Profit and loss - using false weights 
A shopkeeper cheats to the extent of 10% while buying and selling by
  using false weights. His total gain is

This looked as a simple question to me. But when I searched for answers, i could see there are two answers for this question. Majority of sites give the answer as $21\%$ where as some show as $22\dfrac{2}{9}\%$. Hence the doubt came

Source 1
Gain % = 21%. (They use a general formula, not clear how this formula
  is derived)
( This website also shows the same answer - see question 3)

.

Source 2
Here, the answer is given as $22\dfrac{2}{9}\%$
The explanation looks convincing to me. 

Majority of sites show 21% as I explained in first source where the second source shows $22\dfrac{2}{9}\%$. Could anyone help me in deciding the right method and right answer. 
Note: By looking into the explanation given in source2, I feel right answer is $22\dfrac{2}{9}\%$. But, then the first source uses formula which I am not clear and they show answer as 21%. Hence the confusion is. There can be many questions for my exam using the same pattern and it is important for me to be clear on this.
 A: "Cheats to the extent of 10%" is ambiguous. It can either mean that the error is 10% of the true weight of the goods, or that the error is 10% of weight that the shopkeeper pays/charges for.
These two interpretations lead to different results.
Suppose that he buys and sells one kilogram of goods at a market price of \$1000 per kilogram.
Under the first interpretation he would pay his supplier for 900 grams of goods but charge his customer for 1100 grams, making a profit of \$200.
Under the second interpretation he would pay his supplier for (approximately) 909 grams of goods and charge his customer for 1111 grams, making a profit of \$202.
An additional ambiguity creeps in when we try to express the profits \$200 and \$202 as percentages of something -- percentages of what? The merchant's initial outlay? The true value of the kilogram of goods? The retail price? Each of those choices, too, yield different answers.
Thus I think at least the following answers could all be justified:
$$ \frac{200}{1000} = 20\% \qquad \frac{200}{900} = 22.2\% \\[2em]
\frac{202}{1000} = 20.2\% \qquad \frac{202}{909} = 22.2\% $$

We would get 21% by saying that the shopkeeper earns a profit of 10% when buying and again when selling -- and $1.10\times 1.10 = 1.21$ -- but the total profit in dollars doesn't appear to be 21% of any relevant amount that could possibly arise during the transaction.
In order to get 21% we would need to say that "cheats to extent of 10%" means that the error in the shopkeeper's measurements is 10% of the lower of the true weight and what the shopkeeper says the weight is. Then he would buy for \$909 and sell for \$1100. But that would be a very artificial and unlikely interpretation of "cheating by 10%".
A: Suppose that he was supposed to pay $100$. Since he cheated, he paid $90$. When he sold, he charged $110$. So, the profit was $20$. In terms of percent $$\frac{20} {90}=22\dfrac{2}{9}\%$$
Do you agree with me ? I must confess that I am not a very good business man !
A: Source 1 appears to assume that the shopkeeper buys $110$ units of stuff for the price of $100$ units. He could do this by using weights that
are 10 percent heavier than he tells the seller they are.
That does seem to be one way to "cheat to the extent of 10%".
Since the shopkeeper thereby gets $1.1$ units of goods for the price of $1$ unit, the cost to him for each unit is $\frac{1}{1.1}$ of the honest cost.
But what if, when the shopkeeper tells the seller how heavy his weights are,
he names a weight 10 percent lighter than the true value.
Then when he buys $100$ units of goods, he pays for only $90$ units,
so the cost for each unit is $0.9$ of the honest cost.
Since $0.9 < \frac{1}{1.1}$, the second cheat allows the shopkeeper
to pay less for his goods, and it seems therefore that a
clever cheating shopkeeper would choose the second cheat if he can.
So it comes down to how you interpret "cheats to the extent of 10%",
and whether this allows the second cheat. Since none of the sources
are very explicit about exactly how the cheating is done,
I prefer to think that the second cheat is possible and that
the answer $22 \tfrac29\%$ is therefore correct.
A: Lets assume he had 100 Rupees.
By cheating 10% he got 110 worth of material. 
Now again he cheats by 10%. when selling 110 worth of material.
10% more of 110 is 121. Hence finally he is getting Rs 121 for the cost of 100.
G is 21 and G% is (21/100)*100 = 21%.  Hence finally his gain percent is 21%
