A) 9, 11
B) 11, 13
C) 11, 19
D) 9, 13
Correct Answer: C
Solution :
Let one number \[y=2-\frac{3}{2}=\frac{1}{2}.\] Other number\[=2x-3\] \[=\frac{(x+y)({{x}^{2}}+{{y}^{2}}-xy)}{{{(xy)}^{3}}}=\frac{(x+y)\left[ {{(x+y)}^{2}}-3xy \right]}{{{(xy)}^{3}}}\] \[=\frac{a\left[ {{(a)}^{2}}-3(b) \right]}{{{(b)}^{3}}}\] \[\because \] \[x+y=a\] \[\therefore \]The two numbers are 11 and 19.
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