Understanding the Dishonest Dealer Problem

This problem involves a common scenario with dishonest dealers who use false weights to make a profit while claiming to sell goods at the cost price. The key to solving such problems is to understand that the profit is made because the customer pays for a larger quantity than they actually receive.

Calculating Gain Percentage with False Weights

When a dealer uses a false weight, the gain percentage is calculated based on the difference between what the customer pays for and the actual cost of the quantity delivered. The customer pays based on the declared weight (e.g., 1 kg), but the dealer's cost is based on the false weight actually given. The profit is the difference between the selling price (based on declared weight at cost price) and the cost price of the actual weight delivered.

Let's define the terms:

• Cost Price (CP): The price at which the dealer buys the goods.
• Selling Price (SP): The price at which the dealer sells the goods.
• True Weight: The weight the dealer claims to sell (e.g., 1 kg or 1000 gm).
• False Weight: The actual weight the dealer gives to the customer.

The dealer professes to sell at cost price. This means the Selling Price for the claimed quantity is equal to the Cost Price of that claimed quantity.

Let the cost price per gram be $\text{Re } 1$. So, the cost of 1000 gm is $\text{Rs } 1000$.

The dealer claims to sell 1000 gm at cost price. So, the selling price for the customer is $\text{Rs } 1000$.

Let the false weight used for 1000 gm be $W$ grams.

The dealer actually sells $W$ grams. The cost to the dealer for $W$ grams is $W \times \text{CP per gram} = \text{Rs } W$ (assuming CP per gram is Re 1).

The profit made by the dealer on selling $W$ grams is:

\text{Profit} = \text{Selling Price} - \text{Actual Cost} = 1000 - W

The profit percentage is given as 15%. The profit percentage is always calculated on the actual cost of the quantity sold, which is the cost of $W$ grams.

\text{Gain}\% = \frac{\text{Profit}}{\text{Actual Cost}} \times 100

$15 = \frac{1000 - W}{W} \times 100$

Solving for the False Weight

Now, let's solve the equation for $W$, the false weight:

$15 = \frac{1000 - W}{W} \times 100$

Divide both sides by 100:

$\frac{15}{100} = \frac{1000 - W}{W}$

$0.15 = \frac{1000 - W}{W}$

Multiply both sides by $W$:

$0.15W = 1000 - W$

Add $W$ to both sides:

$0.15W + W = 1000$

$(0.15 + 1)W = 1000$

$1.15W = 1000$

Divide both sides by 1.15:

$W = \frac{1000}{1.15}$

Let's perform the calculation:

$W \approx 869.5652...$

The question asks to round off the answer to one digit after the decimal.

Rounding $869.5652...$ to one decimal place gives $869.6$ gm.

Conclusion

For a kilogram (1000 gm), the dishonest dealer uses a false weight of 869.6 gm to gain 15% while selling at the cost price.

Revision Table: Dishonest Dealer Concepts

Additional Information on Profit and Loss with False Weights

Problems involving false weights are common in profit and loss. They test the understanding that profit is always calculated on the actual cost incurred by the seller.

• If a dealer gains $X$% by using a false weight while selling at cost price, it means they are giving less weight than claimed. The ratio of weights is related to the profit percentage.
• The formula used above, $\text{False Weight} = \text{True Weight} \times \frac{100}{100 + \text{Gain}\%}$, is a direct application of the profit percentage formula when the SP of the true weight equals the CP of the false weight.
• Alternatively, you can think of it as the dealer selling $(100 + \text{Gain}\%)$ of the actual weight for the price of 100 units of claimed weight. If the gain is 15%, they sell 115 units' worth of price for 100 units of claimed weight, but actually deliver less than 100 units of actual weight. The cost of the actual weight delivered is what the profit is based on.

Understanding the relationship between the weight difference and the profit percentage is crucial for solving these types of quantitative aptitude questions.