Question

If the average (arithmetic mean) of $t$ and $(t + 2)$ is $x$ and if the average of $t$ and $(t - 2)$ is $y$, What is the average of $x$ and $y$?
${\text{(A) 1}}$
${\text{(B) }}\dfrac{t}{2}$
${\text{(C) t}}$
${\text{(D) t + }}\dfrac{{\text{1}}}{{\text{2}}}$

Answer 1

Here we have to first find the values of $x$ and $y$ using the formula of mean, and once we have the values, we will find the mean of both the values $x$ and $y$ to get the final correct answer.

Formula used: ${\text{Mean = }}\dfrac{{{\text{Sum of terms}}}}{{{\text{number of terms}}}}$

Complete step-by-step solution:
It is given that the problem statement that the average of the terms $t$ and $(t + 2)$ is $x$.
Since there are $2$ terms mathematically using the formula of mean we write the value of $x$ as:
$\Rightarrow x = \dfrac{{t + (t + 2)}}{2}$
On opening the brackets, we get:
$\Rightarrow x = \dfrac{{t + t + 2}}{2}$
On simplifying the numerator, we get:
$\Rightarrow x = \dfrac{{2t + 2}}{2}$
Since the number $2$ is common in both the terms, we can remove out it as common and write:
$\Rightarrow x = \dfrac{{2(t + 1)}}{2}$
Now since $2$ is being multiplied both in the numerator and denominator we can it and write:
$\Rightarrow x = (t + 1)$
Therefore, the value of $x$ is $(t + 1)$
Also, it is given from the problem statement that the average of the terms $t$ and $(t - 2)$ is $y$.
Since there is $2$ terms mathematically using the formula of mean we write the value of $y$ as:
$\Rightarrow y = \dfrac{{t + (t - 2)}}{2}$
On opening the brackets, we get:
$\Rightarrow y = \dfrac{{t + t - 2}}{2}$
On simplifying the numerator, we get:
$\Rightarrow y = \dfrac{{2t - 2}}{2}$
Since the number $2$is common in both the terms, we can remove out it as common and write:
$\Rightarrow y = \dfrac{{2(t - 1)}}{2}$
Now since $2$ is being multiplied both in the numerator and denominator we can it and write:
$\Rightarrow y = (t - 1)$
Therefore, the value of $y$ is $(t - 1)$
Now we have to find the value of the average of $x$ and $y$
Since there are $2$terms,
In mathematically it can be written as:
$Mean = \dfrac{{x + y}}{2}$
On substituting the value of $x$ and $y$ we get:
$\Rightarrow$ $Mean = \dfrac{{(t + 1) + (t - 1)}}{2}$
On opening the brackets, we get:
$\Rightarrow$ $Mean = \dfrac{{t + 1 + t - 1}}{2}$
On simplifying the numerator, we get:
$\Rightarrow$ $Mean = \dfrac{{2t}}{2}$
Now since $2$ is being multiplied both in the numerator and denominator we can it and write:
$\Rightarrow$ $Mean = t$

Therefore, the correct option is option $(C)$.

Note: In this question we have an alternate solution:
Instead of solving the part $x$ and $y$ differently and then finding the average of both the terms we can directly substitute the value of $x$ and $y$ in the formula of mean to get the final required answer,
We know:
$x = \dfrac{{t + (t + 2)}}{2}$ and $y = \dfrac{{t + (t - 2)}}{2}$
On finding the average of $x$ and $y$ we get:
$Mean = \dfrac{{t + (t + 2) + t + (t - 2)}}{4}$
On remove the bracket we get:
$Mean = \dfrac{{t + t + 2 + t + t - 2}}{4}$
On adding the numerator term and we get
$Mean = \dfrac{{4t}}{4}$
Therefore $Mean = t$