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Hint: Find the sum of age of 4 family members 10 years ago by multiplying the given mean by 4. Let us assume the age of the youngest child born be $x$. The value of the $x$ can be calculated as the mean is given as 24. Mean is given by \[\dfrac{{{\text{sum of ages}}}}{{{\text{number of members}}}}\].
Complete step-by-step answer:
Let us calculate the sum of the 4 members of the family 10 years ago. It is given that the mean of the ages of the family members is given as 24.
Mean of the ages of the family members is given as \[\dfrac{{{\text{sum of ages}}}}{{{\text{number of members}}}}\]
Since the number of the members of the family were 4 ten years ago, the sum of their ages can be calculated as:
Sum of the ages ten years ago $4\left( {24} \right)$
$ \Rightarrow 96$
The present sum of the ages of the 4 members can be given by adding 40 to the sum of ages of the members ten years ago, as there are 4 members of the family.
Sum of the present ages of the 4 family members $ = 40 + 96$
$ = 136$
Let us assume the age of the youngest born child be $x$
Thus the age of the child born with a difference of two years from the youngest child will be $x + 2$.
Thus the present sum of ages of the family can be calculated by adding all the ages of the family members.
Total sum of the ages of the family $ = 136 + x + x + 2$
\[ = 138 + 2x\]
The mean of the ages of the family is given as 24.
Substituting 6 for the number of family members and \[138 + 2x\] for sum of ages of members of family in the formula \[\dfrac{{{\text{sum of ages}}}}{{{\text{number of members}}}}\], we can solve for $x$.
$
\Rightarrow 24 = \dfrac{{138 + 2x}}{6} \\
\Rightarrow 144 = 138 + 2x \\
\Rightarrow 6 = 2x \\
\Rightarrow x = 3 \\
$
Thus the age of the youngest child of the member is 3 years.
Hence, option C is correct.
Note: The mean of the series of numbers \[{a_1},{a_2}{\text{ }}....{\text{ }}{a_n}\] is given as \[\dfrac{{{a_1} + {a_2}{\text{ }}.... + {\text{ }}{a_n}}}{n}\] where \[n\] is the number of the values in the series.
Complete step-by-step answer:
Let us calculate the sum of the 4 members of the family 10 years ago. It is given that the mean of the ages of the family members is given as 24.
Mean of the ages of the family members is given as \[\dfrac{{{\text{sum of ages}}}}{{{\text{number of members}}}}\]
Since the number of the members of the family were 4 ten years ago, the sum of their ages can be calculated as:
Sum of the ages ten years ago $4\left( {24} \right)$
$ \Rightarrow 96$
The present sum of the ages of the 4 members can be given by adding 40 to the sum of ages of the members ten years ago, as there are 4 members of the family.
Sum of the present ages of the 4 family members $ = 40 + 96$
$ = 136$
Let us assume the age of the youngest born child be $x$
Thus the age of the child born with a difference of two years from the youngest child will be $x + 2$.
Thus the present sum of ages of the family can be calculated by adding all the ages of the family members.
Total sum of the ages of the family $ = 136 + x + x + 2$
\[ = 138 + 2x\]
The mean of the ages of the family is given as 24.
Substituting 6 for the number of family members and \[138 + 2x\] for sum of ages of members of family in the formula \[\dfrac{{{\text{sum of ages}}}}{{{\text{number of members}}}}\], we can solve for $x$.
$
\Rightarrow 24 = \dfrac{{138 + 2x}}{6} \\
\Rightarrow 144 = 138 + 2x \\
\Rightarrow 6 = 2x \\
\Rightarrow x = 3 \\
$
Thus the age of the youngest child of the member is 3 years.
Hence, option C is correct.
Note: The mean of the series of numbers \[{a_1},{a_2}{\text{ }}....{\text{ }}{a_n}\] is given as \[\dfrac{{{a_1} + {a_2}{\text{ }}.... + {\text{ }}{a_n}}}{n}\] where \[n\] is the number of the values in the series.
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