Answer
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Hint: First of all this is a very simple and a very easy problem. This problem deals with algebraic mathematics. Here given the arithmetic mean of the ages of a teacher and three students. We are asked to find the age of the teacher. Here in any given data the arithmetic mean of the data is given by the ratio of sum of all the given observations in the data to the total number of observations.
The average of the given data ${a_1},{a_2},{a_3},....{a_n}$, is given by:
$ \Rightarrow \dfrac{{{a_1} + {a_2} + {a_3} + .... + {a_n}}}{n}$
Complete step-by-step answer:
Given that there is a teacher and three students.
The three students are of the same age, but the teacher is not.
Given that the average of the ages of the teacher and the three students is 20 years.
Also given that the difference between the age of teacher and the students is 20 years.
Let the age of the teacher = $t$
Let the age of one of the three students = $x$
Now the ages of all the three students is the same, as given.
The average age of the teacher and the three students is 20 years, as mathematically expressed below:
$ \Rightarrow \dfrac{{t + x + x + x}}{4} = 20$
$ \Rightarrow \dfrac{{t + 3x}}{4} = 20$
$ \Rightarrow t + 3x = 80$
The difference between the age of the teacher and that of a student is 20 years, as given below:
$ \Rightarrow t - x = 20$
$ \Rightarrow x = t - 20$
Now substituting the expression for $x$ in the obtained expression $t + 3x = 80$, as given below:
$ \Rightarrow t + 3x = 80$
$ \Rightarrow t + 3(t - 20) = 80$
Expanding the above expression, as given below:
$ \Rightarrow t + 3t - 60 = 80$
$ \Rightarrow 4t = 140$
Dividing the above expression with 4, as given below:
$ \Rightarrow t = \dfrac{{140}}{4}$
$ \Rightarrow t = 35$
Hence the age of the teacher is 35 years.
The age of the teacher is 35 years.
Note:
Please note that this problem can be done in another way which is slightly different from this method, which is described here. As we have obtained 2 equations from the given information, one is $t + 3x = 80$ and $t - x = 20$, here instead of solving for $t$ by substitution, we could rather solve the both equations for $t,x$ and we can proceed forward, where the final answer will be the same as obtained in the previous method before.
The average of the given data ${a_1},{a_2},{a_3},....{a_n}$, is given by:
$ \Rightarrow \dfrac{{{a_1} + {a_2} + {a_3} + .... + {a_n}}}{n}$
Complete step-by-step answer:
Given that there is a teacher and three students.
The three students are of the same age, but the teacher is not.
Given that the average of the ages of the teacher and the three students is 20 years.
Also given that the difference between the age of teacher and the students is 20 years.
Let the age of the teacher = $t$
Let the age of one of the three students = $x$
Now the ages of all the three students is the same, as given.
The average age of the teacher and the three students is 20 years, as mathematically expressed below:
$ \Rightarrow \dfrac{{t + x + x + x}}{4} = 20$
$ \Rightarrow \dfrac{{t + 3x}}{4} = 20$
$ \Rightarrow t + 3x = 80$
The difference between the age of the teacher and that of a student is 20 years, as given below:
$ \Rightarrow t - x = 20$
$ \Rightarrow x = t - 20$
Now substituting the expression for $x$ in the obtained expression $t + 3x = 80$, as given below:
$ \Rightarrow t + 3x = 80$
$ \Rightarrow t + 3(t - 20) = 80$
Expanding the above expression, as given below:
$ \Rightarrow t + 3t - 60 = 80$
$ \Rightarrow 4t = 140$
Dividing the above expression with 4, as given below:
$ \Rightarrow t = \dfrac{{140}}{4}$
$ \Rightarrow t = 35$
Hence the age of the teacher is 35 years.
The age of the teacher is 35 years.
Note:
Please note that this problem can be done in another way which is slightly different from this method, which is described here. As we have obtained 2 equations from the given information, one is $t + 3x = 80$ and $t - x = 20$, here instead of solving for $t$ by substitution, we could rather solve the both equations for $t,x$ and we can proceed forward, where the final answer will be the same as obtained in the previous method before.
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