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Profit, Loss & Discount

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high importance~2 Q in Tier 119 formulas⚡ 14 shortcuts5 subtopics
OverviewLearn (5)Formulas (19)Examples (5)Practice (69)
Subtopic 4 of 5·← Marked Price, Discount and the CP–MP–SP Chain·Same Selling Price: One Profit, One Loss →

Dishonest Dealer, False Weights & Claims

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The dealer cheats on the quantity, not the price. Compare true goods against paid goods.

Sells at CP with false weight: using w g instead of W g gives gain%=W−ww×100\text{gain\%} = \frac{W - w}{w} \times 100

Claims a loss but cheats on weight: claims L% loss but gives c% less weight ⇒ true multiplier =100−L100−c= \frac{100 - L}{100 - c}.

Cheats both ways — takes b% extra while buying and gives s% less while selling: multiplier =100+b100−s= \frac{100 + b}{100 - s}.

Setup a per-1000 g table: what he pays for, what he actually gets, what he actually gives.

📖 Notes🧩 5 question types⚡ Formulas🎯 13 practice Q

Detailed notes

The idea

A dishonest dealer cheats on the quantity, not on the printed price. He charges for 1 kg but hands over 900 g, or uses a short measure. Compare what he charges for against what he actually gives — the gain is always bigger than it looks.

Sells at cost price with a false weight

He uses w grams instead of a true W (usually 1000 g), at the cost price: gain%=W−ww×100\text{gain\%} = \frac{W - w}{w} \times 100 900 g per kg: 100900=1119%\frac{100}{900} = 11\frac{1}{9}\%. 800 g: 200800=25%\frac{200}{800} = 25\%. 750 g: 250750=3313%\frac{250}{750} = 33\frac{1}{3}\%. Why divide by w? His cost is for the w grams he actually handed over; his income is for W grams. Profit% is measured on the goods that left his shop.

Reverse: gain given, find the false weight

Gains 25% at cost price → 1000−ww=25100\frac{1000 - w}{w} = \frac{25}{100} → w=800w = 800 g. In general w=W×100100+gain%w = W \times \frac{100}{100 + \text{gain\%}}. The answer is always below the true weight.

Claims a loss but cheats on weight

"Claims a 10% loss but gives only 800 g per kg." For every 800 g delivered he charges the price of a "10%-loss kilo" = 0.9 of the true kilo price: multiplier=100−L100−c=9080=1.125 ⇒ 1212% gain\text{multiplier} = \frac{100 - L}{100 - c} = \frac{90}{80} = 1.125 \ \Rightarrow\ 12\frac{1}{2}\% \text{ gain} Equal claim and shortfall (20% loss, 20% short) give exactly no profit, no loss — a favourite option.

Cheats at both ends

Takes b% extra while buying and gives s% less while selling (everything at cost price): multiplier=100+b100−s\text{multiplier} = \frac{100 + b}{100 - s} 10% extra, 10% less: 11090=119\frac{110}{90} = \frac{11}{9} → gain 2229%22\frac{2}{9}\%. 10% extra, 20% less: 11080=1.375\frac{110}{80} = 1.375 → 37.5% gain.

Short weight plus a price rise

Uses an 800 g weight AND sells 20% above cost: money per false kilo =1.2×= 1.2 \times the true kilo price, goods given =0.8= 0.8 → 1.20.8=1.5\frac{1.2}{0.8} = 1.5 → 50% gain. Write the two cheats as chips and divide.

How to attack any version

  1. Fix a true unit (1 kg, 1 litre, 1 metre).
  2. Write what he receives and what he gives, in units of that true measure.
  3. Gain% =(receivedgiven−1)×100= \left(\frac{\text{received}}{\text{given}} - 1\right) \times 100.

Worked mini-case: a pulse seller takes 110 kg while paying for 100 kg (₹40 per kg → ₹4,000), and later hands over 90 kg while charging for 100 kg (₹4,000). Per ₹4,000 he receives 110 kg and gives away 90 kg, so the multiplier is 11090=119\frac{110}{90} = \frac{11}{9} → gain 2229%22\frac{2}{9}\% — no rupee arithmetic needed beyond the two counts.

Quick revision

  • False weight at CP: W−ww×100\frac{W - w}{w} \times 100 — divide by what he gives.
  • Claimed loss L% with c% short weight: 100−L100−c−1\frac{100 - L}{100 - c} - 1.
  • Both ends: 100+b100−s−1\frac{100+b}{100-s} - 1.
  • Equal claim and shortfall → 0%.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: False weight sold at cost pricevery common3 practice Q
How to spot it:

'Sells at cost price but uses a weight of 800 g / 900 g for a kilogram' — the gain per cent is asked.

gain%=W−ww×100\text{gain\%} = \frac{W - w}{w} \times 100
  1. Take the true weight W (usually 1000 g) and the false weight w.
  2. Gain% =W−ww×100= \frac{W - w}{w} \times 100 — divide by w, the weight actually given.
  3. Sanity check: smaller w → bigger gain.

Why: his cost is for w grams, his income is for W grams, and profit% is on the goods he actually parted with.

Example: A shopkeeper sells sugar at cost price but uses a weight of 900 g for 1 kg. His gain per cent?

1000−900900×100=1119%\frac{1000 - 900}{900} \times 100 = 11\frac{1}{9}\%.

