Question

Solve by cross multiplication. A part of a monthly hostel charge is fixed and remaining depends on the number of days one has taken food. When student A takes food for 20 days she has to pay Rs. 1000, whereas student B who takes food for 26 days pays Rs. 1180. Find the fixed charge and charge per day.

Answer 1

Dear student,

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{fixed}\mathrm{charges}=\mathrm{Rs}\mathrm{x} \\ \mathrm{Let}\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{food}\mathrm{per}\mathrm{day}=\mathrm{Rs}\mathrm{y} \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{given}\mathrm{conditions}: \\ \mathrm{x}+20\mathrm{y}=1000....\left(1\right) \\ \mathrm{x}+26\mathrm{y}=1080.....\left(2\right) \\ \mathrm{The}\mathrm{above}2\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}: \\ \mathrm{x}+20\mathrm{y}-1000=0....\left(3\right) \\ \mathrm{x}+26\mathrm{y}-1180=0.....\left(4\right) \\ \mathrm{By}\mathrm{cross}\mathrm{multiplication}: \\ \frac{\mathrm{x}}{20\times \left(-1180\right)-\left(26\right)\times \left(-1000\right)}=\frac{\mathrm{y}}{-1000\times 1-\left(-1180\right)}=\frac{1}{26-20} \\ \Rightarrow \frac{\mathrm{x}}{-23600+26000}=\frac{\mathrm{y}}{-1000+1180}=\frac{1}{6} \\ \Rightarrow \frac{\mathrm{x}}{2400}=\frac{\mathrm{y}}{180}=\frac{1}{6} \\ \Rightarrow \frac{\mathrm{x}}{2400}=\frac{1}{6}\mathrm{and}\frac{\mathrm{y}}{180}=\frac{1}{6} \\ \Rightarrow \mathrm{x}=400\mathrm{and}\mathrm{y}=30 \\ \mathrm{So},\mathrm{fixed}\mathrm{charges}=\mathrm{Rs}400 \\ \mathrm{Cost}\mathrm{of}\mathrm{food}\mathrm{per}\mathrm{day}=\mathrm{Rs}30 \end{gathered}$$

Regards