Profit, Loss & Discount

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Profit, Loss and CP–SP Basics

Profit and loss are always calculated on the cost price (CP) unless stated otherwise.

$\text{Profit} = SP - CP, \qquad \text{Loss} = CP - SP$ $\text{Profit\%} = \frac{\text{Profit}}{CP} \times 100, \qquad SP = CP\left(\frac{100 \pm x}{100}\right)$

Multiplier chips: profit of 20% → SP = 1.2 × CP; loss of 15% → SP = 0.85 × CP. Percentages as fractions are faster: 12.5% = $\frac{1}{8}$ , 16⅔% = $\frac{1}{6}$ , 8⅓% = $\frac{1}{12}$ .

To recover CP from SP at x% profit: $CP = \frac{100 \cdot SP}{100 + x}$ — divide by the chip, don't take x% of SP.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 16 practice Q

Detailed notes

The words you need

Cost price (CP): what the seller paid for the article, including any extra costs such as transport or repair ("overheads").
Selling price (SP): what the buyer pays the seller.
Profit (gain) $= SP - CP$ when SP is bigger; loss $= CP - SP$ when SP is smaller. A shopkeeper buys a fan for ₹1,250 and sells it for ₹1,400: profit $= ₹150$ .

Profit and loss per cent are always on CP

$\text{Profit\%} = \frac{\text{Profit}}{CP} \times 100, \qquad \text{Loss\%} = \frac{\text{Loss}}{CP} \times 100$ For the fan: $\frac{150}{1250} \times 100 = 12\%$ . Dividing by SP (giving $10\frac{5}{7}\%$ ) is the most common wrong option. If there are overheads, add them to the CP first: bought ₹2,400 + ₹150 transport → CP ₹2,550.

Multipliers (chips) — the fast way

Profit of x% means $SP = CP \times \frac{100 + x}{100}$ ; loss of x% means $SP = CP \times \frac{100 - x}{100}$ .

Change	Chip
20% profit	$\frac{6}{5}$
25% profit	$\frac{5}{4}$
$16\frac{2}{3}\%$ profit	$\frac{7}{6}$
10% loss	$\frac{9}{10}$
$12\frac{1}{2}\%$ loss	$\frac{7}{8}$
Going back from SP to CP means dividing by the chip: SP ₹1,140 at 5% loss → $CP = 1140 \div \frac{19}{20} = ₹1{,}200$ .

Two selling prices, one cost price

"At 8% loss; had he sold for ₹336 more he would gain 6%." The two SPs differ by $8 + 6 = 14\%$ of CP. So $14\%$ of CP $= 336$ → $CP = ₹2{,}400$ . Rule: difference in rupees $=$ difference in percentages (add them when one is a loss and the other a gain) of CP. Same idea: "selling at ₹720 gives as much profit as the loss when selling at ₹560" → CP is exactly in the middle: $\frac{720 + 560}{2} = ₹640$ .

Counting articles instead of rupees

"CP of 15 articles = SP of 12 articles." Take the price of one article as 1. The seller gets money for 15 by giving only 12 → profit $= \frac{15 - 12}{12} \times 100 = 25\%$ (divide by the goods given, which are the ones that cost him). "Buys 4 for ₹15, sells 5 for ₹24": find per-article prices (₹3.75 and ₹4.80), or use a common count of 20 articles: CP ₹75, SP ₹96 → profit 28%.

Changing CP and SP together

"If he had bought it 20% cheaper and sold it ₹60 cheaper, he would gain 25%." Let CP = 100 units and write both stories as chips: $1.1 \times CP - 60 = 0.8 \times CP \times 1.25$ → $0.1 \times CP = 60$ → $CP = ₹600$ .

Quick revision

Profit% and loss% are on CP (plus overheads).
SP $= CP \times$ chip; CP $= SP \div$ chip.
Two SPs: rupee gap $=$ % gap of CP.
Article counts: profit% $= \frac{\text{extra}}{\text{articles given}} \times 100$ .
Always check the answer by rebuilding SP from CP.

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: Profit or loss per cent from CP and SPvery common2 practice Q

How to spot it:

CP and SP are given (maybe with transport or repair cost) and the profit or loss per cent is asked.

\text{Profit\%} = \frac{SP - CP}{CP} \times 100

Add any overheads to the CP.
Profit (or loss) $= SP - CP$ .
Divide by CP (never by SP) and multiply by 100.

Why: profit is measured against the money the seller invested.

Example: A fan bought for ₹1,250 is sold for ₹1,400. Find the profit per cent.

$\frac{150}{1250} \times 100 = 12\%$ .

Type 2: Find CP or SP from a profit / loss per centvery common2 practice Q

How to spot it:

One price and the profit or loss per cent are given; the other price is asked.

