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17 Jul 2009, 22:42
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N
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C is the finite sequence \(C_1\)=0,\(C_2\)=\(\frac{1}{2}\),\(C_3\)=\(\frac{2}{3}\)...determined by the equation \(C_f\)=\(\frac{f-1}{f}\)where f is a positive integer.D is a similiar finite sequence determined by the equation \(D_g\) = \(\frac{g}{g+1}\) where g is a positive integer,Is the sum of allthe values in C equal to the sum of all the values in D?
1.g \(\neq\)f
2.The difference between the median of C and the median of D is \(\frac{3}{43}\)
Phew this took a while to type!Anybody with a detailed explanation?
1.g \(\neq\)f
2.The difference between the median of C and the median of D is \(\frac{3}{43}\)
Ans:E
Phew this took a while to type!Anybody with a detailed explanation?
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19 Jul 2009, 00:24
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B for me too.
Let's look at the sets when sum of all the values of C is equal to that of D
C: \(0\); \(\frac{1}{2}\); \(\frac{2}{3}\); \(\frac{3}{4}\);
D: \(\frac{1}{2}\); \(\frac{2}{3}\); \(\frac{3}{4}\);
as we can see, f=g+1 that leads to the fact that median of C is less than median of D (as C has more members that D). Let's add one member to each set:
C: \(0\); \(\frac{1}{2}\); \(\frac{2}{3}\); \(\frac{3}{4}\); \(\frac{4}{5}\);
D: \(\frac{1}{2}\); \(\frac{2}{3}\); \(\frac{3}{4}\); \(\frac{4}{5}\);
again, median of C is less than median of D, and it will continue as long as we increase number of members in both sets.
So, if sum of all values of C is equal to that of D, then the difference between the median of C and the median of D is negative. Hence, B is sufficient.
Let's look at the sets when sum of all the values of C is equal to that of D
C: \(0\); \(\frac{1}{2}\); \(\frac{2}{3}\); \(\frac{3}{4}\);
D: \(\frac{1}{2}\); \(\frac{2}{3}\); \(\frac{3}{4}\);
as we can see, f=g+1 that leads to the fact that median of C is less than median of D (as C has more members that D). Let's add one member to each set:
C: \(0\); \(\frac{1}{2}\); \(\frac{2}{3}\); \(\frac{3}{4}\); \(\frac{4}{5}\);
D: \(\frac{1}{2}\); \(\frac{2}{3}\); \(\frac{3}{4}\); \(\frac{4}{5}\);
again, median of C is less than median of D, and it will continue as long as we increase number of members in both sets.
So, if sum of all values of C is equal to that of D, then the difference between the median of C and the median of D is negative. Hence, B is sufficient.
