CBSE Class 12 Computer Science Revision Worksheet Set 03

Q1. Write a function in C++ that counts the number of words in a long string where each word is separated by atleast one blank space and the string terminates with a period.

Q2. Write a function in C++ to find the sum of the following series:
1+2+3+4+5+6+….. upto N terms.

Q3. Write a function in C++ to find the sum of the following series:
1²+3²+5²+7²+9²+….upto n terms.

Q4. Write a function in C++ to find the sum of the following series:
(1)+(1+2)+(1+2+3)+(1+2+3+4)…. Upto n terms.

Q5. Write a function in C++ to find the sum of the following series:
(2²)+(2²+4²)+(2²+4²+6²)+(2²+4²+6²+8²)…. Upto n terms.

Q6. Write a function in C++ to find the sum of the following series:
1/1²+1/3²+1/5²+1/7²+…upto n terms

Q7. Write a C++ function having two value parameters X and N with result type float to find the sum of series given below:
1+X¹/2! + X²/3!+…..+Xⁿ/(N+1)!

Q8. Write a program to print the truth table for XY + Z.

Q9. Write a C++ function that converts a decimal numbers between 0 to 63 into a 8‐bit binary number and prints the binary equivalent.

Q10. Write a program in C++ to print first 10 multiples of an integer N, where N is to be entered by user.

Q1. What is a parameter? Differentiate between an actual and a formal parameter with an example. Name the different types of formal parameter supported by C++. What type of parameter you must be using when passing an array to a function?

Q2. What is a compiler directive? Why do we need #include in a C++ program? Name the include file, to which following built‐in functions belong to:
(a) strcmp (b) randomize (c) setw() (d) isalnum()
(e) sin() (f) gotoxy()

Q3. Find the syntax error(s),if any, in the following program: 
include<iostream.h>
void main()
{
int R; W=90;
while w>60
{
R=W‐50;
switch(w)
{
20:cout<<”Lower Range”<<endl;
30:cout<<”Middle Range”<<endl;
20: cout<<”Higher Range”<<endl;
}
}
}

Q4. What would be output by the following program:
#include<iostream.h>
void Execute (int &B, int C=100)
{
int TEMP=B+C;
B+=TEMP;
if(C==100)
cout<<TEMP<<B<<C<<endl;
}
void main()
{
int M=90, N=10;
Execute(M);
cout<<M<<N<<endl;
Execute(M,N);
cout<<M<<N<<endl;
}

Q5. Give the output of the following program segment:
void main()
{
char *NAME = “CoMPutER”;
for(int x=0;x<strlen(NAME);x++)
if(islower(NAME[x])))
NAME[x]=toupper(NAME[x]);
else
if(isupper(NAME[x]))
if(X%2==0)
NAME[x]=tolower(NAME[x]);
else
NAME[x]=NAME[x‐1];
puts(NAME);
}

