Maths · 1 Number

1.12 Percentages

Before you start

Changing between fractions and decimals, e.g. $\frac{3}{4} = 0.75$ .
Multiplying and dividing by 10 and by 100 without a calculator.
Converting units: minutes and hours, grams and kilograms, centimetres and metres.
What a power means, e.g. ${1.05}^{3} = 1.05 \times 1.05 \times 1.05$ , and where the power key is on your calculator.

Syllabus objectives covered (5)

Calculate a given percentage of a quantity.
Express one quantity as a percentage of another.
Calculate percentage increase or decrease.
Calculate with simple and compound interest.
Calculate using reverse percentages.

Part 1 of 7

Finding a percentage of a quantity

Per cent means 'out of 100'. So $35 %$ means $\frac{35}{100}$ , which is the decimal $0.35$ .
To find a percentage of a quantity, change the percentage to a decimal and multiply. The word 'of' means multiply.
$35 % of 240 = 0.35 \times 240 = 84$
Changing a percentage to a decimal is dividing by 100. The trap is a one-digit percentage: $7 % = 0.07$ , not $0.7$ . The decimal $0.7$ is $70 %$ .
A percentage can be more than 100%. $135 % = 1.35$ , so $135 %$ of $60$ is $1.35 \times 60 = 81$ . More than 100% of something is more than the thing itself, which is a quick check that the answer makes sense.
A deposit is the part of a price you pay at the start. The rest is paid later. A 20% deposit on a $350 sofa is $70, because $0.2 \times 350 = 70$ . That leaves $280 to pay.
Without a calculator, build the percentage from pieces that are easy to find. $10 %$ is divide by 10. $1 %$ is divide by 100. $5 %$ is half of $10 %$ . $50 %$ is half, and $25 %$ is a quarter.
For $35 %$ of $240$ : $10 % = 24$ , so $30 % = 72$ . Half of $10 %$ gives $5 % = 12$ . Then $35 % = 72 + 12 = 84$ , the same as the decimal method.
Write each piece down. A method mark is for a valid method, so if your final answer is wrong, the examiner can only give that mark for working that is on the page.

Part 2 of 7

One quantity as a percentage of another

Write the first quantity as a fraction of the second, 'part over whole'. Then multiply by 100 to turn the fraction into a percentage.
$percentage = \frac{part}{whole} \times 100$
The whole is the quantity after the word 'of'. So $18$ as a percentage of $75$ is $\frac{18}{75} \times 100 = 24 %$ .
Put both quantities in the same units first. 45 minutes as a percentage of 2 hours is $\frac{45}{120} \times 100 = 37.5 %$ , because 2 hours is 120 minutes. Dividing 45 by 2 compares minutes with hours and gives a wrong answer.
The answer is more than 100% when the part is bigger than the whole. $90$ as a percentage of $60$ is $\frac{90}{60} \times 100 = 150 %$ .
Profit is the selling price minus the cost price. If you sell for less than you paid, the loss is the cost price minus the selling price.
Percentage profit is the profit as a percentage of the cost price, the amount the seller paid. Percentage loss works the same way, with the loss on top.
$percentage profit = \frac{profit}{cost price} \times 100$
Example: buy for $40 and sell for $52. The profit is $12, and $\frac{12}{40} \times 100 = 30 %$ . If you divide by the selling price instead, $\frac{12}{52} \times 100 \approx 23.1 %$ , you get a number that is not the percentage profit.

Part 3 of 7

Percentage increase and decrease

An increase of 15% keeps all of the original (100%) and adds 15% more. So the new amount is 115% of the original, which is one multiplication by $1.15$ .
A decrease of 15% leaves $100 % - 15 % = 85 %$ of the original, so you multiply by $0.85$ .
The number you multiply by is the multiplier. For a change of $r %$ :
$increase: \times (1 + \frac{r}{100}) decrease: \times (1 - \frac{r}{100})$
You can also find the 15% and then add it or take it away. Both methods are valid. The multiplier is one step instead of two, and it is the method you need for repeated changes and for reverse percentages.
A discount is a decrease in a price. A pay rise is an increase in earnings. A 4% pay rise on a salary of $31 500 gives a new salary of $32 760, because $31 500 \times 1.04 = 32 760$ .
An increase of more than 100% is possible. After a 150% increase the new amount is $100 % + 150 % = 250 %$ of the original, so the multiplier is $2.5$ .
To find a percentage change, put the change over the original amount, the value before the change:
$percentage change = \frac{new - original}{original} \times 100$
A negative answer means a decrease. A price that falls from $80 to $68 changes by $\frac{68 - 80}{80} \times 100 = - 15 %$ , which is a decrease of 15%.
Always divide by the original. Dividing by the new value finds the change as a percentage of the new value, which is a different number and not what the question asks.

