A) 69
B) 80
C) 81
D) 77
Correct Answer: A
Solution :
(a) Since \[{{x}^{2}}+{{y}^{2}}<25\] and \[x\] and \[y\] are integers, the possible value of \[x\] and \[y\in \left( 0,\pm 1,\pm 2,\pm 3,\pm 4 \right).\]thus \[x\] and can be chosen in 9 ways each and \[\left( x,y \right)\] can be chosen in \[9\times 9=81\] ways. However we have to exclude cases \[\left( \pm 3,\pm 4 \right)\],\[\left( \pm 4,\pm 3 \right)\] and \[\left( \pm 4,\pm 4 \right)\]\[\left( i.e. \right)\]\[3\times 4=12\] cases. Hence, the number of permissible values \[=81-12=69\]
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