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Group A consists of X students and their total age is 221 and their average age is an integer.When group A is merged with Group B with twice the number of students (the number of students between 30 and 40) average age of B is reduced by 1.What is the original average age of B?

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Verified Answer :

johnkellock

mean = sum / count

Group A
m = 221 / c
221 = c x m
221 = 17 x 13 (as it is a whole number under 20)

Group B
m = s / 34

(221 + x) / 51 = x /34 - 1

544/ 34 = 16

16

Answer:

16 years

Step-by-step explanation:

Given,

Number of students in group A = X,

Sum of their ages = 221,

So, the average age of students in group A = \(\frac{\text{Sum of ages}}{\text{Total students}}\)

\(=\frac{221}{X}\)

According to the question,

\(\frac{221}{X}=\text{Integer}\)

∴ X must be a factor of 221,

∵ 221 = 13 × 17,

Now, If X = 13,

Then the number of students in group B = 26 ( NOT POSSIBLE )

If X = 17,

Then the number of students in group B = 34

Which is between 30 and 40,

Now, Let S be the sum of ages of students in group B,

So, the average age in group B = \(\frac{S}{34}\)

Again according to the question,

\(\frac{S+221}{17+34}=\frac{S}{34}-1\)

\(\frac{S+221}{51}=\frac{S-34}{34}\)

\(34S+7514 = 51S - 1734\)

\(34S - 51S = -1734 - 7514\)

\(17S = 9248\)

\(\implies S = 544\)

Therefore, the average age of B = \(\frac{544}{34}\) = 16 years.

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