Question

a certain number of articles are purchased for Rs.1200. If price per article is increased by Rs. 5, then 20 less article can be purchased. what is the original number of articles​

Answer 1

Dear student,

We can solve this question as follows.

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{each}\mathrm{article}=\mathrm{Rs}\mathrm{x} \\ \mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{articles}\mathrm{bought}\mathrm{for}\mathrm{Rs}1200=\mathrm{y} \\ \mathrm{So},\mathrm{xy}=1200 \\ \Rightarrow \mathrm{y}=\frac{1200}{\mathrm{x}}....\left(1\right) \\ \mathrm{Increased}\mathrm{cost}\mathrm{of}1\mathrm{article}=\mathrm{Rs}\left(\mathrm{x}+5\right) \\ \mathrm{So},\mathrm{number}\mathrm{of}\mathrm{articles}\mathrm{bought}\mathrm{for}\mathrm{Rs}1200=\mathrm{y}-20 \\ \mathrm{So},\mathrm{y}-20=\frac{1200}{\mathrm{x}+5} \\ \Rightarrow \mathrm{y}=\frac{1200}{\mathrm{x}+5}+20...\left(2\right) \\ \mathrm{So},\mathrm{from}\left(1\right)\mathrm{and}\left(2\right) \\ \frac{1200}{\mathrm{x}}=\frac{1200}{\mathrm{x}+5}+20 \\ \Rightarrow \frac{1200}{\mathrm{x}}-\frac{1200}{\mathrm{x}+5}=20 \\ \Rightarrow \frac{\mathrm{x}+5-\mathrm{x}}{\mathrm{x}\left(\mathrm{x}+5\right)}=\frac{1}{60} \\ \Rightarrow {\mathrm{x}}^2+5\mathrm{x}-300=0 \\ \Rightarrow {\mathrm{x}}^2+20\mathrm{x}-15\mathrm{x}-300=0 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}+20\right)-15\left(\mathrm{x}+20\right)=0 \\ \Rightarrow \left(\mathrm{x}-15\right)\left(\mathrm{x}+20\right)=0 \\ \Rightarrow \mathrm{x}=15\mathrm{or}\mathrm{x}=-20\left(\mathrm{rejected}\right) \\ \Rightarrow \mathrm{cost}\mathrm{of}1\mathrm{article}=\mathrm{Rs}15 \\ \mathrm{Number}\mathrm{of}\mathrm{articles}\mathrm{bought}=\mathrm{y}=\frac{1200}{15}=80 \end{gathered}$$
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