Question

A dealer bought two tables for Rs 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then, he found that each table was sold for the same price. Find the cost price of each table.

Answer 1

$$\begin{gathered} G\mathrm{iven}\mathrm{that}the\mathrm{selling}\mathrm{price}\mathrm{is}\mathrm{same}\mathrm{for}\mathrm{both}\mathrm{the}\mathrm{tables}. \\ \mathrm{Let}the\mathrm{C}.\mathrm{P}\mathrm{of}1\mathrm{table}\mathrm{be}\mathrm{Rs}.x,\mathrm{then}the\mathrm{C}.\mathrm{P}\mathrm{of}the\mathrm{other}willbe\mathrm{Rs}.\left(3120-x\right). \\ \mathrm{Los}son\mathrm{the}\mathrm{first}\mathrm{table}=15\% \\ \mathrm{Therefore},S.P=C.P\left(\frac{100-l\mathrm{oss}\%}{100}\right) \\ \mathrm{S}.\mathrm{P}=\frac{85x}{100}=\mathrm{Rs}.0.85x \\ \mathrm{Gain}on\mathrm{the}\mathrm{second}\mathrm{table}=36\% \\ \mathrm{Therefore},\mathrm{S}.\mathrm{P}=C.P\left(\frac{100+g\mathrm{ain}\%}{100}\right) \\ S.P=\mathrm{Rs}.1.36\left(3120-x\right) \\ \mathrm{Since}\mathrm{both}tableshavethesameS.P, \\ \mathrm{So},0.85\mathrm{x}=1.36\left(3120-x\right) \\ 0.85x=4243.20-1.36x \\ 2.21x=4243.20 \\ x=\frac{4243.20}{2.21} \\ x=\mathrm{Rs}.1920 \\ \mathrm{So},\mathrm{the}\mathrm{cost}\mathrm{price}\mathrm{of}the\mathrm{first}\mathrm{table}is\mathrm{Rs}.1920. \\ \mathrm{Cost}\mathrm{price}\mathrm{of}the\mathrm{second}\mathrm{table}=\mathrm{Rs}.(3120-1920)=\mathrm{Rs}.1200 \end{gathered}$$