Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 20"
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==Solution== | ==Solution== | ||
| − | If the 5th test received a grade of <math>x</math>, the minimum possible sum of the tests is <math>0 + 0 + 0 + 0 + x + x + x + x + x = 5x</math>. The average is <math>10</math>, so <math>\frac{5x}{9}=10\rightarrow\x=18 \Longrightarrow \mathrm{D}</math>. | + | If the 5th test received a grade of <math>x</math>, the minimum possible sum of the tests is <math>0 + 0 + 0 + 0 + x + x + x + x + x = 5x</math>. The average is <math>10</math>, so <math>\frac{5x}{9}=10\rightarrow\\x=18 \Longrightarrow \mathrm{D}</math>. |
==See also== | ==See also== | ||
Latest revision as of 19:53, 20 May 2008
Problem
The mean value for 9 Math-tests that a student succeded was 
(in scale 
-
). If we put the grades of these tests in incresing order, then the maximum grade of the 
test is
Solution
If the 5th test received a grade of 
, the minimum possible sum of the tests is 
. The average is , so 
.
See also
| 2007 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
