may appear in a number of place in the program. Symbolic names has need to define at the beginning of a program.

Question: State the difference between the declaration

of a variable and the definition of a symbolic name?

Answer:

Variables has need to declare at the beginning of the body but after the main. The syntax for declaring a variable is as follow:

Data-type v1,v2,……..vn;

v1, v2,……..vn are the names of variables. For example, valid declarations are int count;

int number,total; float ratio;

Symbolic names has need to define at the beginning of a program. A symbolic name constants is defined as follows:

#define symbolic-name value of constant

Valid example of constant definations are:

#define STRENGTH 100

#define PASS MARK 5

Question: What is initialization? Why it is important?

Answer:

The process of giving initial values to variables is called initialization. Some examples are

int final_value =100;

char yes =’x’; double balance =75.84;

C permits the initialization of more then one variable using multiple assignment operators.For example the statements

p=q=s=0; x=y=z=MAX;

are valid.

Question: What are the qualifiers that an int can have

at a time?

Answer:

A signed int can take a value between -32768 to 32767 and an unsigned int can take a value 0 to 65535.

Question: Describe the purpose of the qualifiers

constant and volatile

.

Answer:

We may like the value of certain variables to remain constant during the excution of a program. We can achieve this by declaring the variable with the qualifier constant at the time of initialization. Example:

const int class_size=40;

ANSI standard defines another qualifier volatile that could be used to tell explicitly the complier that a variables value may be changed at any time by some external sources ( from outside the program). For example:

volatile int date;

Question: When dealing with very small or very large

numbers, what steps would you like you take to

improve the accurancy of the calculation?

When we are dealing with a very short number we can improve the accurancy of calculation by using a keyword short before the keyword. Example short int.

When we are dealing with a very large number we can improve the accurancy of calculation by using a keyword long before the keyword. Example long int.

Question: Which of the following are invalid constants

and why?

0.0001

Answer: (valid) 5x1.5

Answer: (Invalid)

Reason: Exponent must be an integer. 99999

Answer: Valid

Reason: Long integer.

+100 Answer: ( valid) 75.45E-2 Answer: ( Valid ) -45.6 Answer: ( Valid ) “15.75” Answer: ( Invalid )

Reason: “” sign is not permitted.

-1.79e+4

Answer: (valid) 0.00001234 Answer: ( Valid )

Question: Which of the following are invalid variable

and why?

Minimum Answer: ( valid ) First.name Answer: ( Invalid )

Reason:. Sign is not permitted. N1+n2

Answer: ( Invalid )

Reason: + sign is not permitted. &name

Answer: ( Invalid )

Reason: & is not permitted. Doubles

Answer: ( Valid )

Reason: Keyword may be a part of variable name. 3rd_{_row}

Answer: ( Invalid )

Reason: First character must be a letter or underscore. n$

Answer: ( Invalid )

Reason: Dollar sign is not permitted. Row1 ( Valid )

Float

Answer: ( Invalid )

Reason: float is a keyword. Sum Total

Answer: ( Invalid )

Row Total

Answer: ( Invalid )

Reason: White space is not permitted. Column total

Answer: ( Invalid )

Reason: White space is not permitted.

Question: Find errors, if any, in the following

declaration statements.

{ Intx; float letter,DIGIT; double=p,q; exponent alpha,beta; m,n.z:INTEGER short char c; long int m;count; long float temp; getch();

}

Error1: intx should have a type.

Error2: Line number 6, expression syntax.

Error3: Line number 7, Declaration should be properly. Error4:Line number 9, Unreachable code.

Problem no 2.1: Write a program to determine and

print the sum of the following harmonic series for a

given value of n:

1+1/2+1/3+………+1/n

Solve:

#include<stdio.h> #include<conio.h> Void main() { int n; float I, sum, t; clrscr(); printf(“1+1/2+1/3+………+1/n\n”); printf(“Enter the value of n\n”); scanf(“%d”,&n); sum=0; for(i=1;i<=n;i++) { t=1/i; sum=sum+t; } printf(“%f”,sum); getch(); }

Output:

1+1/2+1/3+………….+1/n

Enter the value of n 4

2.083333

The value of n should be given interactively through

the terminal.

Problem no 2.2: Write a program to read the price of an item in decimal form ( like 15.95 ) and print the output in paisa ( like 1595 paisa) .

Solve:

#include<stdio.h> #include<conio.h> void main() { int b; float a; a=15.95; clrscr(); b=100*a; printf("%d",b); getch(); }

Output:

1595

Problem no 2.3: Write a program that’s prints the even

numbers from 1 to 100.

#include<stdio.h> #include<conio.h> void main() { int i; for(i=1;i<=100;i++) { if(i%2==0) printf(" %d",i); } getch();

}

Output

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 9 8 100

Problem no 2.4: Write a program that request two

float type numbers from the users and then divides the

first number by the second and display the result along

with the numbers.

Solve:

#include<stdio.h> #include<conio.h> void main()

{

float number1, number2, number3; clrscr();

printf("Enter the value of number1 and number2\n"); scanf("%f %f",&number1,&number2);

number3=number1/number2;

printf("%f/%f=%f",number1,number2,number3); getch();

Output:

Enter the value of number1 and number2 15.5

6.6

15.5/6,6=2.348484

Problem no 2.5: The price of one kg of rice is Rs. 16.75

and one kg of sugar is Rs. 15. Write a program to get

these values from the user and display the prices as

follows:

Item Price

Rice Rs 16.75

Sugar Rs 15.00

Solve:

#include<stdio.h> #include<conio.h> void main () { float Rice,Sugar; Rice=16.75; Sugar=15.00; clrscr(); printf("***LIST OF ITEMS***\n"); printf("Item \tPrice\n"); printf("Rice\tRs%.2f\n",Rice); printf("Sugar\tRs%.2f\n",Sugar); getch(); }

Output:

***LIST OF ITEMS*** Item Price Rice Rs16.75

Question: Identify syntax errors in the following

program. After correcting, what output would you

expect when you execute it.

