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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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Respuesta :

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 = [tex]\frac{\text{Sum of ages}}{\text{Total students}}[/tex]

[tex]=\frac{221}{X}[/tex]

According to the question,

[tex]\frac{221}{X}=\text{Integer}[/tex]

∴ 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 = [tex]\frac{S}{34}[/tex]

Again according to the question,

[tex]\frac{S+221}{17+34}=\frac{S}{34}-1[/tex]

[tex]\frac{S+221}{51}=\frac{S-34}{34}[/tex]

[tex]34S+7514 = 51S - 1734[/tex]

[tex]34S - 51S = -1734 - 7514[/tex]

[tex]17S = 9248[/tex]

[tex]\implies S = 544[/tex]

Therefore, the average age of B = [tex]\frac{544}{34}[/tex] = 16 years.

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