375.

A shopkeeper buys some books for Rs. 80. If he had bought 4 more books for the same amount, each book would have cost Rs. 1 less. Find the number of books he bought.

                                            OR

The sum of two number is 9 and the sum of their reciprocals is $\frac{1}{2}$. Find the numbers.

Total cost of books = Rs.80

Let the number of books = x

So the cost of each book = Rs $\frac{80}{\mathrm{x}}$

Cost of each book if he buy 4 more books = Rs $\frac{80}{\mathrm{x} + 4}$

As per given in equations:

$$\begin{gathered} \frac{80}{\mathrm{x}} - \frac{80}{\mathrm{x} + 4} = 1 \\ \Rightarrow \frac{80\mathrm{x} + 320 - 80\mathrm{x}}{\mathrm{x}( \mathrm{x} + 4)} = 1 \\ \Rightarrow \frac{{\displaystyle 320}}{{\displaystyle \mathrm{x}{ }^2+ 4\mathrm{x}}} = 1 \\ \Rightarrow {\mathrm{x}}^2 + 4\mathrm{x} - 320 = 0 \\ \Rightarrow (\mathrm{x} + 20) ( \mathrm{x}- 16) = 0 \\ \Rightarrow \mathrm{x} = -20 \mathrm{or} \mathrm{x} = 16 \end{gathered}$$

Since number of books cannot be negative,

so the number of books he brought is 16.

                                   OR

Let the frist number be x then the second number be 9 - x as the sum of both  number is 9.

Now the sum of their reciprocal is $\frac{1}{2}$, therefor

 $$\begin{gathered} = \frac{1}{\mathrm{x}} + \frac{1}{9 - \mathrm{x}} = \frac{1}{2} \\ \Rightarrow \frac{9 - \mathrm{x} + \mathrm{x}}{\mathrm{x}( 9 - \mathrm{x} )} = \frac{1}{2} \\ \Rightarrow \frac{9}{9\mathrm{x} - {\mathrm{x}}^2} = \frac{{\displaystyle 1}}{{\displaystyle 2}} \\ \Rightarrow 18= 9\mathrm{x} - {\mathrm{x}}^2 \\ \Rightarrow {\mathrm{x}}^2 - 9\mathrm{x} + 18 = 0 \\ \Rightarrow ( \mathrm{x}- 6 ) ( \mathrm{x} - 3 ) = 0 \\ \Rightarrow \mathrm{x} = 6 \mathrm{or} \mathrm{x} = 3 \end{gathered}$$

If  x = 6 then other number is 3.

And, if x = 3 then other number is 6.

Hence numbers are 3 and 6.