2000 AMC 8 Problems/Problem 23

Revision as of 19:57, 28 July 2011 by Mrdavid445 (talk | contribs) (Created page with "There is a list of seven numbers. The average of the first four numbers is <math>5</math>, and the average of the last four numbers is <math>8</math>. If the average of all seven...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

There is a list of seven numbers. The average of the first four numbers is $5$, and the average of the last four numbers is $8$. If the average of all seven numbers is $6\frac{4}{7}$, then the number common to both sets of four numbers is

$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\text{(D)}\ 7\qquad\text{(E)}\ 7\frac{3}{7}$