CBSE Guess > Papers > Important Questions > Class XII > 2012 > Computer Science > Computer Science Mr. Pradumna Singh

Computer Science- CBSE CLASS XII

Q. 9. ) Define a class TAXPAYER in C++ with following description :                             For - 4 Marks
Private members :

1. Name  of type string
2. PanNo of type string
3. Taxabincm (Taxable income) of type float
4. TotTax of type double
5. 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 OUTAX( ) to allow user to view the content of all the data members.

Ans.
class TAXPAYER
{private:
char Name[30],PanNo[30];
float Taxabincm;
double TotTax;
void CompTax()
{ if(Taxabincm >500000)
TotTax= Taxabincm*0.15;
else if(Taxabincm>300000)
TotTax= Taxabincm*0.1;
Else if(Taxabincm>160000)
TotTax= Taxabincm*0.05;
else
TotTax=0.0; }
public:
TAXPAYER(char nm[], char pan[], float tax, double ttax) //parameterized constructor
{ strcpy(Name,nm);
strcpy(PanNo,pan);
Taxabincm=tax;
TotTax=ttax; }

void INTAX()
{ gets(Name);
cin>>PanNo>>Taxabincm;
CompTax(); }

void OUTAX()

{ cout<<Name<<’\n’<<PanNo<<’\n’<<Taxabincm<<’\n’<<TotTax<<endl; } };

Q. 10 .Answer the questions (1) to (4) based on the following :                                 For - 4 Marks
class Student
{ private :
           char Rollno[20], Sname[30];
                             protected :
           auto float marks;
 public:
      Student( );
      void ENROL( );
      void SHOW( );
};
class Graduate: public Student
{    char Fname[30];
protected:
        unsigned int age;
public:
      Graduate( );
      void GENROL( );
      void GSHOW( );
};
      class Pgraduate: private Graduate
      {
       char Mname[25];
      signed int year;
      public:
             Pgraduate( );
void PGENROL( );
             void PGSHOW( );
};

1. Mention the member names that are accessible by an object of Pgraduate class.    

Ans.     PGENROL( ), PGSHOW( )

2. Name the data members which can be accessed by the objects of Graduate class.

Ans.     None

3. Name the data members that can be accessed by  the functions of Pgraduate  class.

Ans.     Mname[25], year, age & marks

4. How many bytes will be occupied by an object of class Pgraduate?

Ans.     113 bytes

Q . 11. Write a function TRANSFERP( 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.                             For - 3 Marks

Ans.:
TRANSFERP( int ALL[ ], int N)
{ int PRIME[100],i,j,tp=0,count;
for(i=0;i<N;i++)
{
count=0;
for(j=0;j<=ALL[i];j++)
if(ALL[i]%j= =0)
count++;
if(count= = 2)
{
PRIME[tp]=ALL[i];
tp++;
}
} //end of for
//displaying all prime numbers of array PRIME[]
cout<<”\nAll prime numbers in resultant array are:\n”;
for(i=0;i<tp;i++)

cout<<PRIME[i]<<’ ‘; }

Q. 12 . b) 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.                             For - 3 Marks
Sol.
Given : B=?, W=10, m=40, n=32, I=7, J=10, PP[I][J]=5000, LBr=0, LBc=0
Row Major:
Address of PP[7][10]=B+W(n(I-LBr)+(J-LBc))
                           5000=B+10(32(7-0)+(10-0))
                           5000=B+10(224+10)
                           5000=B+10*234
                           5000=B+2340
Therefore                B=5000-2340 = 2660
Now Address of PP[18][22] = 2660+10(32(18-0)+(22-0))
                                              =2660+10(576+22)
                                              =2660+10*598
                                              =2660+5980 = 8640

Prepared By: Mr. Pradumna Singh
mail to: [email protected]