The sum and products of two numbers are 12 and 35 respectively. The sum of their reciprocals will be:

1 $\frac{1}{3}$ $\frac{1}{3}$
2 $\frac{1}{5}$ $\frac{1}{5}$
3 $\frac{12}{35}$ $\frac{{12}}{{35}}$
4 $\frac{35}{12}$ $\frac{{35}}{{12}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let two numbers are x and y . \\ Now, Sum of the numbers, \\ x + y = 12 \\ Product of the number, \\ x y = 35 \\ Then sum of the reciprocals \\ = (\frac{1}{x}) + (\frac{1}{y}) \\ = \frac{(x + y)}{x y} \\ = \frac{12}{35} \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{two}}\,{\text{numbers}}\,{\text{are}}\,x\,{\text{and}}\,y. \cr & {\text{Now,}}\,{\text{Sum}}\,{\text{of}}\,{\text{the}}\,{\text{numbers}}, \cr & x + y = 12 \cr & {\text{Product}}\,{\text{of}}\,{\text{the}}\,{\text{number}}, \cr & xy = 35 \cr & {\text{Then}}\,{\text{sum}}\,{\text{of}}\,{\text{the}}\,{\text{reciprocals}} \cr & = \left( {\frac{1}{x}} \right) + \left( {\frac{1}{y}} \right)\, \cr & = \frac{{\left( {x + y} \right)}}{{xy}} \cr & = \frac{{12}}{{35}} \cr}$