$10$ years ago,the average age of a family of $4$ members was $24$ years. Two children have been born (with an age difference of $2$ years),and the present average age of the family remains the same. The present age of the youngest child is (in years):

$A$ group of boys has an average weight of $36 \,kg$. One boy weighing $42 \,kg$ leaves the group and another boy weighing $30 \,kg$ joins the group. If the average now becomes $35.7 \,kg$,then how many boys are there in the group?

The average weight of $3$ men $A, B$ and $C$ is $84 \, kg$. Another man,$D$,joins the group,and the average weight becomes $80 \, kg$. If another man,$E$,whose weight is $3 \, kg$ more than that of $D$,replaces $A$,then the average weight of $B, C, D$ and $E$ becomes $79 \, kg$. The weight of $A$ is (in $kg$):

The average of $3$ numbers is $22$. If the first number is $\frac{3}{8}$ of the sum of the other two numbers,what is the first number?

The average marks of $120$ students are $35.$ If the average of passed students was $39$ and failed students was $15,$ then find the number of students who have passed.

The average of the first $46$ natural numbers is: