## Concept 1: Basic Formulae

If a dishonest dealer professes to sell his goods at cost price, but uses false weight, then :

Gain % = $\frac{(True \hspace{1ex} weight − False \hspace{1ex} weight)}{(False \hspace{1ex} weight)}$ × 100%

or

Gain % = $\frac{Error}{(True \hspace{1ex} value − error)}$ × 100%

Note: If the above formula evaluates to a negative value, then it means that there is a net loss.
• In terms of Multiplying factor:
The Multiplying factor because of cheating on volume = $\frac{Amount \hspace{1ex} charged \hspace{1ex} for}{Actual \hspace{1ex} amount \hspace{1ex} sold}$
(Amount charged for = Reading of the weight scale)
##### Q. A shopkeeper sold an article at cost price, but he uses a weight of 900 gm in place of 1 kg. What must be his profit percentage?

Explanations :

Let quantity purchased be 1 kg for Rs. 1000.

The shopkeeper sold the article at cost price. But since he sold only 900 gm, the 100 gm that he saved will be his profit.

So, C.P. = Rs. 900 and S.P. = Rs. 1000

Profit = S.P. – C.P. = 1000 – 900 = Rs. 100

So, profit percentage = (profit / C.P.) × 100 = (100 / 900) × 100 = (1 / 9) × 100 = 11.11%

Explanation 2: Percentage Method

Shopkeeper uses a weight of 900 gm in place of 1 kg, i.e. a reduction of 10%.

Now, we need to get back to 100. Difference remains the same, i.e. 10, but now the base will be 90.

The corresponding profit percentage = (10/90) × 100 = 11.11%

Explanation 3: Fraction Method

Shopkeeper uses a weight of 900 gm in place of 1 kg, i.e. a reduction of 10%.

10% = 1/10

Difference remains the same, but new base = 10 – 1 = 9

So, profit percentage = (1/9) × 100 = 11.11%

Explanation 4: Formula Method

If a dishonest dealer professes to sell his goods at cost price, but uses false weight, then :

Gain percentage = $\frac{(True \hspace{1ex} weight − False \hspace{1ex} weight)}{(False \hspace{1ex} weight)}$ × 100% = (1000 −900)/900 × 100% = 100/900 × 100% = 11.11%

Explanation 5: Multiplying Factor Method

Let quantity purchased be 1 kg.

But since he sold only 900 gm, the 100 gm that he saved will be his profit.

The Multiplying factor because of cheating on volume = $\frac{Amount \hspace{1ex} charged \hspace{1ex} for}{Actual \hspace{1ex} amount \hspace{1ex} sold}$ = 1000 / 900 = 10/9 = 1.1111

So, profit percentage = (1.1111 – 1) × 100 = 11.11%

## Concept 2

If a dishonest trader sells his goods at x% profit or loss on cost price and uses y% less weight, then his profit percentage or loss percentage will be:

$\frac{(𝑦±𝑥)}{(100−𝑦)}$ × 100%

(Use + sign if goods are sold at x% profit and – sign if they are sold at x% loss.)

## Concept 3

If a dishonest trader uses faulty weight of w units instead of r units and claims to make a profit of p%, then:

Real net profit percentage = $\frac{100 (r - w) + rp}{w}$

(in case the trader claims to make a loss, we will use - sign for p)

Note: if the formula evaluates to a negative value, it means there is a net loss.
##### Q. A shopkeeper seems to sell an article at a profit of 25%. But, he uses a weight which is in reality 20% less than the weight indicated on it. What must be his real profit percentage?

Explanations :

Let quantity purchased be 1 kg for Rs. 1000.

But since he sold only 800 gm, the 200 gm that he saved will be his profit too.

So, C.P. = Rs. 800 and S.P. = Rs. 1000

Profit = S.P. – C.P. = 1000 – 800 = Rs. 200

So, profit percentage = (profit / C.P.) × 100 = (200 / 800) × 100 = (1 / 4) × 100 = 25%

Required profit percentage = 25 + 25 + (25 × 25)/100 = 50 + 6.25 = 56.25% (here we have just applied the formula of successive percentages)

Explanation 2: Percentage Method & Fraction Method

• Percentage Method

Shopkeeper seems to sell an article at a profit of 25%. (this is first profit)

But, he uses a weight which is in reality 20% less than the weight indicated on it.

