If the sum of five consecutive even numbers is 40 more than the average of those numbers, then find the middle number of the series?
Let the first number be x and the consecutive nos be x+1,...,x+4
If the sum of five consecutive even numbers is 40 more than the average of those numbers,
Acc to question,
$$x+(x+1)+(x+2)+(x+3)+(x+4)= 40+\frac{x+(x+1)+(x+2)+(x+3)+(x+4)}{5}$$
$$5x+10 = 42+x$$
$$x=8$$
Therefore the 3rd number (which is the middle number)$$=x+2=10$$
Option B is correct.
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