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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 $40\%$
in Numerical Ability by Veteran (52.8k points)
edited by | 352 views
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cheats to the extent of 20%, what does it mean?
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I'm also stuck with this.
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Yeah  Lakshman, the wordings are bit confusing

However, i add the answer, if  you're still stuck, then let me know.

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Please explain question,i'm confuse in question

1 Answer

+4 votes
Best answer
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\%} $.
by Boss (15.2k points)
selected by
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No

This is wrong

Please see this 27:52

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Now check
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Can I solve this question using effective percentage??
+1
$\text{We know that }$ $\color{maroon}{Profit\% = \dfrac{(100+\text{common gain%})^2}{100}-100}$
$\qquad \qquad  \qquad \qquad \qquad= \dfrac{(100+20)^2}{100} - 100$

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

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

$\qquad \qquad \qquad \qquad \qquad= \color{green}{44\%} $.
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@Sukanya @Lakshman , Could u please explain what is d meaning of effective percentage ?
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effective % means, I think % in total

I think it is nothing but formula one he asked for
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I am getting 45%
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45% is not the answer.
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See,

when cost of buying 80 rs. the selling price 100 rs

                             1 rs.     the selling price $\frac{100}{80}$ rs.

                            100 rs.   "       "        "    $\frac{100}{80}\times 100$=125rs.

So, in time of buying profit is (125-100)=25rs.

Now, in time of selling in 120rs when buying cost 100 rs.

So, at time of selling they gain 20 rs.

So, total profit 20+25=45%

Am I doing any mistake ?

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@srestha 44% (Gain)  is the right answer

+2

Effective %  = $a + b + \frac{a*b}{100}$

Gain %  =  20 +  20 + $\frac{20*20}{100}$

             =  40  + 4

             = 44%

So the answer is (B)

Please see videos in 27:52 I already upload in above comment section.

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ok, understood

But still want to know, where is my mistake?

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