The average of 7 numbers in a certain list is 12. The average of the 4 smallest numbers in this list is 8, while the average of the 4 greatest...

Tuesday, October 29, 2013

The average of 7 numbers in a certain list is 12. The average of the 4 smallest numbers in this list is 8, while the average of the 4 greatest...

Denote the numbers as $\displaystyle{a}_{{1}}\leq{a}_{{2}}\leq{a}_{{3}}\leq{a}_{{4}}\leq{a}_{{5}}\leq{a}_{{6}}\leq{a}_{{7}}.$ It is given that:

$\displaystyle\frac{{{a}_{{1}}+{a}_{{2}}+{a}_{{3}}+{a}_{{4}}+{a}_{{5}}+{a}_{{6}}+{a}_{{7}}}}{{7}}={12},$

$\displaystyle\frac{{{a}_{{1}}+{a}_{{2}}+{a}_{{3}}+{a}_{{4}}}}{{4}}={8},$

$\displaystyle\frac{{{a}_{{4}}+{a}_{{5}}+{a}_{{6}}+{a}_{{7}}}}{{4}}={20}.$

From these equation we can find the sum of the $\displaystyle{3}$ greatest numbers and the sum of the $\displaystyle{3}$ smallest numbers:

$\displaystyle{a}_{{1}}+{a}_{{2}}+{a}_{{3}}={\left({a}_{{1}}+{a}_{{2}}+{a}_{{3}}+{a}_{{4}}+{a}_{{5}}+{a}_{{6}}+{a}_{{7}}\right)}$

$\displaystyle-{\left({a}_{{4}}+{a}_{{5}}+{a}_{{6}}+{a}_{{7}}\right)}={12}\cdot{7}-{20}\cdot{4}={4},$

$\displaystyle{a}_{{5}}+{a}_{{6}}+{a}_{{7}}={\left({a}_{{1}}+{a}_{{2}}+{a}_{{3}}+{a}_{{4}}+{a}_{{5}}+{a}_{{6}}+{a}_{{7}}\right)}$

$\displaystyle-{\left({a}_{{1}}+{a}_{{2}}+{a}_{{3}}+{a}_{{4}}\right)}={12}\cdot{7}-{8}\cdot{4}={52}.$

The difference in question is 52 - 4 = 48.

But actually these conditions are contradictory. It is simple to find $\displaystyle{a}_{{4}},$ it is

$\displaystyle{\left({a}_{{4}}+{a}_{{5}}+{a}_{{6}}+{a}_{{7}}\right)}-{\left({a}_{{5}}+{a}_{{6}}+{a}_{{7}}\right)}={80}-{52}={28}.$

But all next numbers, $\displaystyle{a}_{{5}},$ $\displaystyle{a}_{{6}}$ and $\displaystyle{a}_{{7}},$ must be at least $\displaystyle{28},$ therefore its sum is at least $\displaystyle{84},$ not $\displaystyle{52}.$ So the correct answer is "this is impossible".