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 \[=\frac{1}{6}+\frac{1}{4}=\frac{2+3}{12}=\frac{5}{12}\] |
You need to login to perform this action.
You will be redirected in
3 sec
