A) 49
B) 48
C) 45
D) 40
Correct Answer: B
Solution :
Let remaining two variates are a and b then |
\[\frac{a+b+2+4+10+12+14}{7}=8\] And \[\frac{{{a}^{2}}+{{b}^{2}}+4+16+100+144+196}{7}-{{(8)}^{2}}=16\] |
\[\Rightarrow \] \[a+b=4\] and \[{{a}^{2}}+{{b}^{2}}=100\] \[\Rightarrow \]\[ab=\frac{{{(a+b)}^{2}}-({{a}^{2}}+{{b}^{2}})}{2}=\frac{196-100}{2}=48.\] |
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