Profit, Loss & Discount

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Successive Changes & Equivalent Single Change

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

The idea in one minute

Successive changes happen one after another, and the second change works on the value the first one left. So two changes never simply add: discounts of $20\%$ and $10\%$ equal one discount of $28\%$ .

The safe method: write what fraction of the price each change keeps, and multiply.

What successive means

A shop cuts a bill by $20\%$ , then gives $10\%$ more off the reduced bill. The second $10\%$ falls on the smaller amount left after the first cut.

So the total discount is less than $20 + 10 = 30\%$ . It is actually $28\%$ .

Rule: Each new change works on the value left by the previous change, never on the original price.

Two changes: a + b + ab/100

For two changes of $a\%$ and $b\%$ (use a minus sign for a fall):

\text{net} = a + b + \frac{ab}{100}

For two discounts $d_1$ and $d_2$ the kept-fraction version becomes:

\text{single discount} = d_1 + d_2 - \frac{d_1 d_2}{100}

$20\%$ and $15\%$ : $35 - 3 = 32\%$ . $10\%$ and $20\%$ : $30 - 2 = 28\%$ .

Careful: Never add discounts directly. $10\% + 20\% = 30\%$ is the trap option.

Chips work for any number of changes

A discount of $d\%$ keeps $\dfrac{100-d}{100}$ of the price. Multiply the kept fractions.

Three discounts of $10\%$ , $20\%$ , $25\%$ : $\dfrac{9}{10} \times \dfrac{4}{5} \times \dfrac{3}{4} = \dfrac{27}{50}$ . The customer pays $54\%$ , so the single discount is $46\%$ , not $55\%$ .

A marked price of Rs 5,000 with $10\%$ and $5\%$ off: $5000 \times \dfrac{9}{10} \times \dfrac{19}{20} = 4275$ .

Tip: The order of the changes does not matter. Multiply in whatever order is easiest.

Comparing two offers

"Flat $40\%$ " against " $30\%$ then $10\%$ ": the second is $30 + 10 - 3 = 37\%$ . The flat offer is better by $3\%$ of the bill.

If that difference is Rs 72, the bill is $72 \div \dfrac{3}{100} = 2400$ .

Note: A single discount always beats successive discounts that add up to the same number.

Finding a missing discount

MP Rs 1,600, first discount $15\%$ , final price Rs 1,224.

After $15\%$ : $1600 \times \dfrac{17}{20} = 1360$ .
Second discount: $\dfrac{1360 - 1224}{1360} = \dfrac{136}{1360} = 10\%$ .

Two equal discounts take Rs 6,400 to Rs 5,184: kept $= \dfrac{5184}{6400} = \dfrac{81}{100} = \left(\dfrac{9}{10}\right)^2$ , so each discount is $10\%$ .

Watch: The second discount is a per cent of the price after the first discount, never of the MP.

A rise and then a fall

A price raised $20\%$ then cut $20\%$ ends $4\%$ lower: $\dfrac{6}{5} \times \dfrac{4}{5} = \dfrac{24}{25}$ .

A TV priced Rs 30,000 goes to Rs 36,000, then to Rs 28,800. The final price is Rs 1,200 below the original.

An article marked $30\%$ above CP and sold after two $10\%$ discounts: $\dfrac{13}{10} \times \dfrac{9}{10} \times \dfrac{9}{10} = 1.053$ , a profit of $5.3\%$ .

Example: After a rise of $25\%$ a value was changed again and ended $15\%$ up overall. Second change: $\dfrac{115}{125} = \dfrac{23}{25}$ , a cut of $8\%$ .

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

Equivalent single discount

How to spot it:

'Successive discounts of 20% and 15% equal a single discount of?' Two or three discounts, no rupees.

d = d_1 + d_2 - \frac{d_1 d_2}{100}

Method

For two discounts, use the one-line formula.
For three or more, multiply kept fractions and subtract from 100%.
Check the answer is less than the plain sum.

Why it works:

Each later discount acts on an already reduced price.

Try this

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

Show solution

$20 + 15 - \dfrac{20 \times 15}{100}$ .
$= 35 - 3 = 32$ .

Answer

32%

Type 2very common2 practice Q

Price after successive discounts

How to spot it:

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

SP = MP \times \frac{100 - d_1}{100} \times \frac{100 - d_2}{100}

Method

Write each discount as a kept fraction.
Multiply the MP by all the kept fractions.
Cancel before multiplying.

Why it works:

Every discount scales the running price by its kept fraction.

Try this

The marked price of an article is Rs 5,000 and discounts of 10% and 5% are given. What does the customer pay?

Show solution

Kept fractions: $\dfrac{9}{10}$ and $\dfrac{19}{20}$ .
$5000 \times \dfrac{9}{10} = 4500$ .
$4500 \times \dfrac{19}{20} = 4275$ .

Answer

Rs 4,275

Type 3common2 practice Q

Flat discount versus successive discounts

How to spot it:

A flat per cent is compared with two successive percents, often through a rupee difference.

