10 years ago, the average age of a family of 4 members was 24 years. Two children having been born(with an age difference of 2 years), the present average age of the family is again 24 years. Then the present average age of the family is again 24 years. Then the present age of the youngest child is :
A. 1 year
B. 2 years
C. 3 years
D. 4 years
Answer
597.9k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 8 Social Science: Engaging Questions & Answers for Success

Master Class 8 Science: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 Social Science: Engaging Questions & Answers for Success

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Give me the opposite gender of Duck class 8 english CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Advantages and disadvantages of science

Full form of STD, ISD and PCO

What are gulf countries and why they are called Gulf class 8 social science CBSE

