Answer to Question #166564 in C++ for Kefilwe Mamabolo

Question

2.1 Study the following descriptions and write only the statements to achieve the required 

objectives in both algorithm and C++ syntax.

2.1.1 The array called quantity has 20 elements. Write statements that will help in counting

how many of these elements have a value of less than 50. Also determine the sum of 

the values that are more than or equal to 50. Display these 2 answers.

2.1.2 The array called itemPrice has 150 elements. Write the statements to determine and 

display the highest price. If there is more than one item with this price, the number of 

items must also be displayed. 

Answer 1

Solution 2.1.1

using namespace std;


#define ARRAY_SIZE	20
#define MIN_NUM	0
#define MAX_NUM	100
#define NUM_TSH	50


main(void)
{
	int Quantity[ARRAY_SIZE],n,NumCount=0,Sum=0;
    srand(time(0)); 
    cout<<"\nThe values are: ";
    for(n=0;n<ARRAY_SIZE;n++)
    {
		Quantity[n]=-1;
		while(Quantity[n]<MIN_NUM || Quantity[n]>MAX_NUM) Quantity[n] = rand();
		cout<<Quantity[n]<<", ";
		if(Quantity[n]<NUM_TSH) 	NumCount++;
		else				Sum = Sum + Quantity[n];
	}
	cout<<"\n\nSolution-1: No. of values <  "<<NUM_TSH<<" = "<<NumCount;
	cout<<"\n\nSolution-2: Sum of values >= "<<NUM_TSH<<" = "<<Sum;
	return(0);
}

Output:

Solution 2.1.2

using namespace std;


#define ARRAY_SIZE	150
#define MIN_PRICE	0
#define MAX_PRICE	100


main(void)
{
	int ItemPrice[ARRAY_SIZE],n=0,MaxPr,Count=0;
    srand(time(0)); 
    cout<<"\nThe price values are: ";
    MaxPr = MIN_PRICE;
    for(n=0;n<ARRAY_SIZE;n++)
    {
		ItemPrice[n]=-1;
		while(ItemPrice[n]<MIN_PRICE || ItemPrice[n]>MAX_PRICE) ItemPrice[n] = rand();
		cout<<ItemPrice[n]<<", ";
		if(MaxPr<=ItemPrice[n]) MaxPr=ItemPrice[n];
	}
    for(n=0;n<ARRAY_SIZE;n++)
    {
		if(ItemPrice[n]==MaxPr)			Count++;
	}
	
	cout<<"\n\nThe highest price = "<<MaxPr;
	cout<<"\n\nNo. of Items with highest price (="<<MaxPr<<") are: "<<Count;
	return(0);
}

Output: