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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

  1. $0.7$
  2. $1$
  3. $1.2$
  4. $0.50$
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option (B)
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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

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