Question

Average age of 5 boys is 16 years with that of 4 boys is 16 year 3 months.the age of the 5th boy is what?​

Answer 1

average age of 5 boys is

average=total sum of all ages/total number of boys

6 years=12×16=192months

192=total sum of ages/5

192×5=total sum of ages

960=sum of there ages

average of 4 boys

average=sum of there ages/total number of boys

16years and 3 months=192+3=195 months

195=sum of there ages/4

195×4=sum of there ages

780=sum of there ages

age of 5th boy=sum of age of 5 boys-sum of age of 4 boys

age of 5th boy= 960-780

age of 5th boy= 180 months

age of 5th boy = 15 years old

Answer 2

Let the age of the boys be -

a1 , a2 , a3 , a4 & a5.

Their average age is given 16yrs.

That means ;

$\frac{a1 + a2 + a3 + a4 + a5}{5} = 16$

Now, we can simplify this to find sum of 5 ages.

$\frac{a1 + a2 + a3 + a4 + a5}{5} = 16 \\ \\ \\ \implies a1 + a2 + a3 + a4 + a5 = 16 \times 5 \\ \\ \\ \implies \boxed{a1 + a2 + a3 + a4 + a5 = 80} \: \: (eq.01)$

Now,

16yr 3month = 16 3/12 = 16 1/4yrs

Average of 4 boys is 16 1/4 yrs

$\frac{a1 + a2 + a3 + a4}{4} = 16 \frac{1}{4} \\ \\ \\ \implies a1 + a2 + a3 + a4 = \frac{65}{4} \times 4 \\ \\ \\ \implies \boxed{a1 + a2 + a3 + a4 = 65} \: \: (eq.02)$

Now age of the fifth boy can be found by subtracting eq. 02 from equation 01

$a1 + a2 + a3 + a4 + a5 - (a1 + a2 + a3 + a4 ) = 80 - 65 \\ \\ \\ \\ \implies a1 + a2 + a3 + a4 + a5 - a1 - a2 - a3 - a4 = 15 \\ \\ \\ \implies \boxed{a5 = 15}$

The age of the 5th boy is 15yrs.