A) 5, 3
B) 7, 1
C) 4, 4
D) 2, 6
Correct Answer: A
Solution :
Let the two numbers be x and y. According to the given conditions, \[x+y=8~\] ....(1) and \[\frac{1}{x}+\frac{1}{y}=\frac{8}{15}\] ....(2) Putting value of \[x=8-y\]in (2), we get \[\frac{1}{8-y}+\frac{1}{y}=\frac{8}{15}\] \[\Rightarrow \] \[\frac{y+8-y}{y(8-y)}=\frac{8}{15}\] \[\Rightarrow \]\[{{y}^{2}}-8y+15=0\]\[\Rightarrow \] \[{{y}^{2}}-5y-3y+15=0\] \[\Rightarrow \]\[(y-5)\,(y-3)=0\] \[\Rightarrow \]\[y=5\] or \[y=3\] From (1), \[x=3\]or \[x=5\] Thus, the numbers are 5 and 3.
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