Part 4 of 7

Repeated percentage change

When one change follows another, the second percentage is a percentage of the new amount, not of the original. So multiply by each multiplier in turn.
An increase of 20% followed by a decrease of 20% does not bring you back to the start:
$1.2 \times 0.8 = 0.96$
The overall multiplier $0.96$ means 96% of the original is left, an overall decrease of 4%. The 20% decrease was taken from a bigger amount than the 20% increase was, so it took away more than the increase added.
The same change repeated $n$ times is the multiplier to the power $n$ . A value that falls by 5% each year for 3 years is multiplied by ${0.95}^{3} = 0.857375$ . That is an overall fall of $14.2625 %$ , not $15 %$ .
Never add the percentages. An increase of 10% and then 20% is not 30%: $1.1 \times 1.2 = 1.32$ , which is an overall increase of 32%.

Part 5 of 7

Simple interest

Interest is money paid for the use of money. A bank pays you interest on savings, and you pay interest on a loan. The amount saved or borrowed is the principal.
The rate of interest is a percentage per year, also written 'per annum'.
With simple interest, the interest is the same every year, because it is always calculated on the original principal.
You must remember the formula: the syllabus says percentage formulas are not given. With principal $P$ , rate $R %$ per year and time $T$ years, the interest $I$ is:
$I = \frac{P \times R \times T}{100}$
The formula says ' $R %$ of $P$ , every year, for $T$ years'. If you forget it, you can build it again from that sentence.
Example: $1200 at 2.5% per year simple interest for 4 years earns $120 interest, because $\frac{1200 \times 2.5 \times 4}{100} = 120$ . The account then holds $1200 + 120 = 1320$ dollars.
$T$ must be in years. 18 months is $T = 1.5$ .
A question can give the interest and ask for $R$ , $T$ or $P$ . Put the values you know into the formula, then solve for the missing letter.

Part 6 of 7

Compound interest

With compound interest, the interest for each year is added to the account. The next year's interest is then calculated on the new, larger total.
So each year the amount is multiplied by the same multiplier. At 2.5% per year, the multiplier is $1.025$ .
After $n$ years the multiplier has been used $n$ times. With principal $P$ and rate $r %$ per year, the amount $A$ is given below. This formula is not given on the paper either.
$A = P {(1 + \frac{r}{100})}^{n}$
Example: $1200 at 2.5% per year compound interest for 4 years gives $1200 \times {1.025}^{4} = 1324.575 \dots$ , which is $1324.58 to the nearest cent. Simple interest gave $1320, because simple interest never earns interest on interest.
Read what the question asks for. The formula gives the amount, the value of the investment. The interest is the amount minus the principal: $1324.58 - 1200 = 124.58$ , so $124.58.
Keep the full calculator value until the end, and round only the final answer. Unless the question says otherwise, give a non-exact answer to 3 significant figures. A money question may ask for the nearest cent instead, so read the instruction.
To find how many whole years it takes to pass a target, try whole numbers for $n$ and write each trial down. Show the value for the last year below the target and the first year above it. Those two values prove your answer is the smallest number of years.

Part 7 of 7

Reverse percentages

Sometimes you know the amount after a percentage change and you need the original amount. This is a reverse percentage.
The change was: original $\times$ multiplier $=$ new amount. To undo a multiplication, divide:
$original = \frac{new amount}{multiplier}$
Example: a coat costs $68 after a 15% discount. The sale price is 85% of the original price, so the original price was $68 \div 0.85 = 80$ dollars.
The trap: do not find 15% of $68 and add it on. That gives $78.20, which is wrong. The 15% was 15% of the original price, not 15% of the sale price.
The cost price from the selling price and the percentage profit works the same way. A seller makes a 15% profit by selling at $96.60. The selling price is 115% of the cost price, so the cost price is $96.60 \div 1.15 = 84$ dollars.
To undo a repeated change, divide by the complete multiplier. After 2 years of 5% compound interest an account holds $4630.50, so the principal was $4630.50 \div {1.05}^{2} = 4200$ dollars.
Check by going forwards: for the cost price above, $84 \times 1.15 = 96.60$ . If the forward check does not give the amount in the question, your reverse step is wrong.

Worked examples

Example 1. (a) Without a calculator, work out $17.5 %$ of $480$ . (b) Express 1.8 kg as a percentage of 750 g.

