A) 135
B) 240
C) 73
D) 106
Correct Answer: D
Solution :
Let the numbers be \[x\] and\[y\,\,\text{and}\]\[x>y\]. \[\therefore \] \[xy=45\] and \[x-y=4\] \[\therefore \] \[{{x}^{2}}+{{y}^{2}}={{(x-y)}^{2}}+2xy\] \[={{(4)}^{2}}+2\times 45=16+90\] \[=106\]You need to login to perform this action.
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