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A Shopkeeper cheats to the extends of $20\%$ while buying as well as selling.By using false weight.His gain or loss is?

  1. Loss $50\%$
  2. Gain $44\%$  
  3. Loss $44\%$
  4. Gain $40\%$
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The question is. A shopkeeper cheats to the extents of 20% while buying as well as selling, by using false weight.

Now, $\color{violet}{\text{what is meant by cheats to the extent?}}$

$\color{purple}{\text{Cheats to the extent means He can maximum cheats up to 20%,}}$ i.e

If he actually want to buy $120 gms$ , then he pays for only $100 gms$ i.e. $100 rupees$. $\big[{\text{we're taking 1rs/gm }}\big]$

Because his faulty weight shows $20\%$ increase in weight $\Rightarrow$ if anyone put $100 gms$ on his balance, it will show $120 gms$ [As $100 \times 20\% = 20 , ∴ 100+20 = 120 gm$]

Now, when he sells then also he cheats.

When he sells, that $120 gm$ thing will show up $120 \times 20\% = 24, ∴ 120+24 = 144 gm$], so he get $144 rs$ for $144 gm$.,

But actually it weighs $120 gms$, & the shopkeeper only pays $100$ rupees for it, but when he sell he gets $144 rs$.

$\color{green}{\text{So, There will be a profit%  i.e}}$ $\color{orange}{\dfrac{(144-100)}{100}\% = 44\%}$

$\color{blue}{\text{Using Formula:}}$

$\text{We know that }$ $\color{maroon}{Profit\% = \dfrac{(100+\text{common gain%})^2}{100}-100}$
$\qquad \qquad  \qquad \qquad \qquad= \dfrac{(100+20)^2}{100} - 100$

$\qquad \qquad  \qquad \qquad \qquad= \dfrac{120 \times 120}{100} - 100$

$\qquad \qquad \qquad \qquad \qquad= 144 - 100$

$\qquad \qquad \qquad \qquad \qquad= \color{green}{44\%} $.
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