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Three grades of 55+5/9 and one grade of 100 produce an average of 66+2/3.
Let a,b,c, and d be the grades with average v and 0<=a<=b<=c<=d<=100. Now 4v=a+b+c+d --> 4v/(b+c)=1+(a+d)/(b+c)=2.4 so
b+c = 4v/2.4 (equation 1)
Since a+d+b+c=4v, a+d=4v-(b+c)=4v-4v/2.4=7v/3. So
a+d = 7v/3 (equation 2)
Now d <= 100 --> 7v/3 - a <= 100 --> a >= 7v/3 - 100 and since b >= a, it follows that b >= 7v/3 - 100. So by equation 1, 4v/2.4 - c >= 7v/3 - 100. Solving this inequality for v yields
v <= 1.5(100-c) (inequality 1)
Assume there exist grades such that v > 66+2/3. Then 4v > 266+2/3 so 4v/2.4 > 111+1/9 so by equation 1, b+c > 111+1/9. Since c >= b it follows that c > 55+5/9 --> 100-c < 44+4/9 --> 1.5(100-c) < 66+2/3 so by inequality 1, v < 66+2/3 contradicting our assumption. |
