Answer:
There were 4 numbers in the original integer list.
Step-by-step explanation:
Suppose that x is the sum of the integers in the original list, n is the number of integers in the original list and y is the average of those integers. The next formula represents the average of the numbers:
[tex]y=\frac{x}{n}[/tex]
From this equation, you can tell that x=ny.
The problem gives us the following information:
1. The average of a list of integers goes up by 2 when 23 is added to the list, the next equation describes this situation (call this equation Eq 1)
[tex]\frac{x+23}{n+1}=y+2[/tex]
2. If 9 is added to the new list, then the average reduces by 1, the next equation describes this situation, you have to add 2 to n (call this equation Eq 2)
[tex]\frac{x+23+9}{n+2}=y+2-1[/tex]
Now, use the fact that x=ny and replace it in both equations, the new equations will be:
Eq 1:
[tex]\frac{ny+23}{n+1}=y+2[/tex]
[tex]ny+23=(y+2)(n+1)[/tex]
Eq 2:
[tex]\frac{ny+23+9}{n+2}=y+2-1[/tex]
[tex]ny+32=(y+1)(n+2)[/tex]
Solve y in Eq 1:
[tex]ny+23=ny+y+2n+2\\ny-ny+23-2=y+2n\\y=21-2n[/tex]
Replace y in Eq 2:
[tex]n(21-2n)+32=((21-2n)+1)(n+2)\\21n-2n^2+32=(22-2n)(n+2)\\21n-2n^2+2n^2+32=22n+44-4n\\21n-22n+4n=44-32\\3n=12\\n=\frac{12}{3}\\n=4[/tex]
The number of integers in the original list (n) was 4.