\text{difference} = (\text{single} - \text{equivalent})\% \text{ of the price}

Method

Convert the successive discounts to one equivalent discount.
Subtract it from the flat discount.
That gap, as a per cent of the bill, gives the rupee difference.

Why it works:

Both offers start from the same price, so only the per cent gap matters.

Try this

The difference between a 40% flat discount and successive discounts of 30% and 10% on a bill is Rs 72. Find the bill.

Show solution

Successive $= 30 + 10 - 3 = 37\%$ .
Gap $= 3\%$ of the bill $= 72$ .
Bill $= 72 \times \dfrac{100}{3} = 2400$ .

Answer

Rs 2,400

Type 4common2 practice Q

Find a missing discount

How to spot it:

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

\frac{100 - d_2}{100} = \frac{SP}{MP \times \frac{100 - d_1}{100}}

Method

Apply the known discount to the MP.
Divide the final price by that reduced price to get the second kept fraction.
For two equal discounts, take the square root of SP over MP.

Why it works:

The second discount is a per cent of the price after the first.

Try this

An article marked Rs 1,600 is sold for Rs 1,224 after two discounts, the first being 15%. Find the second discount.

Show solution

After $15\%$ : $1600 \times \dfrac{17}{20} = 1360$ .
$\dfrac{1224}{1360} = \dfrac{9}{10}$ , so the second discount is $10\%$ .

Answer

10%

Type 5common2 practice Q

Markup followed by cuts

How to spot it:

'Marked 30% above CP, then two discounts of 10%' or 'increased 20% then decreased 20%'; profit or net change asked.

\text{net} = a + b + \frac{ab}{100} \ \text{or multiply all chips}

Method

Write every rise and fall as a chip.
Multiply the chips.
Subtract 1 for the net change; same x up and down loses $\dfrac{x^2}{100}$ per cent.

Why it works:

The second change acts on a different base from the first, so chips must multiply.

Try this

An article is marked 30% above its cost price and sold after successive discounts of 10% and 10%. Find the profit per cent.

Show solution

Chips: $\dfrac{13}{10} \times \dfrac{9}{10} \times \dfrac{9}{10}$ .
$= \dfrac{1053}{1000} = 1.053$ .
Profit $= 5.3\%$ .

Answer

5.3%

Formula sheet

Two successive changes

\text{net} = a + b + \frac{ab}{100}

Use negative b for a fall.

Same change twice (rise)

2x + \frac{x^2}{100}

Same change twice (fall)

2x - \frac{x^2}{100}

Equivalent single discount

D = d_1 + d_2 - \frac{d_1 d_2}{100}

For two discounts.

Kept fraction

\text{kept} = \prod \frac{100 - d_i}{100}

Any number of discounts; customer pays this share.

Shortcuts that save time

⚡ Multiply the chips

Chips handle any mix of rises and falls, in any order, and extend to three or more changes.

Example

A price is increased by 20% and the new price is then decreased by 10%. Find the net change.

Show solution

$\dfrac{6}{5} \times \dfrac{9}{10} = \dfrac{27}{25}$ .
$1.08$ means a net rise of $8\%$ .

Answer

8% increase

⚡ One line for two discounts

Two discounts collapse with $d_1 + d_2 - \dfrac{d_1 d_2}{100}$ . One line, no chips.

Example

Find the single discount equal to successive discounts of 20% and 10%.

Show solution

$20 + 10 - \dfrac{20 \times 10}{100}$
$= 30 - 2 = 28$ .

Answer

28%

⚡ Work backwards for the missing change

Net chip divided by the known chip gives the unknown chip.

Example

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

Show solution

Net chip $= \dfrac{23}{20}$ , first chip $= \dfrac{5}{4}$ .
$\dfrac{23}{20} \div \dfrac{5}{4} = \dfrac{23}{25}$ , a fall of $8\%$ .

Answer

8% decrease

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding discounts: $20\% + 10\% = 30\%$ .

Use $d_1 + d_2 - \dfrac{d_1 d_2}{100} = 28\%$ .

Mistake 02

Applying the second discount to the original marked price.

The second discount falls on the already reduced price.

Mistake 03

Writing $+\dfrac{ab}{100}$ for a rise followed by a fall.

Give b a minus sign when it is a fall.

Mistake 04

Using the two-change formula once for three discounts.

Multiply three kept fractions instead.

Mistake 05

Believing the order of discounts changes the answer.

Chips multiply; multiplication order does not matter.

Quick revision

Read this the night before the exam.

Each change acts on what the last one left.
Two changes: $a + b + \dfrac{ab}{100}$ with signs.
Two discounts: $d_1 + d_2 - \dfrac{d_1 d_2}{100}$ .
Any number of changes: multiply the kept fractions.
Up $x\%$ then down $x\%$ : net fall of $\dfrac{x^2}{100}\%$ .
Missing change: net chip $\div$ known chip.

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.

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