There are 6 consecutive even numbers M₁, M₂, M₃, M₄, M₅, M₆ and 5 consecutive odd numbers N₁, N₂, N₃, N₄, N₅. The average of the even numbers is 4 more than the average of the odd numbers. If the sum of the even numbers is 24 more than the sum of the odd numbers, find the average of the odd numbers.

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1. Let the average of the odd numbers be $a$. Then sum of odds = $5a$. 
2. The average of evens is $a+4$, so sum of evens = $6(a+4)$. 
3. Given sum of evens is 24 more than sum of odds: $6(a+4)=5a+24$. 
4. Simplify: $6a+24=5a+24\Rightarrow a=0$. 
FINAL_ANSWER: b

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