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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 2 of 5·← Profit, Loss and CP–SP Basics·Marked Price, Discount and the CP–MP–SP Chain →

Successive Changes & Equivalent Single Change

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Two percentage changes applied one after another do not simply add — the second acts on the already-changed value.

Net change=a+b+ab100(use −b for a decrease)\text{Net change} = a + b + \frac{ab}{100} \quad (\text{use } -b \text{ for a decrease})

  • Same change x twice: increase 2x+x21002x + \frac{x^2}{100}; decrease 2x−x21002x - \frac{x^2}{100}.
  • Two discounts of d1,d2d_1, d_2: equivalent single discount =d1+d2−d1d2100= d_1 + d_2 - \frac{d_1 d_2}{100}.

Chip method: multiply (1±a/100)(1±b/100)(1 \pm a/100)(1 \pm b/100) — this never fails, even for three changes, and reverses cleanly.

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

Detailed notes

What "successive" means

Successive changes happen one after another, and each new change works on the value left by the previous one. A shop gives 20% off, and then a further 10% off on the reduced bill. The second 10% is on the smaller amount, so the total discount is less than 30%.

Two changes: the a+b+ab100a + b + \frac{ab}{100} rule

For two changes of a%a\% and b%b\% (use minus for decreases): Net change=a+b+ab100\text{Net change} = a + b + \frac{ab}{100} For two discounts d1d_1 and d2d_2 this becomes Equivalent single discount=d1+d2−d1d2100\text{Equivalent single discount} = d_1 + d_2 - \frac{d_1 d_2}{100} 20% and 15% → 35−3=32%35 - 3 = 32\%. 10% and 20% → 30−2=28%30 - 2 = 28\%.

Chips — works for any number of changes

Each discount d%d\% keeps 100−d100\frac{100 - d}{100} of the price. Multiply the kept fractions: 10%, 20%, 25% → 0.9×0.8×0.75=0.540.9 \times 0.8 \times 0.75 = 0.54 → the customer pays 54% → single discount =46%= 46\% (not 55%). Price after discounts on a marked price (MP) of ₹5,000 at 10% and 5%: 5000×0.9×0.95=₹4,2755000 \times 0.9 \times 0.95 = ₹4{,}275. The order of the discounts does not matter — multiplication gives the same result either way.

Comparing two offers

"40% flat" versus "30% + 10%": the second is 30+10−3=37%30 + 10 - 3 = 37\%, so the flat offer saves 3% of the bill more. If the difference is ₹72, the bill is 720.03=₹2,400\frac{72}{0.03} = ₹2{,}400. A single discount is always better for the buyer than successive discounts that add up to the same number.

Finding a missing discount

MP ₹1,600, first discount 15%, final price ₹1,224. After the first discount the price is ₹1,360; the second discount =1360−12241360=10%= \frac{1360 - 1224}{1360} = 10\%. Two equal discounts taking ₹6,400 to ₹5,184: kept fraction =51846400=0.81=0.92= \frac{5184}{6400} = 0.81 = 0.9^2 → each discount 10%.

Increase then decrease

A price raised by 20% and then cut by 20% ends 4% lower: 1.2×0.8=0.961.2 \times 0.8 = 0.96. In general, up x%x\% then down x%x\% (either order) gives a loss of x2100%\frac{x^2}{100}\%. A TV priced ₹30,000 → ₹36,000 → ₹28,800: the buyer pays ₹1,200 less than the original. Markup followed by successive discounts: marked 30% above CP, then 10% and 10% off → 1.3×0.81=1.0531.3 \times 0.81 = 1.053 → profit 5.3%.

Common traps

  • Adding discounts directly (10% + 20% = 30% is wrong).
  • Taking the second discount on the original MP.
  • For three discounts, using the two-change formula only once.

Quick revision

  • Two discounts: d1+d2−d1d2100d_1 + d_2 - \frac{d_1 d_2}{100}.
  • Any number: multiply the kept fractions 100−d100\frac{100 - d}{100}.
  • Up x%x\% then down x%x\% → net fall of x2100%\frac{x^2}{100}\%.
  • Missing discount: divide the final price by the price after the known discounts.

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: Equivalent single discountvery common3 practice Q
How to spot it:

'Successive discounts of 20% and 15% are equal to a single discount of?' — two or three discounts, no rupee values.

d=d1+d2−d1d2100d = d_1 + d_2 - \frac{d_1 d_2}{100}
  1. For two discounts, use d1+d2−d1d2100d_1 + d_2 - \frac{d_1 d_2}{100}.
  2. For three or more, multiply the kept fractions and subtract from 100%.
  3. Check: the answer is less than the plain sum of discounts.

Why: each later discount acts on an already reduced price.

