Average

a+b+c = 3*6=18

while any of the two numbers randomly selected have a sum of 10(i.e. 2*5)

Therefore the third number is 18-10=8

Let the 3 numbers be x , y and z. hence by the first condition..

x+y+z/3=6.......

and by second condition............

x+y/2=5

hence by substituting x+y=10 in the first equation the answer is 8...

Suppose the three numbers are a,b,c

You have that the average of 3 numbers = 6 You have that the average of 2 numbers = 5

in equations: (a+b+c)/3 = 6 (a+b)/2 = 5

Multiplying the first equation by 3 gives you
a+b+c = 18 and a+b = 18-c

substituting (a+b) in second equation (18-c)/2 = 5 18-c = 10 c=8