A) \[\frac{1}{2}\]
B) \[\frac{5}{12}\]
C) \[\frac{7}{12}\]
D) \[\frac{12}{5}\]
Correct Answer: B
Solution :
Let the numbers be x and y, respectively. |
\[x+y=10\] ?(i) |
\[xy=24\] |
\[{{(x-y)}^{2}}={{(x+y)}^{2}}-4xy\] |
\[=100-4\times 24=4\] |
\[\Rightarrow \] \[x-y=2\] ...(ii) |
On adding Eqs. (i) and (ii), we get |
\[x+y=10\] |
\[x-y=2\] |
\[2x=12\]\[\Rightarrow \]\[x=6\] |
From Eq. (i), y = 4 |
Sum of reciprocals \[=\frac{1}{6}+\frac{1}{4}=\frac{2+3}{12}=\frac{5}{12}\] |
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