Question

If ${\overline{x}}_1$ and${\overline{x}}_2$ are the means of two series such that ${\overline{x}}_1<{\overline{x}}_2$and$\overline{x}$ is the mean of the combined series, then

• A. $\overline{x}<{\overline{x}}_1$
• B. $\overline{x}>{\overline{x}}_2$
• C. ${\overline{x}}_1<\overline{x}<{\overline{x}}_2$
• D. $\overline{x}=\frac{{\overline{x}}_1+{\overline{x}}_2}{2}$

Answer 1

Answer: (C)

${\overline{x}}_1<\overline{x}<{\overline{x}}_2$

Let the number of terms of two series are $n_1$ and $n_2$ whose means are ${\overline{x}}_1$ and ${\overline{x}}_2$ respectively.
$\because \overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_1}{n_1+n_2}$
Now $\overline{x}-{\overline{x}}_1=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_1}{n_1+n_2}-\overline{x_1}$
$\Rightarrow \overline{x}-\overline{x_1}=\frac{n_2({\overline{x}}_2-{\overline{x}}_1)}{(n_1+n_2)}>0(\because {\overline{x}}_2>{\overline{x}}_1)$
$\Rightarrow \overline{x}>{\overline{x}}_1$
Again $\overline{x}-{\overline{x}}_2=\frac{n_1({\overline{x}}_1-{\overline{x}}_2)}{(n_1+n_2)}<0(\because {\overline{x}}_1<{\overline{x}}_2)$
$\Rightarrow \overline{x}<{\overline{x}}_2$
Hence, $\overline{x_1}<\overline{x}<\overline{x_2}$