Question

The mean of $3,a + 2,8,12,2a - 1$ and $6$ is $7$ . Find the value of $a$ .
A. $2$
B. $3$
C. $4$
D. $5$

Answer 1

Hint: We are given a set of numbers and first we need to calculate the number of observations given. And the mean of these numbers is given by dividing the sum of the numbers by the number of observations. Then, we need to equate it with the given mean and calculate the value of $a$ .

Complete step-by-step answer:
We are given the mean of a few numbers.
At first, we need to find how many numbers are given as the number of observations plays a major role.
The numbers given are $3,a + 2,8,12,2a - 1$ and $6$
From this we get that the number of observations is $6$ .
The mean of the numbers is given by adding all the numbers and dividing them by the number of observations
So first, let's add all the numbers given
The given numbers are $3,a + 2,8,12,2a - 1$ and $6$
Group all the like terms together and proceed with the calculation
$
  3 + a + 2 + 8 + 12 + 2a - 1 + 6 = 3 + 2 + 8 + 12 - 1 + 6 + a + 2a \\
   = 30 + 3a \\
 $
The mean of these numbers is given by dividing the sum the numbers by the number of observations, that is $6$
Hence mean is given is given by
$Mean = \dfrac{{30 + 3a}}{6}$
We are given that the mean of these numbers is $7$
So, now we need to equate the equation of the mean which we found to the given mean, that is $7$ .
$7 = \dfrac{{30 + 3a}}{6}$
Multiplying both sides by $6$ , we get
$
  7*6 = \dfrac{{30 + 3a}}{6}*6 \\
  42 = 30 + 3a \\
 $
Now, subtract $30$ on both sides of the above equation
$
  42 - 30 = 30 + 3a - 30 \\
  12 = 3a \\
 $
Divide by $3$ on both sides
\[
  \dfrac{{12}}{3} = \dfrac{{3a}}{3} \\
  a = 4 \\
 \]
Now we get the value of $a$ to be $4$

So, the correct answer is “Option C”.

Note: In the first option, we are given the value of $a$ to be $2$ . Hence, the option is not correct.
In the second option, we are given the value of $a$ to be $3$ . Hence, the option is not correct.
In the third option, we are given the value of $a$ to be $4$ and this matches our calculated value. Hence, the option is correct.