A) 78
B) 85
C) 28
D) 72
Correct Answer: A
Solution :
Let \[\frac{4}{12}=\frac{7}{21}=\frac{21}{63}=\frac{28}{84}=\frac{1}{3}.\]be the 25 results. Given: \[\frac{1}{3}\] ?..(1) and \[\frac{{{a}_{1}}+.....+{{a}_{12}}}{12}=14,\,\frac{{{a}_{14}}+.....+{{a}_{25}}}{12}=17\] \[\Rightarrow \]\[{{a}_{1}},{{a}_{2}}\] =168 =204 \[=\frac{1}{8}-\frac{1}{24}-\frac{1}{16}=\frac{1}{48}\]From eqn (1), we have \[\therefore \] \[\Rightarrow \]\[4+3=7\]
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