1 1 vote A shopkeeper sells two tables, each procured at cost price $p,$ to Amal and Asim at a profit of $20\%$ and at a loss of $20\%$, respectively. Amal sells his table to Bimal at a profit of $30\%$, while Asim sells his table to Barun at a loss of $30\%$. If the amounts paid by Bimal and Barun are $x$ and $y$, respectively, then $(x −y) / p$ equals $0.7$ $1$ $1.2$ $0.50$ Quantitative Aptitude cat2019-2 quantitative-aptitude profit-loss + – go_editor 14.2k points 2.5k views answer comment Share Follow Print See 1 comment 1 1 comment reply haralk10 838 points commented Apr 25, 2021 reply Follow flag option (B) 0 0 replyShare Please log in or register to add a comment.
1 1 vote Let the cost price of table $1=p=100$ $100 \overset{+20 \%} {\longrightarrow} 100 \times \frac{120}{100} = 120\;(\text{Amal}) \overset{+30\%}{\longrightarrow} 120 \times\frac{130}{100} = 156 = x \;(\text{Bimal)} $ Let the cost price of table $2 = p = 100$ $100 \overset {-20\%}{\longrightarrow} 100 \times \frac {80}{100}=80\; (\text{Asim}) \overset{-30\%}{\longrightarrow} 80 \times \frac{70}{100 } = 56 = y \;(\text{Barun})$ Therefore$, \frac{x-y}{p} = \frac{156-56}{100} = \frac{100}{100} = 1$ Correct Answer: B Anjana5051 answered Jul 28, 2021 • edited Jul 28, 2021 by Anjana5051 Anjana5051 12.1k points comment Share Follow 0 reply Please log in or register to add a comment.