Understanding Dishonest Trader Problems with False Weights

This problem involves a dishonest trader who claims to sell goods at the cost price but makes a profit by using a false weight. The key to solving such problems is understanding that the profit comes from the difference between the weight the trader claims to sell and the weight they actually sell.

Let's break down the situation:

• The trader claims to sell 1 kg (which is 1000 grams).
• The trader uses a false weight, let's call it 'x' grams, instead of 1000 grams.
• The trader charges the customer for 1000 grams at the cost price.
• The trader actually sells only 'x' grams, incurring the cost price for only 'x' grams.
• The profit is the difference between the money received (cost of 1000g) and the actual cost incurred (cost of x g).

Calculating Profit Percentage with False Weights

When a trader uses a false weight, the profit percentage can be calculated using the following relationship:

$\text{Profit} \% = \frac{\text{True Weight} - \text{False Weight}}{\text{False Weight}} \times 100$

In this problem:

• The True Weight (what should be given) = 1 kg = 1000 grams.
• The False Weight (what is actually given) = x grams.
• The Profit % = 12.5%.

Solving for the False Weight

Now, we can plug the given values into the formula and solve for 'x', the false weight used by the dishonest trader:

$12.5 = \frac{1000 - x}{x} \times 100$

Divide both sides by 100:

$\frac{12.5}{100} = \frac{1000 - x}{x}$

$0.125 = \frac{1000 - x}{x}$

Multiply both sides by 'x':

$0.125x = 1000 - x$

Add 'x' to both sides:

$0.125x + x = 1000$

$1.125x = 1000$

Now, solve for 'x':

$x = \frac{1000}{1.125}$

To simplify the division, we can write 1.125 as a fraction or multiply the numerator and denominator by 1000 to remove the decimal:

$x = \frac{1000}{1.125} = \frac{1000 \times 1000}{1.125 \times 1000} = \frac{1000000}{1125}$

Alternatively, $1.125 = 1 + 0.125 = 1 + \frac{1}{8} = \frac{8+1}{8} = \frac{9}{8}$.

So, $x = \frac{1000}{9/8} = 1000 \times \frac{8}{9} = \frac{8000}{9}$

Now, calculate the value:

$x = \frac{8000}{9} \approx 888.888...$

Rounding to one decimal place, the false weight used is approximately 888.9 grams. However, looking at the options, 888.8 g is provided, which is likely the intended precision ($\frac{8000}{9}$ truncated).

Conclusion on False Weight Used

The dishonest trader uses approximately 888.8 grams instead of 1000 grams to gain a profit of 12.5% while claiming to sell at the cost price.

Revision Table: Key Concepts

Additional Information: Solving False Weight Problems

False weight problems are common in profit and loss topics. They test your understanding that profit can be made by manipulating either price or quantity.

• Understanding the Setup: Always identify the claimed quantity (True Weight) and the actual quantity given (False Weight). The cost incurred is for the False Weight, while revenue is based on the True Weight (often calculated at the claimed Cost Price per unit of True Weight).
• Alternative Approach (Ratio): If the profit is P%, it means that the ratio of the cost of the quantity sold to the selling price received is 100 : (100+P). In the case of selling at cost price with a false weight, the selling price received for the false weight is equivalent to the cost of the true weight. So, Cost of False Weight : Cost of True Weight = 100 : (100+P).
  Let CP per gram be $C$. Cost of False Weight ($x$) = $C \times x$. Cost of True Weight (1000g) = $C \times 1000$.
  So, $C \times x : C \times 1000 = 100 : (100+12.5)$
  $x : 1000 = 100 : 112.5$
  $\frac{x}{1000} = \frac{100}{112.5}$
  $x = 1000 \times \frac{100}{112.5} = 1000 \times \frac{1000}{1125} = 1000 \times \frac{8}{9} = \frac{8000}{9}$
  $x \approx 888.8$ grams.
  This ratio method gives the same result and can be a quick way to solve these problems.
• General Formula: If a trader sells at a markup/markdown (M%) on CP and uses a false weight W' instead of W, the overall profit percentage is given by:
  Overall Profit % = $\left( \frac{\text{True Weight}}{\text{False Weight}} \times \frac{100 + \text{Markup } \%}{100} - 1 \right) \times 100$.
  In this problem, Markup % = 0 (sells at CP). So, Overall Profit % = $\left( \frac{1000}{x} \times \frac{100 + 0}{100} - 1 \right) \times 100 = \left( \frac{1000}{x} - 1 \right) \times 100$.
  $12.5 = \left( \frac{1000}{x} - 1 \right) \times 100$
  $0.125 = \frac{1000}{x} - 1$
  $1.125 = \frac{1000}{x}$
  $x = \frac{1000}{1.125} = \frac{8000}{9} \approx 888.8$ grams.
  All methods lead to the same result, reinforcing the relationship between profit and the quantity difference.