cfc198 wrote:
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.
A. 8
B. 9
C. 10
D. 12
E. 15
(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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