Ask Question
25 January, 12:46

Declare a variable named pump number as an integer type. Test if it is within the following range: Greater and equal to 1 and smaller and equal to 10, if it is, multiple its value by 13. Now test if it is an even number, if it is, deliver a message to the user, "it is a bad number!". If the pump number is not in the range described above, then deliver a message to user, "out of range value!"

+2
Answers (1)
  1. 25 January, 13:08
    0
    Following are the program in C+ + Language.

    #include / / header file

    using namespace std; / / namespace

    int main () / / main function

    {

    int pump; / / variable declaration

    cout<<"Enter the number: ";

    cin>>pump; / / Read the value by the user

    if ((pump > = 1) && (pump < = 10)) / / checking condition

    pump=pump * 13;

    if (pump % 2 = = 0) / / checking condition

    cout<<" it is a bad number!";

    else

    cout<<" out of range value!";

    return (0);

    }

    Output:

    Enter the number:4

    it is a bad number!

    Explanation:

    Following are the Description of the following statements.

    Declared a variable "pump" vaiable of int type. Read the number by the user by using cin function. Check the condition if ((pump > = 1) && (pump < = 10)) then it executed the if statements block. Inside this if there is also one another i. e nested if if (pump % 2 = = 0) if this statement then it executed the statement inside if otherwise else block will be executed
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Declare a variable named pump number as an integer type. Test if it is within the following range: Greater and equal to 1 and smaller and ...” in 📗 Computers & Technology if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers