Question

Which of the following statements is always true?

• A. $\frac{(x-y)}{2}$is a rational number between $x$ and$y$.
• B. $\frac{(x+y)}{2}$ is a rational number between $x$ and$y$.
• C. $\frac{(x\times y)}{2}$is a rational number between $x$ and$y$.
• D. $\frac{(x÷y)}{2}$is a rational number between $x$ and$y$.

Answer 1

The correct option is B

$\frac{(x+y)}{2}$ is a rational number between $x$ and$y$.

Determine the statement that is true.

The Explanation for the correct option:

Option (B): $\frac{(x+y)}{2}$ is a rational number between $x$ and$y$.

The mean of a data set is the ratio of the sum of all observations to the total number of observations.

Put $x=2\mathrm{and}y=8$:

$$\begin{gathered} \frac{(2+8)}{2}=\frac{10}{2} \\ =5 \end{gathered}$$

The value $5$ lies between $2$ and $8$.

Hence, “Option B” is the correct option

The Explanation for the incorrect options:

Option (A): $\frac{(x-y)}{2}$is a rational number between $x$ and$y$.

Put $x=2\mathrm{and}y=8$:

$$\begin{gathered} \frac{(2-8)}{2}=\frac{-6}{2} \\ =-3 \end{gathered}$$

The value $-3$ doesn't lie between $2$ and $8$.

Option (C): $\frac{(x\times y)}{2}$is a rational number between $x$ and$y$.

Put $x=2\mathrm{and}y=8$

Then,

$$\begin{gathered} \frac{(x\times y)}{2}=\frac{(2\times 8)}{2} \\ \Rightarrow =\frac{16}{2} \\ \Rightarrow =8 \end{gathered}$$

The value $8$ lies between $2$ and $8$.

Option (C): $\frac{(x÷y)}{2}$is a rational number between $x$ and$y$.

Put $x=2\mathrm{and}y=8$

$$\begin{gathered} \frac{(x÷y)}{2}=\frac{(2÷8)}{2} \\ =\frac{{\displaystyle \frac{2}{8}}}{{\displaystyle \frac{2}{1}}} \\ =\frac{2}{8}\times \frac{{\displaystyle 1}}{2} \\ =\frac{{\displaystyle 1}}{8} \end{gathered}$$

The value $\frac{1}{8}$ doesn't lie between $2$ and $8$.

Hence, Option A, option C, Option D are incorrect options

Hence, “Option B” is the correct statement.