$16$ children are to be divided into two groups $A$ and $B$ of $10$ and $6$ children. The average percent marks obtained by the children of group $A$ is $75$ and the average percent marks of all the $16$ children is $76$. What is the average percent marks of children of group $B$?

Using the method of alligation, group A has an average of 75 and all 16 children have an overall average of 76, so the difference is 1 for the 10 children in group A (giving a total of 10). Let x be the average of group B, so the difference between x and 76 is multiplied by the 6 children in group B. Equating the deviations gives 10 equals 6 times the quantity x minus 76, so x minus 76 equals 10 divided by 6, which equals 1 and two-thirds. Adding 76 to 1 and two-thirds gives an average of 77 and two-thirds for group B.

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