Profit, Loss & Discount

high importance~2 Q in Tier 124 formulas⚡ 14 shortcuts5 subtopics

Overview Learn (5)Formulas (24)Examples (5)Practice (69)

Dishonest Dealer, False Weights & Claims

⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

A dishonest dealer cheats on the quantity, not the printed price. He charges for a full kilogram but hands over less.

Compare what he charges for with what he actually gives. Gain per cent $=$ (goods charged $-$ goods given) $\div$ goods given $\times 100$ .

The idea

A dealer sells rice "at cost price" but uses a 900 g weight for every kilogram. His price looks honest; his scale is not.

To judge him, put two numbers side by side: what he charges for (1000 g) and what he gives (900 g).

Rule: Gain per cent $= \dfrac{\text{charged} - \text{given}}{\text{given}} \times 100$ . The base is what he gives.

False weight at cost price

He uses $w$ grams in place of a true $W$ grams (usually $W = 1000$ g):

\text{gain\%} = \frac{W - w}{w} \times 100

900 g per kg: $\dfrac{100}{900} = 11\dfrac{1}{9}\%$
800 g per kg: $\dfrac{200}{800} = 25\%$
750 g per kg: $\dfrac{250}{750} = 33\dfrac{1}{3}\%$

His cost covers only the 900 g that left the shop, while he was paid for 1000 g. That is why the base is $w$ .

Watch: Dividing by $W$ instead of $w$ gives $10\%$ instead of $11\dfrac{1}{9}\%$ . That wrong answer is always sitting in the options.

The reverse: gain given, find the weight

A dealer gains $25\%$ while selling at cost price. Then $\dfrac{1000 - w}{w} = \dfrac{1}{4}$ , so $w = 800$ g.

In general $w = W \times \dfrac{100}{100 + \text{gain\%}}$ . The answer always lands below the true weight.

Tip: If your $w$ comes out above 1000 g, you flipped the fraction.

Claims a loss, gives short weight

"Claims to sell at a $10\%$ loss but gives only 800 g per kg."

He charges $90\%$ of the true kilo price for every 800 g he hands over:

\frac{100 - L}{100 - c} = \frac{90}{80} = 1.125

So he truly gains $12.5\%$ . An equal claim and shortfall, like a claimed $20\%$ loss with $20\%$ short weight, gives exactly no profit and no loss.

Careful: A claimed loss can hide a real gain. Always check the weight.

Cheating at both ends

He takes $b\%$ extra goods while buying and gives $s\%$ less while selling, all at cost price:

\text{multiplier} = \frac{100 + b}{100 - s}

$10\%$ extra in, $10\%$ short out: $\dfrac{110}{90} = \dfrac{11}{9}$ , a gain of $22\dfrac{2}{9}\%$ .

Pulse seller's version: he pays for 100 kg but takes 110 kg; later he charges for 100 kg but hands over 90 kg. Per rupee paid, received $\div$ given $= \dfrac{110}{90}$ . Same multiplier, no rupee arithmetic needed.

Short weight plus a price rise

He uses an 800 g weight and sells $20\%$ above cost. Price chip $\dfrac{6}{5}$ , weight chip $\dfrac{1000}{800} = \dfrac{5}{4}$ :

\frac{6}{5} \times \frac{5}{4} = \frac{3}{2} \Rightarrow 50\% \text{ gain}

Example: Two cheats acting together multiply. Charging more per false kilo and giving less per true kilo are independent.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

False weight sold at cost price

How to spot it:

'Sells at cost price but uses a weight of 900 g for 1 kg.' The gain per cent is asked.

\text{gain\%} = \frac{W - w}{w} \times 100

Method

Take the true weight W and the false weight w.
Gain = (W - w) over w, times 100.
Sanity check: smaller w means bigger gain.

Why it works:

His cost is for the w grams he gave; his income is for W grams.

Try this

A shopkeeper sells sugar at cost price but uses a weight of 900 g in place of 1 kg. Find his gain per cent.

Show solution

$\dfrac{1000 - 900}{900} \times 100 = \dfrac{100}{900} \times 100$ .
$= 11\dfrac{1}{9}$ .

Answer

11 1/9%

Type 2common2 practice Q

Gain given, find the false weight

How to spot it:

'He gains 25% while selling at cost price. What weight does he give for a kilogram?'

w = W \times \frac{100}{100 + \text{gain\%}}

Method

Write the gain as a fraction: 25% = 1/4.
Solve (1000 - w) over w equals 1/4.
The answer must be below the true weight.

Why it works:

The gain fraction fixes the ratio of true to false weight.

Try this

A dishonest dealer sells at cost price and still gains 25%. What weight does he give in place of 1 kg?

Show solution

$w = 1000 \times \dfrac{100}{125}$ .
$= 800$ g.

Answer

800 g

Type 3common2 practice Q

Claimed loss with short weight

How to spot it:

'Claims a loss of L% but gives c% less weight.' The true gain or loss is asked.