Question 1. Define a class TAXPAYER in C++ with following description :
Private members :
a. Name of type string
b. PanNo of type string
c. Taxabinc (Taxable income) of type float
d. TotTax of type double
e. A function CompTax( ) to calculate tax according to the following slab:
Taxable Income | Tax %
Up to 160000 | 0
>160000 and <=300000 | 5
>300000 and <=500000 | 10
>500000 | 15
Public members :
- A parameterized constructor to initialize all the members
- A function INTAX( ) to enter data for the tax payer and call function CompTax( ) to assign TotTax.
- A function OUTTAX( ) to allow user to view the content of all the data members.
Answer:
```cpp #include #include using namespace std; class TAXPAYER { string Name; string PanNo; float Taxabinc; double TotTax; void CompTax() { if (Taxabinc <= 160000) { TotTax = 0; } else if (Taxabinc > 160000 && Taxabinc <= 300000) { TotTax = (Taxabinc - 160000) * 0.05; } else if (Taxabinc > 300000 && Taxabinc <= 500000) { TotTax = (140000 * 0.05) + (Taxabinc - 300000) * 0.10; } else { TotTax = (140000 * 0.05) + (200000 * 0.10) + (Taxabinc - 500000) * 0.15; } } public: TAXPAYER(string name, string pan, float income) { Name = name; PanNo = pan; Taxabinc = income; CompTax(); } void INTAX() { cout << "Enter Name: "; getline(cin, Name); cout << "Enter PAN Number: "; cin >> PanNo; cout << "Enter Taxable Income: "; cin >> Taxabinc; CompTax(); } void OUTTAX() { cout << "Name: " << Name << endl; cout << "PAN Number: " << PanNo << endl; cout << "Taxable Income: " << Taxabinc << endl; cout << "Total Tax: " << TotTax << endl; } }; ```
In simple words: We define a class TAXPAYER that holds a taxpayer's personal details and calculates their tax amount based on dynamic slab ranges.

Exam Tip: When designing a class in C++, make sure to declare helper functions like CompTax() under the private access specifier if they are only intended for internal calculations.

Question 2. Give the output of the following program ( Assuming that all required header files are included in the program ) :
#include<iostream.h>
#include<stdio.h>
#include<conio.h>
void TRANSFER(char *s1,char *s2)
{ int n,j=0;
for(int i=0;*(s1+i)!='\0';i++)
{
n=*(s1+i);
if(n%2==0)
*(s2+j++)=*(s1+i);
} }
void main()
{ char *p="CharLesBabBaGe",q[80];
TRANSFER(p,q);
cout<<q<<endl;}
Answer:
The program will output the following result:
**hrLbB**

Tracing Details:
The program loops through each character in the string "CharLesBabBaGe" and checks if the ASCII value of the character is even (i.e., `n % 2 == 0`):
- 'C' (67) is Odd
- 'h' (104) is Even - Copied
- 'a' (97) is Odd
- 'r' (114) is Even - Copied
- 'L' (76) is Even - Copied
- 'e' (101) is Odd
- 's' (115) is Odd
- 'B' (66) is Even - Copied
- 'a' (97) is Odd
- 'b' (98) is Even - Copied
- 'B' (66) is Even - Copied
- 'a' (97) is Odd
- 'G' (71) is Odd
- 'e' (101) is Odd
Characters with even ASCII values are sequentially copied, resulting in "hrLbB".
In simple words: This program processes a string and copies only the characters that have an even ASCII value into a new array.

Exam Tip: When tracing string manipulation code, look up the ASCII values of individual characters carefully (uppercase and lowercase letters have different values) and keep a pointer/counter index trace.

Question 3. Answer the questions (i)to (iv) based on the following: 4
class FacetoFace {
char CenterCode[10];
public:
void Input( );
void Output( ); };
class Online {
char Website[50];
public:
void Sitein( );
void Siteout( );};
class Training : public FacetoFace, private Online {
long Tcode;
float Charge;
int Period;
public:
void Register( );
void Show ( ); };
Answer:
(i) **Multiple Inheritance** is shown in the above example, as class `Training` inherits from more than one base class (`FacetoFace` and `Online`).

(ii) Member functions accessible from `Show()` function of class `Training`:
- From `Training`: `Register()`, `Show()`
- From `FacetoFace`: `Input()`, `Output()`
- From `Online`: `Sitein()`, `Siteout()`

(iii) Members accessible through an object of class `Training`:
- Only public member functions can be accessed: `Register()`, `Show()`, `Input()`, `Output()`.

(iv) **No**, the function `Output()` is not accessible inside the function `Siteout()`.
*Justification:* `Output()` is a member of the class `FacetoFace` and `Siteout()` is a member of the class `Online`. Since there is no inheritance relationship between `FacetoFace` and `Online`, they cannot access each other's members.
In simple words: We analyze a multi-level class hierarchy to identify inheritance types, accessible member functions, and variable visibility boundaries.

Exam Tip: In multiple inheritance, public base class member functions are accessible via the derived class object, while private base class functions are completely hidden outside.

Question 4. Write a function TRANSFER( int ALL[ ], int N) , to transfer all the prime numbers from a one dimensional array ALL[ ] to another one dimensional array PRIME[ ]. The resultant array PRIME[ ] must be displayed on screen.
Answer:
```cpp #include using namespace std; bool isPrime(int num) { if (num <= 1) return false; for (int i = 2; i * i <= num; i++) { if (num % i == 0) return false; } return true; } void TRANSFER(int ALL[], int N) { int PRIME[100]; // Assuming maximum size of 100 for destination array int count = 0; for (int i = 0; i < N; i++) { if (isPrime(ALL[i])) { PRIME[count++] = ALL[i]; } } cout << "Prime numbers in the array: "; for (int i = 0; i < count; i++) { cout << PRIME[i] << " "; } cout << endl; } ```
In simple words: This function scans an input array, identifies all prime numbers using a helper test, and copies them into a second array.

Exam Tip: Always implement a robust prime checking logic (usually testing divisors from 2 up to the square root of the number) to ensure correct filtering.

Question 5. An array PP[40][32] is stored in the memory along the row with each of the elements occupying 10 bytes. Find out the memory location for the element PP[18][22], if the element PP[7][10] is stored at memory location 5000.
Answer:
Since the array is stored along the row, we use the Row-Major formula (assuming 0-based indexing):
$$\text{Address}(PP[i][j]) = \text{Base Address} + W \times [ i \times C + j ]$$
Where:
- Number of columns ($C$) = 32
- Element size ($W$) = 10 bytes
- $\text{Address}(PP[7][10]) = 5000$

Step 1: Calculate Base Address $$5000 = \text{Base Address} + 10 \times [ 7 \times 32 + 10 ]$$ $$5000 = \text{Base Address} + 10 \times [ 224 + 10 ]$$ $$5000 = \text{Base Address} + 10 \times [ 234 ]$$ $$5000 = \text{Base Address} + 2340$$ $$\text{Base Address} = 5000 - 2340 = 2660$$
Step 2: Find Address of $PP[18][22]$ $$\text{Address}(PP[18][22]) = 2660 + 10 \times [ 18 \times 32 + 22 ]$$ $$\text{Address}(PP[18][22]) = 2660 + 10 \times [ 576 + 22 ]$$ $$\text{Address}(PP[18][22]) = 2660 + 10 \times [ 598 ]$$ $$\text{Address}(PP[18][22]) = 2660 + 5980 = 8640$$
In simple words: We use the row-major memory mapping formula to find the base memory address of the 2D array and then locate another specific element's address.

