A problem by Princess Vanshika

Being The Average of 11 Numbers : $\frac{a+b+c+d+e+f+g+h+i+j+k}{11} = 60$

$a+b+c+d+e+f+g+h+i+j+k = 660$

Being The Average of 6 First Numbers : $\frac{a+b+c+d+e+f}{6} = 58$

$a+b+c+d+e+f = 348$

Being The Average of 6 Last Numbers : $\frac{f+g+h+i+j+k}{6} = 63$

$f+g+h+i+j+k = 378$

Adding the Two Sequences :

$\Rightarrow a+b+c+d+e+f+g+h+i+j+k+f= 726$

We know already that $a+b+c+d+e+f+g+h+i+j+k = 660$

Thus We Have : $660+f=726 \Rightarrow f=66$

We get three equations from this problem. Let the numbers be $a_{1},a_{2},a_{3},.....,a_{10}$ and $a_{11}$ . Then,

$\frac{a_{1}+a_{2}+.....+a_{11}}{11}=60 \implies a_{1}+a_{2}+.....+a_{11}=660$ .....(i)

$\frac{a_{1}+a_{2}+.....+a_{6}}{6}=58 \implies a_{1}+a_{2}+.....+a_{6}=348$ .....(ii)

$\frac{(a_{6}+a_{7}+.....+a_{11}}{6}=63 \implies a_{6}+a_{7}+.....+a_{11}=378$ .....(iii)

Adding (ii) and (iii), we get---

$(a_{1}+a_{2}+.....+a_{6})+(a_{6}+a{7}+....+a_{11})=726$

$\implies (a_{1}+a_{2}+.....+a_{11})+a_{6}=726$

$\implies 660+a_{6}=726$ [From (i)] $\implies a_{6}=\boxed{66}$

If the average of the eleven numbers is 60 then the total of them is 660, 60x11. If the average of the first six is 58 then the total of those six is 348. If the average of the last six is 63 the total is 378. Now you add 348 and 378 to get 726. 726=number 1+number 2...+2number 6..., 660=number 1+number 2+ number 3... You have all the same numbers except the first equation has one additional number 6 so we can find the value of number 6 if we subtract 726-660 which ends up being 66.