2.6 False Weight / Dishonest Dealer Problems

Figure: A false weight means the dealer hands over less than what he charges for.

A very popular SSC/RRB question type: a shopkeeper claims to sell at cost price (no profit, no loss) but actually uses a faulty (under-sized) weight to cheat the customer, thereby making a hidden profit.

Setting up the derivation: Suppose the dealer claims to give 1000 g (i.e., 1 kg) but his faulty weight actually dispenses only (1000 − x) g, where x is the shortfall (error). Assume the rate is Re 1 per gram for simplicity.

• Actual cost incurred by the dealer for the goods he really handed over = (1000 − x) × 1 = (1000 − x)
• Money he receives from the customer = price of 1000 g (since he claims/charges for a full kg) = 1000 × 1 = 1000
• Profit = Revenue − Actual Cost = 1000 − (1000 − x) = x
• Profit% (always on his actual cost) = [x / (1000 − x)] × 100

Generalising, if the dealer claims a “true value” y (the weight/quantity he charges for) but actually gives (y − x), with error = x:

Profit% = [x / (y − x)] × 100 = [Error / (True Value − Error)] × 100

An equivalent, faster way to remember this for weights: if a dealer uses a weight of w grams in place of 1000 g while selling at cost price,

Profit% = [(1000 − w) / w] × 100

This is exactly the same formula with x = 1000 − w and (y − x) = w.

Worked Example 13: A dishonest shopkeeper professes to sell his goods at cost price but uses a weight of 900 g for a 1 kg weight. Find his profit percentage.

Solution: Error = 1000 − 900 = 100 g; base = 900 g. Profit% = (100/900) × 100 = 100/9 % = 11 1/9 % ≈ 11.11%.

Worked Example 14: A trader uses a weight of 750 g in place of 1 kg while claiming to sell at cost price. Find his gain percentage.

Solution: Error = 1000 − 750 = 250 g; base = 750 g. Profit% = (250/750) × 100 = 33.33% (100/3 %).

Worked Example 15 (combined markup + false weight): A dealer marks up his goods by 10% above cost price, and additionally uses a weight of 800 g in place of 1 kg. Find his overall (actual) profit percentage.

Solution: Take true weight = 1000 g, rate = Re 1/g, so CP of 1000 g = Rs 1000. - Actual cost of goods physically handed over (800 g) = Rs 800. - Selling price charged = price of 1000 g marked up by 10% = 1000 × 1.10 = Rs 1100. - Profit = 1100 − 800 = Rs 300. - Profit% (on actual CP of 800) = (300/800) × 100 = 37.5%.

This example shows why combined markup-and-false-weight questions (common in Tier-II) can never be solved by simply adding the markup% and the weight-error%— the two effects compound multiplicatively, not additively. Always go back to actual-cost-vs-actual-revenue reasoning as above.

Worked Example 15A (reverse direction — find the weight used): A dishonest dealer, while professing to sell his goods at cost price, still manages a profit of 25% by using a false weight in place of 1 kg. Find the weight he actually uses.

Solution: Using Profit% = [(1000−w)/w] × 100: 25 = [(1000−w)/w] × 100 ⇒ 25w = 100000 − 100w ⇒ 125w = 100000 ⇒ w = 800 g. This “work backwards from the profit%” variant is common when the question gives the gain% and asks for the faulty weight instead of the other way around — simply solve the same formula for w.

Worked Example 15B (double false weight — buys short-changed, sells short-changed, edge case): A dishonest dealer, while purchasing goods, uses a tampered weight so that he actually receives 1200 g every time he pays for a claimed 1000 g (i.e., the supplier is cheated in the dealer’s favour). While selling, the dealer uses a weight of 800 g for a claimed 1000 g (i.e., the customer is now cheated). If the price quoted per “kg” is the same at both purchase and sale, find the dealer’s overall profit percentage.

Solution: Let the quoted price per claimed kg = Rs P (same figure at both ends, representing “professes to sell at cost price”). True cost per actual gram to the dealer = P/1200 (since he pays P but actually receives 1200 g). For every sale of a claimed kg, he hands over only 800 g of actual goods, so the true cost of goods given per sale = 800 × (P/1200) = (2/3)P. Revenue per sale = P. Profit = P − (2/3)P = P/3. Profit% = (P/3) ÷ (2P/3) × 100 = 50%. This “double false weight” (cheats both while buying and while selling) is a genuine edge case of the false-weight family: the two shortfalls do not add (400 g “total error” would wrongly suggest 40%) — they must be combined through actual-cost-vs-actual-revenue reasoning, exactly as with markup-plus-false-weight problems.

Worked Example 15C (false weight while also professing an honest loss): A trader tells customers he is selling at a loss of 6% (to appear sympathetic), but he actually uses a weight of 800 g in place of 1 kg. Find his true profit or loss percentage.

Solution: True weight = 1000 g, CP = Rs 1000 (rate Re 1/g). Actual cost of the 800 g he really hands over = Rs 800. SP charged = price of the claimed 1000 g at a “6% loss” on the claimed CP of 1000 = 1000 × 0.94 = Rs 940. Actual profit = 940 − 800 = Rs 140. Profit% (on actual cost of 800) = (140/800) × 100 = 17.5% profit. This is a valuable trap to recognise: even though the dealer claims and genuinely believes he is selling at a loss, the false weight converts this into a real profit of 17.5% — always compute actual cost vs actual revenue, never trust the dealer’s stated profit/loss% at face value once a false weight is involved.