Example: Find a single discount equivalent to successive discounts of 20% and 15%.

20+15−20×15100=32%20 + 15 - \frac{20 \times 15}{100} = 32\%.

Type 2: Selling price after successive discountsvery common2 practice Q
How to spot it:

A marked price and two or three discounts are given; the price paid is asked.

SP=MP×100−d1100×100−d2100SP = MP \times \frac{100 - d_1}{100} \times \frac{100 - d_2}{100}
  1. Write each discount as a kept fraction (10% → 0.9, 12.5% → 7/8).
  2. Multiply the MP by all kept fractions.

Why: every discount scales the current price by its kept fraction.

Example: The marked price is ₹5,000; discounts of 10% and 5% are given. Find the price paid.

5000×0.9×0.95=₹4,2755000 \times 0.9 \times 0.95 = ₹4{,}275.

Type 3: Comparing a single discount with successive discountscommon2 practice Q
How to spot it:

'The difference between a flat 40% discount and successive 30% and 10% is ₹72' — find the bill or the difference.

difference=(single−equivalent)% of MP\text{difference} = (\text{single} - \text{equivalent}) \% \text{ of MP}
  1. Convert the successive discounts to one equivalent discount.
  2. Subtract from the single discount to get the percentage gap.
  3. That gap is a percentage of the MP; use it to find either value.

Why: both offers start from the same MP, so only the percentages differ.

Example: Difference between a 40% discount and successive discounts of 30% and 10% on a bill is ₹72. Find the bill.

Successive =37%= 37\% → gap 3%3\% of bill =72= 72 → bill =₹2,400= ₹2{,}400.

Type 4: Find a missing discountcommon2 practice Q
How to spot it:

MP, final price and one discount are given (or 'two equal discounts'); the other discount is asked.

100−d2100=SPMP×100−d1100\frac{100 - d_2}{100} = \frac{SP}{MP \times \frac{100 - d_1}{100}}
  1. Apply the known discount to MP.
  2. Compare the final price with that reduced price to get the second kept fraction.
  3. For two equal discounts, take the square root of SPMP\frac{SP}{MP}.

Why: the second discount is a percentage of the price after the first.

Example: MP ₹1,600; after two discounts, the first being 15%, the price is ₹1,224. Find the second discount.

After 15%: ₹1,360. 12241360=0.9\frac{1224}{1360} = 0.9 → second discount 10%10\%.

Type 5: Raise then reduce (markup and successive cuts)common2 practice Q
How to spot it:

'Price increased by 20% and then decreased by 20%' or 'marked 30% above CP, then two discounts of 10%'.

net=a+b+ab100,or multiply all chips\text{net} = a + b + \frac{ab}{100}, \quad \text{or multiply all chips}
  1. Write every rise and fall as a chip (1.3, 0.9, 0.9 …).
  2. Multiply the chips; subtract 1 for the net change.
  3. Same % up and down always ends below the start by x2100%\frac{x^2}{100}\%.

Why: the second change acts on a different base from the first.

Example: An article is marked 30% above CP, then sold after successive discounts of 10% and 10%. Profit%?

1.3×0.9×0.9=1.0531.3 \times 0.9 \times 0.9 = 1.053 → profit 5.3%5.3\%.

Formulas

Two successive changes
net=a+b+ab100\text{net} = a + b + \frac{ab}{100}
Same change twice (increase)
2x+x21002x + \frac{x^2}{100}
Same change twice (decrease)
2x−x21002x - \frac{x^2}{100}
Equivalent single discount
D=d1+d2−d1d2100D = d_1 + d_2 - \frac{d_1 d_2}{100}

Shortcut tricks

⚡ Multiply the chips

Chips work for any mix of increases and decreases, in any order.

Example: A price is increased by 20% and then the new price is decreased by 10%. Net change?

1.20×0.90=1.081.20 \times 0.90 = 1.08 ⇒ net +8% (not +10%).

⚡ The d₁ + d₂ − d₁d₂/100 formula

Two discounts collapse in one line.

Example: Find the single discount equivalent to two successive discounts of 20% and 10%.

20+10−200100=28%20 + 10 - \frac{200}{100} = 28\%.

⚡ Work backwards through chips

For 'find the second change given the net', divide the net chip by the first chip.

Example: After a rise of 25% a value was changed again and ended 15% up overall. Find the second change.

1.15/1.25=0.921.15 / 1.25 = 0.92 ⇒ a decrease of 8%.

Where students lose marks

  • Adding successive discounts directly (20% + 10% = 30% is wrong; the answer is 28%).

  • For a profit then a loss, using +ab/100 where the sign must be negative.

  • Applying the second discount on the original marked price instead of the reduced price.

  • Assuming the order of changes matters — chip multiplication is commutative.

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.