The ratio of the average age of the class without the teacher to the average age including the teacher is 8 : 9. If the ratio of the teacher’s age and the average age of the students is 3 : 1, find the number of students.

Let number of students in the class = x
Let the average age of the students = y
Let the age of the teacher = z

Given,
1) ratio of the average age of the class without the teacher to the average age including the teacher = 8 : 9
Average age of students and teacher = \((No.Of Students*Avg Age + Age of teacher)/(No.of students + 1)\)
= \((x*y + z)/(x+1)\)
Given this ratio = 8:9
\(y/((x*y + z)/(x+1)) = 8/9\)
\(y(x+1)/(x*y + z) = 8/9\)
\(xy + 9y = 8z\) -----------------------equation 1

2) Ratio of the teacher’s age and the average age of the students is 3 : 1
\(z/y = 3/1\)
\(z = 3y\) --------------------------------- equation 2
equation 2 in equation 1 implies,
\(xy + 9y = 8(3z)\)
\(xy = 15y\)
\(y(x-15) =0\)
Since y avg age can't be zero
\(x = 15\)
Number of students = 15

Option E is correct

(A)A logical way..

(I) If s students are there, the strength with the addition of a teacher becomes s+1..
(II) However, teacher's age is 3 times the average of students average age, we can take instead of teacher we are adding the 3 students to total students, say s... so s+3

Therefore, final ratio 8:9 will be similar to ratio s=1 : s+3..
\(\frac{8}{9}=\frac{s+1}{s+3}.....9(s+1)=8(s+3)......9s+9=8s+24......s=24-9=15\)


(B)If you do not understand the reasoning behind the above solution..
# of students = s and average age = a... so total age = as
# after addition of teacher = s+1, and age of teacher = 3a...so total age = as+3a... average age now = \(\frac{as+3a}{s+1}=\frac{a(s+3)}{s+1}\)

Thus initial average age : average age after addition = a : \(\frac{a(s+3)}{s+1}\) = 8:9.....1 : \(\frac{(s+3)}{s+1}\) = 8:9..
\(9*1=\frac{(s+3)}{s+1}*8....9(s+1)=8(s+3)......9s+9=8s+24......s=24-9=15.\)

E
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Hey Bharat, I did the problem basically on the paper first So I wrote the final step after resolving the equation as equation 1 ...you are right I missed avg of students.. I arrived at the right answer..including the avg of students. I have edited it, inline in the first post with the correct steps and thanks for pointing out the mistake.