Exam Tip: In row-major formula, remember to multiply the row index by the number of columns ($C = 32$) and scale by the element size ($W = 10$ bytes).

Question 6. Convert the following infix expression to its equivalent postfix expression Showing stack contents for the conversion:
(A+B)*(C^(D-E)+F)-G
Answer:
Let us trace the conversion step-by-step using an operator stack:

The equivalent postfix expression is: **AB+CDE-^F+*G-**
In simple words: We convert an arithmetic infix expression into its postfix equivalent by utilizing an operator stack to maintain operator precedence.

Exam Tip: The exponential operator ^ has the highest precedence; ensure it is popped off the stack when an operator of lower precedence (like + or -) is encountered.

Question 7. Assume a text file “coordinate.txt” is already created. Using this file create a C++ function to count the number of words having first character capital .Also count the presence of a word ‘Do’.
Answer:
```cpp #include #include #include #include using namespace std; void countWords() { ifstream fin("coordinate.txt"); if (!fin) { cout << "Error opening file.\n"; return; } string word; int capitalCount = 0; int doCount = 0; while (fin >> word) { // Count words starting with a capital letter if (isupper(word[0])) { capitalCount++; } // Count instances of the exact word "Do" if (word == "Do") { doCount++; } } cout << "Number of words starting with a capital letter: " << capitalCount << endl; cout << "Occurrences of the word 'Do': " << doCount << endl; fin.close(); } ```
In simple words: This program opens a text file, reads it word-by-word, and counts how many words start with a capital letter or match the exact word "Do".

Exam Tip: Utilize the standard input file stream ifstream and check characters using the isupper() function for cleaner, faster code.

Question 8. Write function in C++ to perform Insert operation in a dynamically allocated Queue containing names of employees
Answer:
```cpp #include #include using namespace std; struct NODE { char Name[50]; NODE *Link; }; void InsertQueue(NODE *&front, NODE *&rear, char empName[]) { NODE *temp = new NODE; if (temp == NULL) { cout << "Queue Overflow! Out of memory.\n"; return; } strcpy(temp->Name, empName); temp->Link = NULL; if (rear == NULL) { front = rear = temp; } else { rear->Link = temp; rear = temp; } } ```
In simple words: We create a dynamically allocated queue (using a linked list) and add new employee names at the rear end of the queue.

Exam Tip: Remember to set the next pointer (Link) of the newly created node to NULL to properly mark the new end of the dynamic queue.

Question 9. Write a function in C++ to display object from the binary file “PRODUCT.Dat” whose product priceis more than Rs 200. Assuming that binary file is containing the objects of the following class:
class PRODUCT {
int PRODUCT_no;
char PRODUCT_name[20];
float PRODUCT_price;
public:
void enter( ) {
cin>> PRODUCT_no ; gets(PRODUCT_name) ;
cin >> PRODUCT_price; }
void display() {
cout<< PRODUCT_no ; cout<<PRODUCT_name ;cout<< PRODUCT_price; }
int ret_Price( ) {
return PRODUCT_price;
} };
Answer:
```cpp #include #include using namespace std; void displayHighPriceProducts() { ifstream file("PRODUCT.Dat", ios::binary); if (!file) { cout << "Error opening binary file.\n"; return; } PRODUCT obj; while (file.read((char*)&obj, sizeof(obj))) { if (obj.ret_Price() > 200) { obj.display(); cout << endl; } } file.close(); } ```
In simple words: This function reads a binary file containing PRODUCT records and prints the details of any product whose price exceeds Rs 200.

Exam Tip: Always use file.read((char*)&obj, sizeof(obj)) within a loop to read binary objects safely until the end of the file is reached.

Question 10. Write a function Get1from2() function in C++ to transfer the content from two arrays First[ ] and Second[ ] to array All[ ]. The even places (0,2,4…..) of array All[] should get the contents from the array First[ ] and odd places (1,3,5….)of the array All[ ] should get the contents from the array Second[ ]
Eg:
If the First [ ] array contains 30, 60,90,
And the Second [ ] array contains 10, 50,80,
Then All [ ] array should contain 30, 10, 60,50,90,80.
Answer:
```cpp void Get1from2(int First[], int Second[], int All[], int N) { for (int i = 0; i < N; i++) { All[2 * i] = First[i]; // Transfers elements to even indexes All[2 * i + 1] = Second[i]; // Transfers elements to odd indexes } } ```
In simple words: This function takes two arrays of equal size and merges them into a single target array by alternating their elements.

Exam Tip: For any index i in the source arrays, the formula 2i and 2i+1 perfectly places the elements at even and odd indexes of the destination array.