A) 12,27
B) 24,15
C) 15,24
D) 27,12
Correct Answer: C
Solution :
Let the numbers be x and y. According to the equation, \[x+y=39\] ?(1) \[\frac{x}{5}=\frac{y}{8}\]\[=3{{a}^{2}}-29a+40\]\[y=\frac{8}{5}x\] From (1), \[-9,\frac{15}{4},\frac{-13}{3}\] \[\Rightarrow \]\[f(x)=a{{x}^{2}}+bx+c\ne 0\]\[3{{a}^{2}}-29a+40-{{a}^{2}}-13a-40=0\]\[x-y\] Again from (1), \[x+y=39\]\[\left( 3+\frac{5}{x} \right)\left( 9-\frac{15}{x}+\frac{25}{{{x}^{2}}} \right)\]\[y=39-15=24.\]
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