A) 4
B) 5
C) 6
D) 7
Correct Answer: D
Solution :
Mean of 1, 3, 4, 5, 7 and 4 is m. \[\Rightarrow \] \[\frac{1+3+4+5+7+4}{6}=m\Rightarrow m=4\] Now, mean of 3, 2, 2, 4, 3, 3 and p is \[m-1\] \[\Rightarrow \] \[\frac{3+2+2+4+3+3+p}{7}=3\] \[(\because \,\,\,m=4)\] \[\Rightarrow \] \[17+p=21\,\,\,\Rightarrow \,\,\,p=4\] Arranging 3, 2, 2., 4, 3, 3, and 4 in ascending order, we get 2, 2.. 3., 3, 3. 4, 4 \[\therefore \]Median \[(q)={{\left( \frac{7+1}{2} \right)}^{m}}\] term - 4th term = 3 \[\therefore \] \[p+q=4+3=7\]You need to login to perform this action.
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