#include<stdio.h> #include<conio.h> #define PI 3.14159 void main()

{ int R,C; float perimeter; float area; C=PI; R=5; perimeter=2.0*C*R; Area = C*R*R; printf("%f", "%d",&perimeter,&area) }

Errors:

Cpp 10: Undefined symbol Area.

Cpp 12:statement missing,in function{} Cpp12: compound statement missing. Solve: #include<stdio.h> #include<conio.h> #define PI 3.14159 void main() { int R,C; float perimeter,Area; C=PI; R=5; perimeter=2.0*C*R; Area = C*R*R; printf("%f %f",perimeter,Area); getch();

}

CHAPTER-3

EXERCISE NO.3.1: Given the values of the variables X,Y and Z write a program to rotate their values such that X has the value of Y,Y has the value of Z and Z has the value of X. SOLUTION: #include<stdio.h> #include<conio.h> void main() { int x,y,z,temp; clrscr();

printf("Enter the value of x,y,z\n"); scanf("%d %d %d",&x,&y,&z); temp=x; x=y; y=z; z=temp; printf("%d %d %d",x,y,z); getch(); }

EXERCISE NO.3.2: write a program that reads a floating-point number and then displays right-most digit of the integral part of the number.

SOLUTION: #include<stdio.h> #include<conio.h> void main()

{ int a,e; float p; clrscr();

printf("Enter the value of p\n"); scanf("%f",&p); a=int(p); printf("%d\n",a); e=a%10; if(a>10) printf("%d\n",e); getch(); }

EXERCISE NO.3.3. Modify the above program to display to right-most digits of the integral part of the number.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int a,e; clrscr();

printf("Enter the value of a\n"); scanf("%d",&a);

e=a%100;

if(a>100)

getch(); }

3.4: Write a program that will obtain the length and width of a rectangle from the user and compute its area and perimeter.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int length,width; clrscr(); float area,perimeter;

printf("Enter the value of length,width\n"); scanf("%d %d",&length,&width); area=(length*width); perimeter=2*(length+width); printf("%f %f",area,perimeter); getch(); }

EXERCISE NO.3.5: Given an integer number, write a program that displays the number as follows:

First line: all digits

Second line: all except first digit Third line: all except first two digits

…………

Last line: The last digit

For example the number 5678 will be displayed as: 5 6 7 8 6 7 8 8 SOLUTION: #include<stdio.h> #include<conio.h> void main() { int a,b,c,e,x; float p; clrscr();

printf("Enter the value of p\n"); scanf("%f",&p); a=int(p); printf("%d\n",a); e=a%10000; b=e%1000; c=b%100; x=c%10; if(a>10000) printf("%d\n%d\n%d\n%d\n",a,e,b,c,x); else if(a>1000) printf("%d\n%d\n%d\n",a,b,c,x); else if(a>100) printf("%d\n%d\n",a,c,x); else if(a>10)

printf("%d\n%d\n",a,x);

getch(); }

EXERCISE NO.3.6: The straight-line method of computing the yearly depreciation of the value of an item is given by

Depreciation=

Write a program to determine the salvage value of an item when the purchase price , years of service, and the annual depreciation are given.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int years; float s, d,p; clrscr();

printf("Enter the value of years,d,p\n"); scanf("%d %f %f",&years,&d,&p); s=p-(years*d);

printf("%f",s); getch(); }

EXERCISE NO.3.7: Write the program that will read a real number from the keyboard and print the following output in one line:

Smallest integer The given Largest integer not less then number not greater than the number the number SOLUTION: #include<stdio.h> #include<conio.h> void main() { float m; int n,p; clrscr();

printf("Enter the value of m\n"); scanf("%f",&m); n=(m/1)+1; p=m; printf("%d %f %d",n,m,p); getch(); }

EXERCISE NO.3.8: The total distance travelled by a vehicle in t seconds is given by Distance= ut+(at2_)/2

Where u is the initial velocity( meter per second),a is the acceleration (meter per second2_).

Write a program to evaluate the distance travelled at intrevales of time, give the value of u and a. the program should provide the flexibility to the user to select his own time intervals and repeat the calculation for different value of u and a.

SOLUTION: #include<conio.h> void main()

{

int a,u,t; float distance; clrscr();

printf("Enter the value of a,u,t\n"); scanf("%d %d %d",&a,&u,&t); distance=u*t+(a*t*t)/2; printf("%f",distance); getch();

}

EXERCISE NO.3.9: In inventory management ,the Economic Order Quantity for a single item is given by

EOQ=sqrt { ( 2*demand rate*setup rate ) / ( holding cost per item per unit time ) }

And the Time Between Orders

TBO =sqrt { ( 2* setup cost ) / (demand rate * holding cost per item per unit time ) }

SOLUTION 1: #include<stdio.h> #include<conio.h> #include<math.h> void main() { float EOQ,d,s,h,x; Clrscr();

printf("Enter the value of d,s,h\n"); scanf("%f %f %f",&d,&s,&h); x=(2*d*s)/h;

EOQ=sqrt(x); printf("%f",EOQ); getch(); } SOLUTION 2: #include<stdio.h> #include<conio.h> #include<math.h> void main() { float x,s,d,h,TOB; clrscr();

printf("Enter the value of s,d,h\n"); scanf("%f%f%f",&s,&d,&h); x=(2*s)/(d*h); TOB=sqrt(x); printf("%f",TOB); getch(); }

EXERCISE NO.3.10: For a certain electrical circuit with an inductance L and resistance R,the damped natural frequency is given by

Frequency =sqrt { (1/L*C ) - ( R*R/4*C*C ) }

It is desired to study the variation of this frequency with C(capacitance).Write a program to calculate the frequency for different values of C starting from 0.01 to 0.1 in steps of 0.01.

