The average of 7 consecutive whole numbers is 7. Then 1 is added to the first number, 2 is added to the second number, 3 is added to the third number, and so on. What is the average of the seven new numbers? Use the variable n to solve.
Let's use the variable $n$ to represent the first number in the original set of consecutive whole numbers. Then, the other six numbers are $n+1$, $n+2$, $n+3$, $n+4$, $n+5$, and $n+6$.
Since the average of the seven consecutive whole numbers is 7, we can set up the equation:
$$\frac{n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6)}{7} = 7$$
This simplifies to:
$$\frac{7n + 21}{7} = 7$$
Multiplying both sides of the equation by 7 gives:
$$7n + 21 = 49$$
Subtracting 21 from both sides of the equation gives:
$$7n = 28$$
Dividing both sides of the equation by 7 gives:
$$n = 4$$
So the first number in the original set of consecutive whole numbers is 4.
Adding 1, 2, 3, 4, 5, 6, and 7 to the respective numbers in the original set of consecutive whole numbers gives:
4+1 = 5
4+2 = 6
4+3 = 7
4+4 = 8
4+5 = 9
4+6 = 10
4+7 = 11
The average of these seven new numbers is:
$$\frac{5 + 6 + 7 + 8 + 9 + 10 + 11}{7} = \frac{56}{7} = 8$$
Therefore, the average of the seven new numbers is 8. Answer: \boxed{8}.