SP = CP \times \frac{100 \pm x}{100}, \qquad CP = SP \times \frac{100}{100 \pm x}

Write the chip: $\frac{100 + x}{100}$ for profit, $\frac{100 - x}{100}$ for loss.
CP → SP: multiply. SP → CP: divide.
Reduce the chip to a small fraction for speed ( $\frac{19}{20}$ , $\frac{7}{6}$ …).

Why: SP is CP scaled by the chip, so reversing it means dividing.

Example: A table sold for ₹1,140 gives a loss of 5%. Find its CP.

$1140 \times \frac{20}{19} = ₹1{,}200$ .

Type 3: Two selling prices — 'had he sold for ₹x more'very common2 practice Q

How to spot it:

'Sold at a% loss; had he sold for ₹k more he would gain b%' or two SPs giving equal profit and loss.

CP = \frac{k \times 100}{a + b} \ \ (\text{loss } a\%, \text{ gain } b\%)

Both SPs are chips of the same CP.
Their rupee difference equals the difference of the percentages of CP (loss and gain → add; two gains → subtract).
Divide the rupees by that percentage of CP.

Why: the CP is common, so only the percentage gap moves the SP.

Example: Sold at 8% loss; had it been sold for ₹336 more, the gain would be 6%. Find the CP.

$14\%$ of CP $= 336$ → $CP = ₹2{,}400$ .

Type 4: Article-count questionscommon3 practice Q

How to spot it:

'CP of 15 articles = SP of 12', 'buys 4 for ₹15 and sells 5 for ₹24', or lemons sold at so many per rupee.

\text{Profit\%} = \frac{\text{articles bought} - \text{articles sold}}{\text{articles sold}} \times 100

For 'CP of a = SP of b': profit% $= \frac{a - b}{b} \times 100$ (loss if b > a).
For rates like '4 for ₹15', use the LCM of the counts so both prices are for the same number of articles.
For 'how many for ₹k', find the required SP of one article, then divide.

Why: fixing a common count turns the question into a normal CP–SP comparison.

Example: The CP of 15 articles equals the SP of 12. Find the profit per cent.

$\frac{15 - 12}{12} \times 100 = 25\%$ .

Type 5: New SP for a different gain / changed CP and SPcommon2 practice Q

How to spot it:

'Sold at ₹1,530 at a loss of 10%; at what price to gain 10%?' or 'bought 20% cheaper and sold ₹60 less, gain 25%'.

SP_2 = SP_1 \times \frac{100 + b}{100 - a}

Go back to CP by dividing by the first chip.
Multiply by the new chip.
For a changed CP and SP, write both stories with the same CP as a variable and equate.

Why: CP is the anchor that links every selling scenario.

Example: An article sold for ₹1,530 gives a loss of 10%. At what price should it be sold to gain 10%?

$CP = 1530 \times \frac{10}{9} = 1700$ ; $SP = 1700 \times \frac{11}{10} = ₹1{,}870$ .

Formulas

Profit / loss per cent

\text{P\%} = \frac{SP - CP}{CP} \times 100

Selling price

SP = CP\left(1 \pm \frac{x}{100}\right)

Cost price from SP

CP = \frac{SP}{1 \pm \frac{x}{100}} = \frac{100 \cdot SP}{100 \pm x}

No profit, no loss

SP = CP

Shortcut tricks

⚡ Fraction ⇄ percentage swap

Convert the profit % to a fraction; SP/CP becomes a clean ratio.

Example: A trader sells a cycle for ₹1,020 that cost him ₹850. Find the profit per cent.

Profit = 170; $\frac{170}{850} = \frac{1}{5} = 20\%$ .

⚡ Divide by the chip to get CP

SP given with a profit/loss %: undo the multiplier.

Example: A machine is sold for ₹624 at a loss of 4%. Its cost price is:

$CP = \frac{624}{0.96} = 650$ , i.e. ₹650.

⚡ Two prices, same % — scale linearly

If the same article at another CP/SP keeps the same profit %, everything scales by the same factor.

Example: By selling 15 pens a man recovers the cost of 12 pens. His gain per cent is:

Gain on 12 pens' cost = 3 pens ⇒ $\frac{3}{12} = 25\%$ .

Where students lose marks

Calculating profit % on SP instead of CP (a ₹20 profit on SP 120 is NOT 16⅔% gain).
Recovering CP as $SP - x\% \text{ of } SP$ — that's the SP-based error again.
Mixing up profit and loss chips (0.9 is a 10% loss, not a 10% discount on profit).
Ignoring that 'gain of 25%' and 'marked 25% up' refer to different bases (CP vs MP).

Practice sets — 16 questions

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

Topic test · 10 questions

Suggested time 5 min · wrong answers go to your mistake notebook automatically.