#include<stdio.h> #include<conio.h> #include<math.h> void main() { float L,R,C,x,a,b,F; clrscr();

printf("Enter the value of L,R,C\n"); scanf("%f %f %f",&L,&R,&C); a={ (1/L*C) – (R*R/4*C*C) }; F=sqrt(a); Printf(“%f”,F); getch(); }

EXERCISE NO.3.11: Write program to read a four digit integer and print the sum of its digit. Hints: Use / and % operators.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int num,a,b,c,d,x,y,result; clrscr(); printf("Enter a number"); scanf("%d",&num); a=num%10; x=num/10;

b=x%10; y=x/10; c=y%10; d=y/10; result=a+b+c+d; printf("%d",result); getch(); }

EXERCISE NO. 3.12: Write a program to print the size of various data types in C.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int m; clrscr(); m=sizeof(10); printf("Size=%d",m); getch(); }

EXERCISE NO.3.13: Given three values, write a program to read three values from keyboard and print out the largest of them without using if statement.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int x,y,z,a,b;

printf("Enter the value of x,y,z\n"); scanf("%d%d%d",&x,&y,&z); printf("largest\n"); a=(x>y)?x:y b=(a>z)?a:z printf("%d",b); }

EXERCISE NO.3.14: Write a program to read two integer values m and n and to decide and print whether m is multiple of n.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int m,n; printf("Enter m & n,m>=n:"); scanf("%d %d",&m,&n); if(m%n==0) printf("m is a multiple of n");

else

printf("m is not a multiple of n"); getch();

}

EXERCISE N0.3.15: Write a program to read three values using scanf statement and print the following results:

(a)Sum of the values

(b) Average of the three values (c) Largest of the three

(d) Smallest of the three.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int a,b,c,x,y; float sum, average; clrscr();

printf("Enter the value of a,b,c\n"); scanf("%d%d%d",&a,&b,&c); sum=(a+b+c); printf("sum=%f\n",sum); { average=sum/3; printf("average=%f\n",average);

} { printf("Largest\n"); x=(a>b)?a:b; y=(x>c)?x:c; printf("%d\n",y); } { printf("Smallest\n"); x=(a<b)?a:b; y=(x<c)?x:c; printf("%d\n",y); } getch(); }

EXERCISE NO.3.16: The cost of one type of mobile service is Rs.250 plus Rs.1.25 for each call made over and above 100 calls. Write a program to read customer codes and calls made and print the bill for each customer.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int code,call;

float bill; clrscr();

printf("Enter customer code and number of calls made:"); scanf("%d %d",&code,&call);

bill=250+(call*1.25); printf("Bill=%f",bill); getch();

}

EXERCISE NO.3.17: Write a program to print a table of sin and cos functions for the interval 0 180 degrees in increments of 15 as shown below.

---x(degees) sin(x) cos(x) O ……. ……. 15 .…… ……. ….. ……. ……. SOLUTION: #include<stdio.h> #include<conio.h> #include<math.h> #define p1 3.1416 #define MAX 180 void main() { int i;

float x,y,z; clrscr(); i=0;

printf("x(degree) sin(x) cos(x)\n"); while(i<=MAX) { x=(p1/MAX)*i; y=sin(x); z=cos(x); printf("%d\n %f\n %f\n",i,y,z); i=i+15; } getch(); }

EXERCISE NO.3.18: Write a program to compute the values of square-roots and squares of the number 0 to 100 in steps 10 print the output in a tabular form as shown below.

---number Square-root square 0 0 0 100 10 10000 SOLUTION: #include<stdio.h> #include<conio.h> #include<math.h> void main()

{

/*...square root and square of numbers 0 to 100...*/ int i,y; float x; clrscr(); printf("Number\tSquare root\tSquare\n\n"); for(i=0;i<=100;i++) { x=sqrt(i); y=i*i; printf("%d\t%f\t%d\n",i,x,y); } getch(); }

EXERCISE NO.3.19: Write a program that determines whether a given integer is odd or even and displays the number and description on the same line.

SOLUTION: #include<stdio.h> #include<conio.h> void main() { int x; clrscr();

printf("Enter the integer number:"); scanf("%d",&x);

if(x%2==0)

else

printf("The number %d is odd",x); getch();

}

EXERCISE NO.3.20: Write a program to illustrate the use of cast operator in a real life situation. SOLUTION: include<stdio.h> #include<conio.h> void main() { float sum; int n; clrscr(); sum=0; for(n=1;n<=10;++n) { sum=sum+1/(float)n; printf("%2d %6.4f\n",n,sum); } getch(); } Posted by hstu at 9:46:00 PM

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4 comments:

1.

Thank you for your tutorial and lectures. I am just posting a more simple code to

your question 3.2

EXERCISE NO.3.2: Write a program that reads a floating-point number and then displays right-most digit of the integral part of the number.

#include main() {

float a;

int num;

printf("Enter a real number:\t");

scanf("%f", &a);

num = (int)a%10;

printf("Rightmost integer\t %d", num); }

Reply

2.

Niteesh August 1, 2012 at 2:31 AM

Code edit for solution 3.10. Please edit the given values appropriately to suit your solution needs. The code written below is edited and is more in accordance to the

question asked. #include #include main() { float l = 1000.0, r = 500.0;

float step = 0.01, limit = 0.1, c = 0.01;

float frequency = 0;

while(c<=limit) {

frequency = sqrt( (1/l*c) - ((r*r)/(4*c*c)) ); printf("Step - %f\tFrequency - %f\n", step, frequency);

c = c+step; } } Reply 3. Niteesh August 1, 2012 at 2:50 AM

3.11 A more elegant and easy to understand code for beginners...

#include main() {

int a;

printf("Enter four digit integer: \t");

scanf("%d", &a);

//Remember always division operator gives quotient wheras the modulo operator

gives remainder

int a1 = a/1000; // a = 2356, a1 = 2

int a2 = a%1000; // a2 = 356

int a3 = a2/100; // a3 = 3

int a4 = a2%100; //a4 = 56

int a5 = a4/10; //a5 = 5;

int a6 = a4%10; //a6 = 6;

printf("\n\n%d", a1); printf("\n%d", a3); printf("\n%d", a5); printf("\n%d", a6); printf("\n---\ n\n");

int sum = a1+a3+a5+a6;

printf("Sum of Digits - %d", sum); }

3.12 - Write a program to print the size of various data types in C. Commentor's note - The code published above by the blogger is flawed. Below I

present the correct solution.

