Question 323608
Let the numbers = a, b, c, d, e, and f
given:
{{{(a+b+c+d+e+f)/6 = 10}}}
{{{(a+b+c+d)/4 = 12}}}
{{{a + b + c + d = 48}}}
by substitution:
{{{(48 + e + f)/6 = 10}}}
{{{48 + e + f = 60}}}
{{{e + f = 12}}}
{{{(e+f)/2 = 12/2}}}
{{{(e + f)/2 = 6}}}
the average of the remaining 2 numbers is 6