Question

. The mean age of combined group of men and women is $35$ years. If the mean age of men is $36$ years and that of women is $32$ years. The percentage of men and women in the group is respectively given by:

• A. Men = $75\%$, women = $25\%$
• B. Men = $70\%$, women = $30\%$
• C. Men = $50\%$, women = $50\%$
• D. Men = $25\%$, women = $75\%$

Answer 1

The correct option is A

Men = $75\%$, women = $25\%$

Explanation for the correct option:

Find the percentage composition of men and women in the group:

Mean is the ratio of sum of the observations to the total number of observations considered.

Mathematically, $mean=\frac{sumofobservations}{totalnumberofobservations}$

Let, $a,b,c$ be the mean of combined group of men and women, mean age of men and mean age of women respectively.

Let $N,n_1,n_2$ be the total count of combined group of men and women, men only and women only respectively.

Given that, $a=35years,b=36years,c=32years$

$$\begin{gathered} N=n_1+n_2 \\ \Rightarrow n_1+n_2=100\left(say\right) \end{gathered}$$

By substituting the values we get,

$$\begin{gathered} a=\frac{36n_1+32n_2}{100} \\ \Rightarrow \frac{36n_1+32(100-n_1)}{100}=35 \\ \Rightarrow 36n_1+32(100-n_1)=3500 \\ \Rightarrow 36n_1+3200-32n_1=3500 \\ \Rightarrow 36n_1-32n_1=3500-3200 \\ \Rightarrow 4n_1=300 \\ \Rightarrow n_1=75 \end{gathered}$$

Now, $$\begin{gathered} n_2=100-n_1 \\ {n}_2=100-75=25 \end{gathered}$$

Therefore, the percentage composition of men and women in the group is $75\%and25\%$ respectively.

Hence, option (A) is correct.