#include main() { int i = sizeof(int); int f = sizeof(float); int d = sizeof(double); int c = sizeof(char);

printf("Size of data type integer - %d", i); printf("\nSize of data type float - %d", f); printf("\nSize of data type double - %d", d); printf("\nSize of data type char - %d\n\n\n", c); }

E.Balagurusamy C PROGRAMMING

:CHAPTER-4

4.6 STATES ERRORS,IF ANY,IN THE

FOLLOWING STATEMENTS.

[A] :scanf(“%c %f %d”,city,&price,&year);

=NO ERROR.

= THERE WILL BE A & BEFORE AMOUNT.

[C] :scanf(“%f %d”,&amount,&year);

=NO ERROR.

[D] :scanf(\n”%f”,root);

=\n will remain into double quote.

[E] :scanf(“%c %d %ld”,*code,&count,root);

=* IS NOT ALLOWED BEFORE CODE AND &WILL STAY

BEFORE ROOT.

4.7 WHAT WILL BE THE VALUES STORED OF

THE VARIABLES YEAR AND CODE WHEN THE

DATA 1988,X ?

[A] :scanf(“%d %c”,&year,&code);

=YEAR STORS 1988 AND CODE STORS X.

[B] :scanf(“%c %d”,&year,&code);

= YEAR STORS X AND CODE STORS 1988.

[C] :scnaf(“%d %c”,&code,&year);

=CODE STORS 1988 AND YEAR STORS X.

4.8 COUNT,PRICE,CITY HAVE VALUES:

COUNT=1275,

PRICE=235.74,

WHAT WILL BE THE OUTPUT THE

STATEMENT ?

[A] :printf(“%d %f”,count,price);

OUTPUT=1275 235.75.

[B] :printf(“%d %f”,price,count);

OUTPUT=36576 790980

[C] :printf(“%c”,city);

OUTPUT=CAMBRIDGE

4.9 SHOW THE WRONG OF THE OUTPUT

STATEMENTS.

[A] :printf(“%d.7.2%f”,year,amount);

WRONG=7.2 SHOULD REMAIN AFTER THE %

[B] :printf(“%-s,%c”\n,city,code);

WRONG=COMMA IS NOT ALLOWED AND \n SHOULD STAY

INTO QUOTATION.

[C] :printf(“%f %d %s”,price,count,city);

=NO WRONG.

4.10 WHAT VALUES DOSE THE COMPUTER

ASSIGN OF THIS INPUT STATEMENTS?

Scanf(“%4d %*d”,&year,&code,&count);

IF DATA KYED IN 19883745

4.11 HOW CAN WE USE getcher() FUNCTION TO

MULTICHARACTER STRINGS?

=

BY INCLUDING SINGLE QUOTATION OVER

MULTICHARACTER WE

CAN USE getchar() FUNCTION.

4.12 HOW CAN WE USE putchar() FUNCTION

TO MULTICHARACTER STRINGS?

=

BY INCLUDING SINGLE QUOTATION OVER

MULTICHARACTER WE

CAN USE putchar() FUNCTION.

4.13 WHAT IS THE PURPOSE OF scanf()

FUNCTION?

=

IF WE WANT TO TAKE DATA AFTAR RUNNING THE

PROGRAMM THEN

WE USE scanf FUNCTION.

4.14 DESCRIBE THE PURPOSE OF COMMONLY

USED CONVERSION CHARACTERS IN A scanf()

FUNCTION ?

=IT INDICATES WHAT TYPES OF DATA WE TAKE AS

INPUT.

4.15 WHAT HAPPENS WHEN AN INPUT DATA

ITEM CONTAIN ?

[

A

]

MORE CHARACTERS THAN SPECIFIED

FIELD WIDTH.

=

VALUE WILL BE RIGHT-JUSTIFIED.

[B] FEWER CHARACTER THAN SPECIFIED

FIELD WIDTH.

=

VALUE WILL BE LEFT-JUSTEFIED.

4.16 WHAT IS THE PURPOSE OF printf()

FUNCTION ?

=

IT IS USED TO SHOW ANYTHIG ON OUTPUT.

4.17 DESCRIBE THE PURPOSE OF COMMONLY

USED CONVERSION CHARACTERS IN A

printf() FUNCTION ?

= IT INDICATES WHAT TYPES OF DATA WE WANT TO

SHOW ON

OUTPUT.

4.18 WHAT HAPPENS WHEN AN 0UTPUT

DATA ITEM CONTAIN ?

[

A

]

MORE CHARACTERS THAN SPECIFIED

FIELD WIDTH.

=

VALUE WILL BE RIGHT-JUSTIFIED.

[B] FEWER CHARACTER THAN SPECIFIED

FIELD WIDTH.

=

VALUE WILL BE LEFT-JUSTEFIED.

Problem no. 4.1:Given the string“WORDPROCESSING”,

Write a program to read the string from the terminal and

Display the same in the following format:

(a)WORD PROCESSING

(b)WORD

PROCESSING

(c) W.P.

Solution:

#include<stdio.h>

#include<conio.h>

void main()

{

char s[10],d[11];

clrscr();

printf("Enter the string:

");

scanf("%4s%10s",s,d);

printf("(a)%s %s\n",s,d);

printf("(b)%s\n%s\n",s,d);

printf("(c)%.1s.%.1s",s,d);

getch();

}

Output:

Enter the string: WORDPROCESSING

(a) WORDPROCESSING

(b) WORD

PROCESSING

(c) W.P.

