1.

In my class, the average age of 29 boys is 15 years. But by mistake, my friend, aged 20 years, was counted twice and two boys were missed in that count, whose ages differ by 5 years. If even after correction, the average age of all boys in the class remains 15 years, the age of the younger boy, who was missed in counting, is1). 20 years2). 15 years3). 10 years4). 8 years

Answer»

Given that, average age of 29 boys is 15 YEARS. But by mistake, my FRIEND, aged 20 years, counted twice and two boys were missed in that count, whose ages differ by 5 years.

Average age of n PERSONS = (Sum of age of n persons)/(Number of persons)

⇒ Incorrect sum of ages of 29 boys = Average age of 29 boys × 29

⇒ Incorrect sum of ages of 29 boys = 15 × 29 = 435 years

Since one boy, of age 20 years, was counted twice,

Sum of 28 boys = 435 – 20 = 415 years.

Now, let the age of YOUNGER boy among the two, who were missed in counting be x years and that of the older boy is (x + 5) years.

⇒ Correct sum of ages of boys of the class = 415 + x + (x + 5) = 420 + 2x

Since, average age of the boys of the class, after correction is still 15 years, so

Average age of 30 boys = $(\frac{{420 + 2x}}{{30}})$ 

$(\Rightarrow \frac{{420 + 2x}}{{30}} = 15)$

⇒ 420 + 2x = 15 × 30

⇒ 420 + 2x = 450

⇒ 2x = 450 – 420 = 30

⇒ x = 30/2

⇒ x = 15 years


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