Answer:
PRINT "Values for Principal (P), Rate (A) and Time (T)"
INPUT P, A, T
I = P * A * T/100
Amount = P + I
PRINT "Interest: ", I
PRINT "Amount: ", Amount
Explanation:
This prompts the user for values for Principal, Rate and Time
PRINT "Values for Principal (P), Rate (A) and Time (T)"
This gets values for Principal (P), Rate (R) and Time (T)
INPUT P, A, T
This calculates the interest (I)
I = P * A * T/100
This calculates the amount (A)
Amount = P + I
This prints the interest (I)
PRINT "Interest: ", I
This prints the amount (A)
PRINT "Amount: ", Amount
Answer:
All of these statements are true.
Explanation:
Since the while loop is reversing the integer number and leaving the highest order digit in the num and stores the reversed number in the newNum variable.
It skips one digit so if the num is in the range of [100,1000] it will result in a number between 10 and 100.
This loop can never go in infinite loop for any initial value of num because the loop will run as many times as the number of digits.
and if the value of the num is <=10 the while loop will never run and the value of newNum will be 0.
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
B. False, it doesnt only have to be applied word by word