Problem no. 4.2: Write a program to read the

values of x and y and print the results of the

following expression in one line:

(a)(x+y)/(x-y) (b)(x+y)/2 (c)(x+y)*(x-y)

Solution:

#include<stdio.h>

#include<conio.h>

void main()

{

float x,y,a,b,c;

clrscr();

printf("Enter the value of x &

y:

");

scanf("%f%f",&x,&y);

if(x-y==0)

printf("(a)=imagine");

else

{

a=(x+y)/(x-y);

printf("(a)=%.2f",a);

}

b=(x+y)/2;

c=(x+y)*(x-y);

printf("

(b)=%.2f

(c)=%.2f",b,c);

getch();

}

Output:

Enter the value of x &

y:

4

3

(a)=7.00

(b)=3.50

(c)=12.00

Enter the value of x &

y:

7

7

(a)=

imagine

(b)=7.00

(c)=0.00

Problem no. 4.3: Write a program to read the following

numbers, round them off to the nearest integers and

print out

the results in integer form:

Solution:

#include<stdio.h> #include<conio.h> void main() { int p,i; float a; clrscr();

printf("ENTER REAL NUMBER FOR GET NEAREST INTEGER NUMBER\n"); for(i=1;i<=4;i++) { scanf("%f",&a); if(a>=0) p=a+0.5; else p=a-0.5;

printf("\nNEAREST INTEGER NUMBER OF %f IS= %d\n",a,(int)p); }

getch(); }

Output:

ENTER REAL NUMBER FOR GET NEAREST INTEGER NUMBER 35.7 NEAREST INTEGER NUMBER OF 35.7 IS= 36

ENTER REAL NUMBER FOR GET NEAREST INTEGER NUMBER 50.21 NEAREST INTEGER NUMBER OF 50.21 IS=50

ENTER REAL NUMBER FOR GET NEAREST INTEGER NUMBER -23.73 NEAREST INTEGER NUMBER OF -23.73 IS= -24

ENTER REAL NUMBER FOR GET NEAREST INTEGER NUMBER -46.45 NEAREST INTEGER NUMBER OF -46.45 IS= -46

Problem no. 4.4:write a program that read 4 floating

values in the range, 0.0 to20.0, and prints a

horizontal bar chart to represent these values using

the character * as the fill character. For the

purpose of the chart, the values may be rounded off to

the nearest integer . For the example ,

the value 4.36 should be represented as follos,

* * * *

* * * * 4.36

* * * *

Solution:

#include<stdio.h> #include<conio.h> void main() { float a1,a2,a3,a4; int x,y,z,t,i; clrscr();

printf("Enter four float number:"); scanf("%f%f%f%f",&a1,&a2,&a3,&a4); x=a1+0.5;y=a2+0.5;z=a3+0.5;t=a4+0.5; printf("The horizontal bar chard is:\n"); for(i=0;i<x;i++) printf("* "); printf("%.2f\n",a1); for(i=0;i<y;i++) printf("* "); printf("%.2f\n",a2); for(i=0;i<z;i++) printf("* "); printf("%.2f\n",a3); for(i=0;i<t;i++) printf("* "); printf("%.2f\n",a4); getch(); }

Output:

Enter four float number: 4.85 4.36 3.12 5.47 The horizontal bar chard is:

* * * * * 4.85 * * * * 4.36 * * * 3.12 * * * * * 5.47

Problem no.4.5: Write a program to demonstrate the

process of multiplication. The program should ask the

user to enter two two digit integer and print the product

of integers as shown bellow.

45

X

37

7x45 is 315

3x45is 135

Add them 1665

Solution:

#include<stdio.h> #include<conio.h> void main() { int a,b,c,p; clrscr();

printf("Enter 2 two digits number:"); scanf("%d%d",&a,&b); printf(" \t%4d\n\tx%3d\n",a,b); printf("\t---\n"); p=b/10; c=b%10; printf("%dx%dis%6d\n",c,a,c*a); printf("%dx%dis%5d\n",p,a,p*a); printf("\t---\n");

printf("\t---"); getch(); }

Output:

45

X

37

7x45 is

315

3x45is

135

Add them

1665

Problem no.4.6: Write a program to read three

integers from the keyboard using one scanf

statement and output them on one line using:

(a)three printf statements,

(b)only one printf with conversion specifiers and

(c) only one printf without conversion specifiers.

Solution:

#include<stdio.h>

#include<conio.h>

void main()

{

int x,y,z;

clrscr();

printf("Enter three integer value of x,y,&z:");

scanf("%d%d%d",&x,&y,&z);

printf("(a)

X=%d,",x);

printf("Y=%d,",y);

printf("Z=%d\n",z);

printf("(b) X=%3d, Y=%2d, Z=%2d\n",x,y,z);

printf("(c) X= %d, Y=%d, Z=

%d",x,y,z);

getch();

}

Output:

Enter three integer value of x,y,&z: 45 27 89 (a) X=45, Y=27, Z=89

(b) X=45, Y=27, Z=89 (c) X=45, Y=27, Z=89

Problem no.4.7: Write a program that prints the

value 10.45678 in exponential format with the following

specifications:

(a)correct to two decimal place,

(b)correct to four decimal place and

(c)correct to eight decimal place.

Solution:

#include <stdio.h>

#include<conio.h>

int main(void)

{

float a=10.45678,x,y,z;

clrscr();

printf("%8.2e\n%10.4e\n%10.8e",a,a,a);

getch();

return 0;

}

Output:

1.04e+01

1.0456e+01

1.04567804e+01

Problem no.4.98 Write a program to print the

value 345.6789 in fixed-point format with the

following specifications:

(a)correct to two decimal place,

(b)correct to four decimal place and

(c)correct to zero decimal place.

Solution:

#include <stdio.h>

#include<conio.h>

void main()

{

float a=345.6789;

clrscr();

printf("The two decimal place is: %.2f\n",a);

printf("The five decimal place is: %.5f\n",a);

printf("The zero decimal place is: %.0f",a);

getch();

}

Output:

The two decimal place is: 345.67

The five decimal place is: 345.67889

The two decimal place is: 345

Problem no.4.9: Write a program to read the name

ANIL KUMAR GUPTA in three parts using the scanf

statement and to display the same in the following

format using the printf statement.

(a)

ANIL K. GUPTA

(b)

A. K. GUPTA

(c)

GUPTA A. K.

Solution:

#include<stdio.h>

#include<conio.h>

void main()

{

char s[6],d[6],c[6];

clrscr();

printf("Enter the string:");

scanf("%5s%5s%5s",s,d,c);

printf("(a)

%s %.1s. %s\n",s,d,c);

printf("(b)

%.1s.%.1s.%s\n",s,d,c);

printf("(c)

%.1s.%.1s.\n",c,s,d);

getch();

}

Output:

Enter the string:

ANIL KUMAR GUPTA

(a)

ANIL

K. GUPTA

( b)

A. K. GUPTA

(d)

GUPTA A. K.

Problem no.4.10: Write a program to read and disply

the following table of data

Name Code Price

Fan 67831 1234.50

Motor 450 5786.70

The name and code must be left-justified and

price must be right-justified .

Solution:

#include<stdio.h> #include<conio.h> void main() { int code1,code2; float price1,price2; char name1[10],name2[10]; clrscr();

printf("Enter first name ,code and price :"); scanf("%s%d%f",name1,&code1,&price1); printf("Enter second name ,code and price :"); scanf("%s%d%f",name2,&code2,&price2); printf("Name\tCode\tPrice\n"); printf("%-s\t%-d\t%.2f\n",name1,code1,price1); printf("%-s\t%-d\t%.2f\n",name2,code2,price2); getch(); }

Output:

Enter first name ,code and price :

Fan

67831

1234.50

Enter second name ,code and

price :

Motor

450 5786.70

Name

Code

Price

Fan

67831

1234.50

Motor

450

5786.70

Chapter 5

DECISION MAKING AND BRANCHING

REVIEW QUESTION:

RQ-5.1:State whether the following are true or false :

(a)When if statements are nested , the last else gets associated with the nearest

if without an else. Ans: False.

(b)One if can have more than one else clause. Ans: False.

(c)A switch statement can always be replaced by a series of if..else statements. Ans: False.

(d)A switch expression can be of any type. Ans: False.

(e)A program stops its execution when a break statement is encountered. Ans: False.

(f)Each expression in the else if must test the same variable. Ans: True.

(g)Any expression can be used for the if expression. Ans: True.

(h)Each case label can have only one statement. Ans: True.

(i)The default case is required in the switch statement. Ans: True.

(j)The predicate !( (x>=10) (y==5) ) is equivalent to (x<10) && (y!=5 ). Ans: True.

RQ-5.2:Fill in the blanks in the following statements:

(a)The ……….operator is true only when both the operands are true. Ans: logical AND (&&).

(b)Multiway section can be accomplished using an else if statement or the . ………statement.

Ans: switch.

(c)The……….. statement when executed in a switch statement causes. immediate exit from the structure

Ans: break.

(d)The ternary conditional expression using the operator ?: code be easily coded using ………..statement.

Ans: if…else.

(e)The expression !(x!=y)can be replaced by the expression………… Ans: x==y.

RQ-5.3:Find errors, if any, in each of the following segments: Solution:

(a)if((x+y=z) && (y>0) ) printf(" ");

Ans: Error.

printf(" "); (b) if (code >1) a= b+c else a=0 Ans: Error.

Correct ans: if (code >1) a= b+c; else a=0; (c) if(p>0) || (q <0) printf("Sign is negative”); Ans: Error. Correct ans:if((p>0) || (q <0)) printf("Sign is negative”);

RQ-5.4:The following is a segment of a program:

x=1; y=1; if(n>0) x=x+1; y=y-1; printf("%d %d", x,y);

what will be the values of x and y if n assumes a value of (a) 1and (b) 0.

Solution:

(a)The value of x is 2 & y is 0. (b)The value of x & y is imaginary.

RQ-5.5:Rewrite each of the following without using compound relations:

(a) if(grade<=59&&grade>=50) second=second+1; Solution: if(grade<=59) second=second+1; if(grade>=50) second=second+1; (b) if ( number>100||number<0) printf(“Out of range”);

else sum=sum+number; Solution: if ( number>100) printf(“Out of range”); else if(number<0) printf(“Out of range”); else sum=sum+number; (c) if (M1>60&&M2>60||T>200) printf(“Admitted\n”); else

printf (“Not admitted”); Solution: if (M1>60) printf (“Admitted\n”); if (M2>60) printf (“Admitted\n”); else if(T>200) printf (“Admitted\n”); else

printf (“Not admitted”);

RQ-5.6:Assuming x=10 ,state whether the following logical expressions are true or false:

(a)x==10 && x>10 && !x Ans:False. (b)x==10 || x> 10 && !x Ans:True. (c)x==10 && x>10 ||!x Ans:False. (d)x==10 ||x>10 || !x Ans:True.

RQ-5.7:Find errors,if any, in the following switch related statements.Assume that the variables x and y are of int type and x=1 and y=2.

Solution:

(a)switch(y); Ans: Error.

Correct ans: switch(y) (b)case 10;

Ans: Error.

Correct ans: case 10: (c)switch(x+y)

Ans:No error.

(d)switch(x) {Case 2: y= x+y; break}; Ans: Error.

Correct ans: switch(x) {Case 2: y= x+y; break;}

RQ-5.8:Simplify the following compound logical expressions:

(a) !(x<=10) (b)!(x==10)||!((y==5)||(z<0)) Ans:(x>10) Ans: (x>0)

(c)!((x+y==z)&&!(z>5)) (d)!((x<=5)&&(y==10)&&(z<5)) Ans: (x<z) Ans: (x>5)

RQ-5.9:Assuming that x=5, y=0,and z=1 initially ,what will be their

values after executing the following code segments?

