
The mean of the factors of 24 is
A) \[\dfrac{{10}}{3}\]
B) \[\dfrac{9}{4}\]
C) \[\dfrac{{15}}{2}\]
D) \[\dfrac{{17}}{3}\]
Answer
611.4k+ views
Hint: Try to find the factors of the number 24 and then for finding the mean add all those numbers and then divide it by the total number of numbers, i.e., count the number of factors of 24 and divide it.
Complete step by step Solution:
A factor is a number or quantity that when multiplied with another produces a given number or expression. So when we try to factorize 24 we get the following factors
\[1,2,3,4,6,8,12,24\]
Now to find the mean which is also termed as average we have to add all the factors and then divide it by total number of factors
The total number of factors \[ = 8\]
Now Let us denote mean by \[\mu \]
\[\begin{array}{l}
\therefore \mu = \dfrac{{1 + 2 + 3 + 4 + 6 + 8 + 12 + 24}}{8}\\
\Rightarrow \mu = \dfrac{{60}}{8}\\
\Rightarrow \mu = \dfrac{{15}}{2}
\end{array}\]
Therefore option C is the correct option.
Note: 1 is a factor of all real numbers and the number itself is always a factor of itself because when we factorize say any number x, we can always get \[x \times 1 = x\] which is also true for any prime number students often make mistakes here they forget to count 1 and the number itself as a factor.
Complete step by step Solution:
A factor is a number or quantity that when multiplied with another produces a given number or expression. So when we try to factorize 24 we get the following factors
\[1,2,3,4,6,8,12,24\]
Now to find the mean which is also termed as average we have to add all the factors and then divide it by total number of factors
The total number of factors \[ = 8\]
Now Let us denote mean by \[\mu \]
\[\begin{array}{l}
\therefore \mu = \dfrac{{1 + 2 + 3 + 4 + 6 + 8 + 12 + 24}}{8}\\
\Rightarrow \mu = \dfrac{{60}}{8}\\
\Rightarrow \mu = \dfrac{{15}}{2}
\end{array}\]
Therefore option C is the correct option.
Note: 1 is a factor of all real numbers and the number itself is always a factor of itself because when we factorize say any number x, we can always get \[x \times 1 = x\] which is also true for any prime number students often make mistakes here they forget to count 1 and the number itself as a factor.
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