Show the working

(a) Build $17.5 %$ from easy pieces: $17.5 % = 10 % + 5 % + 2.5 %$ . Each piece is half of the one before it.
$10 %$ of $480$ is $48$ . Half of that, $5 %$ , is $24$ . Half again, $2.5 %$ , is $12$ .
$17.5 % of 480 = 48 + 24 + 12 = 84$
Check with the decimal method: $0.175 \times 480 = 84$ .
(b) The two quantities are in different units. Change both to grams: 1.8 kg is 1800 g.
The quantity after 'of' is 750 g, so it goes on the bottom of the fraction.
$\frac{1800}{750} \times 100 = 240 %$
The answer is more than 100% because 1.8 kg is more than 750 g. That is a useful check.

Answer: (a) $84$ (b) $240 %$

Example 2. A pair of trainers costs $80. The price increases by 15%. Later, the new price decreases by 15%. (a) Find the final price. (b) Find the overall percentage change from the original price.

Show the working

(a) The multiplier for a 15% increase is $1.15$ . The multiplier for a 15% decrease is $0.85$ .
The decrease acts on the increased price, so multiply by each multiplier in turn.
$80 \times 1.15 = 92, 92 \times 0.85 = 78.2$
Write money with two decimal places: $78.20.
(b) The original price, $80, goes on the bottom of the fraction.
$\frac{78.2 - 80}{80} \times 100 = - 2.25 %$
The negative sign means a decrease. Check with the overall multiplier: $1.15 \times 0.85 = 0.9775$ , which is $97.75 %$ of the original, a decrease of $2.25 %$ .
This is why +15% and −15% do not cancel: the 15% decrease was taken from $92, not from $80.

Answer: (a) $78.20 (b) a decrease of $2.25 %$

Example 3. Kofi invests $2500 at 3.2% per year. (a) Calculate the simple interest after 5 years. (b) Calculate the value of the investment after 5 years with compound interest. Give your answer correct to the nearest cent. (c) With compound interest, find the smallest whole number of years after which the investment is worth more than $3000.

Show the working

(a) Recall the simple interest formula. Substitute $P = 2500$ , $R = 3.2$ and $T = 5$ .
$I = \frac{2500 \times 3.2 \times 5}{100} = 400$
So the simple interest is $400. That is $80 each year for 5 years.
(b) With compound interest the multiplier each year is $1 + \frac{3.2}{100} = 1.032$ , used 5 times.
$2500 \times {1.032}^{5} = 2926.432 \dots$
To the nearest cent, the value is $2926.43. The interest is $426.43, more than the $400 in (a), because each year's interest also earns interest.
(c) Try whole numbers of years and write each value down.
$2500 \times {1.032}^{5} \approx 2926.43, 2500 \times {1.032}^{6} \approx 3020.08$
After 5 years the value is less than $3000. After 6 years it is more. Showing both values proves that 6 is the smallest number of years.

Answer: (a) $400 (b) $2926.43 (c) 6 years

Example 4. In a sale, a laptop costs $572 after a 12% discount. (a) Find the original price of the laptop. (b) At the original price, the shop makes a 30% profit on the amount it paid for the laptop. Find the amount the shop paid. (c) Find the percentage profit the shop makes when it sells the laptop at the sale price.

Show the working

(a) A 12% discount leaves 88% of the original price, so sale price = original price $\times 0.88$ .
Undo the multiplication by dividing by the multiplier.
$572 \div 0.88 = 650$
Check forwards: $650 \times 0.88 = 572$ , which matches the sale price. Note that 12% of $572 added on gives $640.64, not $650. That is the trap.
(b) A 30% profit means the original price is 130% of the amount the shop paid. Divide by $1.3$ .
$650 \div 1.3 = 500$
(c) At the sale price the profit is $572 - 500 = 72$ dollars. Compare it with the cost price, $500, not with a selling price.
$\frac{72}{500} \times 100 = 14.4 %$

Answer: (a) $650 (b) $500 (c) $14.4 %$

Where people go wrong

Mistake: Finding a reverse percentage by taking the percentage of the new amount and adding it back, e.g. a $68 sale price after a 15% discount turned into $68 + 0.15 \times 68 = 78.20$ .

Instead: Divide the new amount by the multiplier: $68 \div 0.85 = 80$ .

The 15% was 15% of the original price. 15% of the smaller sale price is a smaller amount, so adding it on does not get back to the original.

Mistake: Dividing by the new value when finding a percentage change, or by the selling price when finding a percentage profit.