Now, we need to get back to 100. Difference remains the same, i.e. 20, but now the base will be 80.

The corresponding profit percentage = (20/80) × 100 = 25%

So, 20% less weight is equivalent to 25% profit. (this is second profit)

Required profit percentage = Resultant of two profits = 25 + 25 + (25 × 25)/100 = 50 + 6.25 = 56.25%
(here we have just applied the formula of successive percentages)

• Fraction Method

Shopkeeper seems to sell an article at a profit of 25%. (this is first profit)

But, he uses a weight which is in reality 20% less than the weight indicated on it.

20% = 1/5

Difference remains the same, but new base = 5 – 1 = 4

So, profit percentage = (1/4) × 100 = 25% (this is second profit)

Required profit percentage = Resultant of two profits = 25 + 25 + (25 × 25)/100 = 50 + 6.25 = 56.25%
(here we have just applied the formula of successive percentages)

Explanation 3: Formula Method

• Using formula 1

If a dishonest trader sells his goods at x% profit or loss on cost price and uses y% less weight, then his profit percentage or loss percentage will be = $\frac{(𝑦±𝑥)}{(100−𝑦)}$ × 100%

= ((20 + 25))/((100 −20)) × 100% = 45/80 × 100% = 9/16 × 100% = 56.25%

• Using formula 2

If a dishonest trader uses faulty weight of w units instead of r units and claims to make a profit of p%, then:

Real net profit percentage = $\frac{100 (r - w) + rp}{w}$

= [100 (100 - 80) + 100 × 25] / 80 = [2000 + 2500] / 80 = 4500/80 = 450/8 = 56.25%

Explanation 4: Multiplying Factor Method

Let quantity purchased be 1 kg for Rs. 1000.

But since he sold only 800 gm, the 200 gm that he saved will be his profit.

C.P. of 800 gm = Rs. 800 and S.P. = 1000 + 25% of 1000 = 1000 + 250 = Rs. 1250

The Multiplying factor because of cheating on volume = S.P./C.P. = 1250 / 800 = 125/80 = 1.5625

So, profit percentage = (1.5625 – 1) × 100 = 56.25%

Explanation 5: Short Trick Method

Let C.P. = Rs. 100

Profit percentage = [(125 - 80) / 80] × 100% = (45/80) × 100% = 56.25%

The approaches and concepts discussed in this article are not limited to only problems with faulty weights, but in any problem where the amount sold and that charged for is different.

## Some More Questions

##### Q. Which of the following trick would maximize the profit of a businessman, when he sells a certain item?

I. Sell that item at 8% profit
II. Use an incorrect weight which is 100 grams less than 1 kg, but is marked as 1 kg
III. Mix 10% impurities in that item and sell that item at cost price
IV. Increase the price by 5%, and use an incorrect weight whose real value is 5% less than what is marked on it.

Select the answer using the correct code.

(a) I   (b) II    (c) III   (d) IV

Explanation:

Case I:
In the first case, it is directly given that the profit is 8%.

Case II:
For second case, let the CP of 1 kg of item be Rs. 100
Then CP of 900 g of item= (100/1000) x 900 = Rs. 90
Hence, profit percentage = {(100 - 90)/90} x 100 = 11.11%

Case III:
Let the CP of 1 kg of pure item be Rs. 100
If he adds 10% impurity, then CP of 1 kg = {(1000/1100) x 100} = Rs. 90.90
Hence, profit percent = {(100 - 90.90)/90.90} x 100 = 10.01%

Case IV:
If he reduces weight by 5%, then cost price of 950g = {(100/1000) x 950} = Rs. 95 and SP = Rs. 105
Hence, profit percent = {(105 – 95)/95} X 100 = 10.52%

Hence, the profit is maximum in second case.