\text{gain\%} = \left(\frac{100 - L}{100 - c} - 1\right) \times 100

Method

Write the price he charges per true kilo: (100 - L)%.
Write the worth of goods he gives: (100 - c)%.
Divide the first by the second.

Why it works:

He collects the loss-price of a full kilo but supplies only part of it.

Try this

A merchant claims to sell at a 10% loss but gives only 800 g per kg. What is his true gain or loss per cent?

Show solution

$\dfrac{90}{80} = \dfrac{9}{8}$ .
$\dfrac{1}{8} = 12.5\%$ gain.

Answer

12.5% gain

Type 4common2 practice Q

Cheating while buying and selling

How to spot it:

'Takes 10% extra while buying and gives 10% less while selling, at cost price.'

\text{gain\%} = \left(\frac{100 + b}{100 - s} - 1\right) \times 100

Method

Buying b% extra means he owns (100 + b) units per 100 paid for.
Selling s% short means each 100 units sold cost him (100 - s).
Divide and subtract 1.

Why it works:

The two cheats act at different stages, so they multiply.

Try this

A dealer takes 10% extra goods while buying and gives 10% less while selling, all at cost price. Find his gain per cent.

Show solution

$\dfrac{110}{90} = \dfrac{11}{9}$ .
$\dfrac{2}{9} \times 100 = 22\dfrac{2}{9}$ .

Answer

22 2/9%

Type 5occasional2 practice Q

False weight plus a higher price

How to spot it:

The dealer both uses a short weight and sells above cost price.

\text{gain\%} = \left(\text{price chip} \times \frac{W}{w} - 1\right) \times 100

Method

Write the price chip for the markup.
Write the weight ratio W over w.
Multiply the two and subtract 1.

Why it works:

Charging more per false kilo and giving less per true kilo are independent chips.

Try this

A seller uses an 800 g weight in place of 1 kg and sells 20% above cost price. Find his gain per cent.

Show solution

Price chip $\dfrac{6}{5}$ , weight ratio $\dfrac{5}{4}$ .
$\dfrac{6}{5} \times \dfrac{5}{4} = \dfrac{3}{2}$ .
Gain $= 50\%$ .

Answer

50%

Formula sheet

False weight gain

\text{gain\%} = \frac{W - w}{w} \times 100

W = true weight, w = weight used; divide by w.

Weight from a given gain

w = W \times \frac{100}{100 + \text{gain\%}}

Claimed loss with short weight

\text{multiplier} = \frac{100 - L}{100 - c}

L = claimed loss%, c = short weight%.

Cheat at both ends

\text{multiplier} = \frac{100 + b}{100 - s}

b = extra taken while buying, s = shortfall while selling.

True gain from multiplier

\text{gain\%} = (\text{multiplier} - 1) \times 100

Shortcuts that save time

⚡ The W minus w over w rule

At cost price, the only question is which weight goes below the line. It is the weight he gives.

Example

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

Show solution

$\dfrac{1000 - 900}{900} \times 100$ .
$= \dfrac{100}{900} \times 100 = 11\dfrac{1}{9}\%$ .

Answer

11 1/9%

⚡ A claimed loss can still be a gain

Judge 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 true result.

Show solution

Per true kg he charges $0.9$ of the kilo price.
Per true kg he gives goods worth $0.8$ .
$\dfrac{0.9}{0.8} = 1.125$ , a $12.5\%$ gain.

Answer

12.5% gain

⚡ Both-ends multiplier

Buy-side chip over sell-side chip. Two fractions, one division.

Example

A dealer takes 10% extra goods while buying and gives 10% less than the true weight while selling. Find his gain per cent.

Show solution

$\dfrac{110}{90} = \dfrac{11}{9}$ .
$\dfrac{2}{9} \times 100 = 22\dfrac{2}{9}$ .

Answer

22 2/9% gain

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Dividing by the true weight: $\dfrac{100}{1000} = 10\%$ .

Divide by the weight given: $\dfrac{100}{900} = 11\dfrac{1}{9}\%$ .

Mistake 02

Believing a claimed loss without checking the weight.

Compute the multiplier $\dfrac{100-L}{100-c}$ first.

Mistake 03

Adding the two cheats: $10\% + 10\% = 20\%$ .

Multiply chips: $\dfrac{110}{90} \Rightarrow 22\dfrac{2}{9}\%$ .

Mistake 04

Calling an equal claim and shortfall a profit.

A claimed loss of L% with exactly L% short weight is exactly break-even.

Mistake 05

Losing track of which side each cheat helps.

He gains at both ends: extra goods coming in, short goods going out.

Quick revision

Read this the night before the exam.

False weight at CP: gain $= \dfrac{W-w}{w} \times 100$ , base $w$ .
From gain to weight: $w = W \times \dfrac{100}{100 + g}$ .
Claimed loss with short weight: $\dfrac{100-L}{100-c} - 1$ .
Both ends: $\dfrac{100+b}{100-s} - 1$ .
Equal claim and shortfall $\Rightarrow$ zero profit.
Weight cheat and price cheat multiply as chips.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

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

Continue learning