(a)if(x && y) x=10; else y=10; Output: 10 10 (b)if(x|| y ||z) y=10; else z=0; Output: 1 0 (c)if(x) if(y) z=10; else z=0; Output: 10 0 (d)if(x ==0 || x && y) if(!y)

z=0; else y=1; Output: 0 1

RQ-5.10:Assuming that x=2,y=1 and z=0 initially ,what will be their values after executing the following code segments?

(a) switch(x) { case 2: x=1; y=x+1; case 1: x=0; break; default: x=1; y=0; } Output: 1 0 (b) switch(y) { case 0: x=0; y=0; case 2: x=2; z=2; default: x=1; y=2; } Output: 0 0 0

RQ-5.11:Find the error ,if any,in the following statements: Solution: (a)if(x>=10) printf("\n"); Ans: No error. (b)if(x>=10) printf("OK"); Ans: No error. (c)if(x==10) printf ("Good"); Ans : No error. (d)if(x=<10) printf("Welcome"); Ans : Error.

Correct ans: if(x<=10)

Printf(“Welcome”);

RQ-5.12:What is the output of the following program? Program: main() { int m=5; if(m<3) printf("%d", m+1); else if (m<5) printf("%d", m+2); else if (m<7) printf("%d", m+3); else printf("%d", m+4); getch(); } Output: 8

RQ-5.13:What is the output of the following program? Program: main () { int m=1; if( m==1) { printf ("Delhi");

if(m==2) printf("Chennai"); else printf("Banglore"); } else Printf("END"); getch(); } Output: 1 Delhi 2 Chennai 3 Banglore

RQ-5.14:What is the output of the following program? Program: main() { int m; for(m=1; m<5; m++) printf("%d\n",(m%2) ? m : m*2); getch(); } Output: 1 4 3 8

RQ-5.15:What is the output of following program? Program: main() { int m,n,p; for(m=0; m<3;m++) for(n=0;n<3;n++) for(p=0;p<3;p++) if(m+n+p==2) goto print;

print: printf("%d %d %d",m,n,p); getch(); } Output: 0 0 2

RQ-5.16:What will be the value of x when the following segment is executed?

int x=10,y=15;

x= (x<y)? (y+x) : (y-x) ;

Solution:

The value of x after execution is :-25.

RQ-5.17:What will be the output when the following segment is executed?

int x=0; if(x>=0) if(x>0) printf("Number is positive"); else printf("Number is negative"); Output: 0 Number is positive 1 Number is negative

RQ-5.18: What will be the output when the following segment is executed? Program: char ch = ‘a’ switch(ch) { case ‘a’: printf(“A”); case ‘b’: printf(“B”); case ‘c’: printf(“C”); }

Output: a A b B c C

RQ-5.19:What will be the output of the following segment when executed? Program: main() { int x=10,y=20; if(( x<y)|| (x+5)>10) printf("%d",x); else printf("%d",y); getch(); } Output: 10

RQ-5.20:What will be the output of the following segment when executed? Program: main() { int a=10, b=5; if(a>b) { if(b>5) printf("%d",b); } else printf("%d",a); getch(); } Output: 10

CHAPTER 5

Decision Making and Branching

EXERCISE-5.1 Write a program to determine whether a given number is odd or even and

print the message:

NUMBER IS EVEN or NUMBER IS ODD

(a) without using else option, and (b) with using else option.

Solution:

(a) without using else option:

/*………even or odd………*/ #include<stdio.h> #include<conio.h> void main() { int n; clrscr(); printf(“Enter a number\n”) scanf("%d",&n); if(n%2==0) printf("NUMBER IS EVEN "); if(n%2==1) printf("NUMBER IS ODD "); getch(); }

(b) with else option: /*………even or odd………*/ #include<stdio.h> #include<conio.h> void main() { int n; clrscr(); printf(“Enter a number\n”) scanf("%d",&n); if(n%2==0) printf("Even"); else printf("Odd"); getch(); }

EXERCISE-5.2 Write a program to find the number of and sum of all integers greater than

100 and less than 200 that are divisible by 7.

Solution:

/*…..number between 100-200 divisible by 7……*/

#include<stdio.h> #include<conio.h> void main() { int i,n,r,sum; sum=0; clrscr(); for(i=100;i<=200;i++) { r=i%7; if(r==0) { printf(" %d",i); sum=sum+i; } printf(“Sum=%d”,sum); } getch(); }

EXERCISE-5.3 A set of two linear equations with two unknowns x1 and x2 is given below:

ax1 +bx2=m and cx1+dx2=n The set has unique solution

x1= and x2= provided the determinate ad-cb is not equal to zero.

Write a program that will read the values of constants a,b,c,d,m and n and compute the values of x1 and x2 .An appropriate message should be printed if ad-cb=0.

Solution:

/*…….two linear equation………*/

#include<stdio.h> #include<conio.h> void main() { float a,b,c,d,m,n,x1,x2; clrscr(); printf("Input a,b,c,d,m,n:\n"); scanf("a=%f b=%f c=%f d=%f m=%f n=%f",&a,&b,&c,&d,&m,&n); x1=(m*d-b*n)/(a*d-c*b); x2=(n*a-m*c)/(a*d-c*b); if((a*d-c*b)!=0) printf("x1=%f x2= %f",x1,x2); else

printf("The value is infinity.\n"); getch();

EXERCISE-5.4 Given a list of marks ranging from 0 to 100, write a program to print

number of students:

(a)Who have obtained more than 80 marks, (b) who have obtained more than 60 marks, (c)Who have obtained more than 40 marks, (d) who have obtained 40 or less marks, (e)In the range 81 to 100, (f) in the range 61 to 80,

(g)in the range 41 to 60, and (h) in the range 0 to 40. The program should use a minimum numbers of if statements.

