Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 20"

Latest revision as of 19:53, 20 May 2008

Problem

The mean value for 9 Math-tests that a student succeded was $10$ (in scale $0$ - $20$ ). If we put the grades of these tests in incresing order, then the maximum grade of the $5^{th}$ test is

$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 16\qquad \mathrm{(C) \ } 17\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } 19$

Solution

If the 5th test received a grade of $x$ , the minimum possible sum of the tests is $0 + 0 + 0 + 0 + x + x + x + x + x = 5x$ . The average is $10$ , so $\frac{5x}{9}=10\rightarrow\\x=18 \Longrightarrow \mathrm{D}$ .

@@ Line 5: / Line 5: @@
 ==Solution==
-If the 5th test received a grade of <math>x</math>, the minimum possible sum of the tests is <math>\displaystyle 0 + 0 + 0 + 0 + x + x + x + x + x + x = 6x</math>. The average is <math>\frac{6}{9}x = \frac 23x \le 10</math>, so <math>x \le 15 \Longrightarrow \mathrm{A}</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==

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

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Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 20"

Latest revision as of 19:53, 20 May 2008

Problem

Solution

See also