 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Contestants in a Circle (Posted on 2011-01-31) The eight contestants were seated at a circular table in alphabetic order clockwise: Alice, Bob, Carol, Dave, Eve, Frank, Gisele and Harry. Some points had already been awarded by the judges, in varying amounts to the different competitors, but the competitors didn't know what their totals were. As a last competition, they were to deduce all their scores thus far based upon the announcement of eight totals of three contestants' scores, each total being the sum of three successive people around the table, so for example, the total for Harry, Alice and Bob (the triplet centered on Alice) was given as 38.

The successive triplets centered on Bob through Harry totaled as follows: 49, 45, 48, 52, 44, 39 and 48.

What were in fact the individual scores to that point in the competition?

 See The Solution Submitted by Charlie No Rating Comments: ( Back to comment list | You must be logged in to post comments.) easy way out | Comment 2 of 3 | Equations:

a+b+c=49,==>  A

b+c+d=45,==>  B

c+d+e=48,==>  C

d+e+f=52,==>  D

e+f+g=44,==>  E

f+g+h=39, ==>  F

g+h+a=48,==>  G

h+a+b=38:==>  H

Solution:  TOTAL= 1/3*SUM=121

A+D+G-TOTAL=a=149-121=28

B+E+H-TOTAL=b=127-121=6

c=49-28-6=15

d=45-15-6=24

e=48-24-15=9

f=52-9-24=19

g=44-19-9=16

h=39-16-19=4

Edited on January 31, 2011, 6:23 pm
 Posted by Ady TZIDON on 2011-01-31 18:21:11 Please log in:

 Search: Search body:
Forums (1)