Solution: /*….marks obtain……*/ #include<stdio.h> #include<conio.h> void main() { int marks,count,a,b,c,d,i; a=0; b=0; c=0;d=0; clrscr();

printf("Input 20 boy's marks\n"); for(i=1;i<=20;i++) { scanf("%d",&marks); if(marks>80) a++; else if(marks>60) b++; else if(marks>40) c++; else if(marks<=40) d++; }

printf("Number of students who have obtained more than 80 marks=%d\nNumber

of students who have obtained more than 60 marks=%d\n Number of students who have obtained more than 40 marks=%d\n Number of students who have obtained 40 or less marks=%d",a,b,c,d);

getch(); }

EXERCISE-5.5 Admission to a professional course is subjects to the following conditions:

(a) Marks in Mathematics>=60 (b) Marks in Physics>=50 (c) Marks in Chemistry>=40

(d) Total in all three subjects>=200 or Total in Mathematics and Physics>=150

Given the marks in the three subjects, write a program to process the applications to list the eligible candidates.

Solution:

/*……..admission for a professional course……*/

#include<stdio.h> #include<conio.h> void main() { int r,m,c,p,b; clrscr();

printf("Input Mathmatics,Physics and Chemistry"); scanf("%d%d%d",&m,&p,&c);

r=m+p+c; b=m+p;

if(m>=60&&p>=50&&c>=40&&r>=200&&b>=150) printf("The candidate is eligible");

else

printf("The candidate is not eligible"); getch();

EXERCISE-5.7: Shown below is a Floyd’s triangle . 1 2 3 4 5 6 7 8 9 10 11………..15 79... .. .. .. .. ..91 (a) Write a program to print this triangle.

Solution: /*……….Floyd’s triangle………..*/ #include<stdio.h> #include<conio.h> void main() { int i,j,count,n; clrscr(); count=0;

printf("\n\nHow many rows of Floyd triangle: "); scanf("%d",&n); for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { count++; printf("%d",count); printf(" "); } printf("\n"); }

getch(); }

(b) Modify the program the following from of Floyd’s triangle. 1 0 1 1 0 1 0 1 0 1 1 0 1 0 1 Solution: /*……….Floyd’s triangle………..*/ #include<stdio.h> #include<conio.h> void main() { int i,j,count,n; clrscr(); count=0;

printf("\n\nHow many rows of Floyd triangle: "); scanf("%d",&n); for(i=1;i<=n;i++) { for(j=2;j<=i+1;j++) { printf("%d",(i+j)%2); printf(" "); } printf("\n"); } getch(); }

EXERCISE-5.8 A cloth showroom has announced the following seasonal discounts on

purchase of items:

Purchase amount Discount _{Mill cloth Handloom items} 0-100 5% 101-200 5% 7.5% 201-300 7.5% 10.0% Above300 10.0% 15.0% Write a program using switch and if statements to compute the net amount to be paid by a coustomer. Solution: /*………….marketing of a showroom………*/ #define MC1 0 #define MC2 0.05 #define MC3 0.075 #define MC4 0.10 #define HI1 0.05 #define HI2 0.075 #define HI3 0.10 #define HI4 0.15 #include<stdio.h> #include<conio.h> void main() { float price,net,discount; int level,jobnumber;

clrscr(); input:

printf("Enter level jobnumber and purchase amount\n"); printf("Enter zero for level to End\n");

scanf("%d%d%f",&level,&jobnumber,&price); if(level==0) goto stop;

if(0<=price<=100) level=1; else if(101<=price<=200) level=2; else if(201<=price<=300) level=3; else level=4; switch(level) { case 1: discount=MC1+HI1; break; case 2: discount=MC2+HI2; break; case 3: discount=MC3+HI3; break; case 4: discount=MC4+HI4; break; default:

printf("Error in level code\n"); goto stop;

}

net=price-(price*discount); printf("Net amount=%f\n",net); goto input;

stop:printf("\n\nEND OF THE PROGRAM"); getch();

EXERCISE-5.9 Write a program that will read the value of x and evaluate the following

function y= using

(a) nested if statements. (b) else if statements and (c) conditional operator ?

Solution:

/*…………evaluate the equation………..*/

(a)nested if statements: #include<stdio.h> #include<conio.h> void main() { float x,y; clrscr(); printf("Input x\n"); scanf("%f",&x); if(x!=0) { if(x>0) printf("y=1"); if(x<0) printf("y=-1"); } if(x==0) printf("y=0"); getch(); }

(b)else if statements: #include<stdio.h> #include<conio.h> void main() { float x,y; clrscr(); printf("Input x\n"); scanf("%f",x); if(x!=0) { if(x>0) { printf("1"); } else printf("-1"); } else printf("0"); getch(); } (c)conditional operator: #include<stdio.h> #include<conio.h> void main() { clrscr(); float y,x; printf("Input x\n"); scanf("%f",&x); y=(x!=0)?((x>0)?1:-1):0; printf("%d",y); getch(); }

EXERCISE-5.10 Write a program to compute the real roots of a quadratic equation

ax2+bx2+c=0

The roots are given by the equtions: x1 and x2

The program should request for the values of the constants a,b and c print the values of x1 and x2.Use the following:

(a) No solution, if both a and b are zero (b) There is only one root if a=0(x=-c/b)

(c) There are no real roots, if b2-4ac is negative (d) Otherwise, there no real roots

Test your program with appropriate data so that all logical paths are working as per your design. Incorporate appropriate output messages.

Solution:

/*…….roots of quadratic equation ….*/

#include<stdio.h> #include<conio.h> #include<math.h> void main() { float a,b,c,x,discriminant,root1,root2; clrscr();

printf("Input values of a, b and c\n"); scanf("%f %f %f",&a,&b,&c); discriminant=b*b-4*a*c; if(a==0&&b==0) printf("No solution\n"); else if(a==0) { x=-(c/b); printf("x=%f",x);