Type 2: Reverse: gain per cent given, find the false weightcommon2 practice Q
How to spot it:

'Gains 25% while selling at cost price — what weight does he use instead of 1 kg?'

w=W×100100+gain%w = W \times \frac{100}{100 + \text{gain\%}}
  1. Write W−ww=g100\frac{W - w}{w} = \frac{g}{100}.
  2. Solve: w=W×100100+gw = W \times \frac{100}{100 + g}.
  3. The false weight is always less than W.

Why: the gain fraction fixes the ratio of true to false weight.

Example: A dishonest dealer sells at cost price and gains 25%. What weight does he give for a kilogram?

w=1000×100125=800w = 1000 \times \frac{100}{125} = 800 g.

Type 3: Claimed loss with short weightcommon2 practice Q
How to spot it:

'Claims to sell at a loss of L% but gives c% less weight' — the true gain or loss is asked.

multiplier=100−L100−c,gain%=(multiplier−1)×100\text{multiplier} = \frac{100 - L}{100 - c}, \qquad \text{gain\%} = (\text{multiplier} - 1) \times 100
  1. Write the price he charges per true kg: (100−L)%(100 - L)\% of the kilo price.
  2. Write what the goods he gives should have cost: (100−c)%(100 - c)\%.
  3. Divide the two; above 1 means a gain.

Why: he collects the 'loss-price' of a full kilo but supplies only part of it.

Example: A merchant claims a 10% loss but gives only 800 g per kg. His true result?

9080=1.125\frac{90}{80} = 1.125 → 1212%12\frac{1}{2}\% gain.

Type 4: Cheats while buying and sellingcommon2 practice Q
How to spot it:

'Takes 10% extra while buying and gives 10% less while selling, at cost price' — overall gain per cent.

gain%=(100+b100−s−1)×100\text{gain\%} = \left(\frac{100 + b}{100 - s} - 1\right) \times 100
  1. Buying b% extra means he owns 100+b100\frac{100+b}{100} units per unit paid.
  2. Selling s% short means each unit sold costs him only 100−s100\frac{100-s}{100} units.
  3. Divide the chips; the excess over 1 is his gain.

Why: the two cheats multiply because they act at different stages.

Example: A dealer takes 10% extra goods while buying and gives 10% less while selling, all at cost price. His gain per cent?

11090=119\frac{110}{90} = \frac{11}{9} → gain =29×100=2229%= \frac{2}{9} \times 100 = 22\frac{2}{9}\%.

Type 5: False weight plus a higher priceoccasional2 practice Q
How to spot it:

The dealer both uses a short weight AND sells above cost price — a two-cheat multiplier.

gain%=(price chip×Ww−1)×100\text{gain\%} = \left(\frac{\text{price chip} \times W}{w} - 1\right) \times 100
  1. Price chip: 1+m1001 + \frac{m}{100} for a markup m%.
  2. Weight kept: w/W as a fraction.
  3. Gain% =(price chip×Ww−1)×100= \left(\frac{\text{price chip} \times W}{w} - 1\right) \times 100.

Why: charging more per false kilo and giving less per true kilo are independent chips.

Example: A seller uses an 800 g weight and sells 20% above cost price. His gain per cent?

1.2×1000800=1.5\frac{1.2 \times 1000}{800} = 1.5 → 50% gain.

Formulas

False weight gain
gain%=true weight−used weightused weight×100\text{gain\%} = \frac{\text{true weight} - \text{used weight}}{\text{used weight}} \times 100
Claimed loss + short weight
multiplier=100−L100−c\text{multiplier} = \frac{100 - L}{100 - c}
Cheat at both ends
multiplier=100+b100−s\text{multiplier} = \frac{100 + b}{100 - s}
True gain from multiplier
gain%=(multiplier−1)×100\text{gain\%} = (\text{multiplier} - 1) \times 100

Shortcut tricks

⚡ The W−w over w rule

Loss to the customer is measured against what the dealer actually gave.

Example: A shopkeeper sells rice at cost price but uses a weight of 900 g for 1 kg. Find his gain per cent.

1000−900900×100=1119%\frac{1000 - 900}{900} \times 100 = 11\frac{1}{9}\%.

⚡ Claimed loss can still be a gain

Compare the money per TRUE gram, not per claimed gram.

Example: A merchant claims to sell at a 10% loss but gives only 800 g per kg. Find his actual gain or loss.

Per true kg he receives 0.900.90 and pays 0.800.80: gain =0.9−0.80.8=12.5%= \frac{0.9 - 0.8}{0.8} = 12.5\%.

⚡ Both-ends multiplier

Multiply the buy-side gain chip by the sell-side chip.

Example: A dishonest dealer takes 10% more goods than he pays for and sells 10% less than the true weight. His gain per cent is:

11090−1=29=2229%\frac{110}{90} - 1 = \frac{2}{9} = 22\frac{2}{9}\% gain.

Where students lose marks

  • Using W−wW\frac{W-w}{W} instead of W−ww\frac{W-w}{w} (the base is what the buyer actually received).

  • Believing a '10% loss' claim without checking the weight.

  • In both-ends cheating, adding the two percentages instead of multiplying chips.

  • Mixing up which side the cheat favours — dealer gains when buying AND when selling.

Practice sets — 13 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.