Instead: Put the change over the original value: the old price, or the cost price.

A percentage change is a change compared with where you started. A different bottom number answers a different question.

Mistake: Adding the percentages in a repeated change, e.g. saying +10% then +20% is +30%, or that 3 years at 4% compound interest is 12% interest.

Instead: Multiply the multipliers: $1.1 \times 1.2 = 1.32$ , so +32%. Three years at 4% compound interest is $\times {1.04}^{3}$ .

Each change acts on the result of the change before it, not on the original amount.

Mistake: Giving the amount when the question asks for the interest, or the interest when it asks for the value of the investment.

Instead: The compound interest formula gives the amount. Interest = amount − principal. Simple interest $\frac{P R T}{100}$ gives the interest. Total = principal + interest.

The two formulas give different things, and an answer of the wrong quantity is a wrong answer.

Mistake: Writing one quantity as a percentage of another when they are in different units, e.g. 45 minutes of 2 hours as $\frac{45}{2} \times 100$ .

Instead: Change to the same units first: $\frac{45}{120} \times 100 = 37.5 %$ .

A percentage compares two amounts of the same thing. Minutes over hours is not a fraction of anything.

Check yourself

Without a calculator, work out $15 %$ of $360$ .
Answer
$54$ . $10 % = 36$ and $5 % = 18$ , so $15 % = 36 + 18 = 54$ .
A scooter costs $640. You pay a 15% deposit, then the rest in 8 equal monthly payments. Find each monthly payment.
Answer
$68. The deposit is $0.15 \times 640 = 96$ dollars. That leaves $640 - 96 = 544$ dollars, and $544 \div 8 = 68$ .
Express 40 cm as a percentage of 2.5 m.
Answer
$16 %$ . In the same units, 2.5 m is 250 cm, and $\frac{40}{250} \times 100 = 16$ .
Write down the multiplier for (a) an increase of 7%, (b) a decrease of 7%, (c) an increase of 150%.
Answer
(a) $1.07$ (b) $0.93$ (c) $2.5$
A trader buys 40 bags at $12.50 each. She sells 30 of them at $18 each and the other 10 at $9 each. Calculate her percentage profit.
Answer
$26 %$ . Cost: $40 \times 12.50 = 500$ dollars. Income: $30 \times 18 + 10 \times 9 = 630$ dollars. Profit: $130, and $\frac{130}{500} \times 100 = 26$ .
$1500 earns $180 simple interest in 4 years. Find the rate of interest per year.
Answer
3% per year. $180 = \frac{1500 \times R \times 4}{100}$ , so $180 = 60 R$ and $R = 3$ .
$800 is invested at 4% per year compound interest for 3 years. Find the interest earned, correct to the nearest cent.
Answer
$99.89. The amount is $800 \times {1.04}^{3} = 899.8912$ , so the interest is $899.8912 - 800 = 99.8912$ , which is $99.89.
After a 20% pay rise, Ana earns $33 600 a year. Find her salary before the rise.
Answer
$28 000, because her new salary is 120% of the old one and $33 600 \div 1.2 = 28 000$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A shop reduces all of its prices by 15% in a sale.

(a) The normal price of a phone is $340. Calculate the sale price of the phone. 2 marks

Mark scheme

M1 340 × 0.85 oe, or 340 − 0.15 × 340
A1 289 cao

(b) The shop paid $250 for the phone. Calculate the percentage profit the shop makes when it sells the phone at the sale price. 2 marks

Mark scheme

M1 (their (a) − 250) ÷ 250 × 100 oe; FT their (a)
A1 15.6

(c) The sale price of a tablet is $221. Calculate the normal price of the tablet. 3 marks

Mark scheme

M2 221 ÷ 0.85 oe, or M1 for 221 associated with 85% seen, e.g. 0.85x = 221
A1 260 cao

Question 2. Mai has $4000 to invest for 6 years. Option A pays simple interest at 3.5% per year. Option B pays compound interest at 3.2% per year.

(a) Calculate the total interest Mai earns with Option A. 2 marks

Mark scheme

M1 4000 × 3.5 × 6 ÷ 100 oe
A1 840 cao

(b) Calculate the value of Mai's investment after 6 years with Option B. Give your answer correct to the nearest cent. 2 marks

Mark scheme

M1 4000 × 1.032⁶ oe
A1 4832.13 cao

(c) Mai says, 'Compound interest always earns more than simple interest.' Use your answers to parts (a) and (b) to explain whether Mai is right for these two options. 2 marks

Mark scheme

M1 their (b) − 4000, i.e. 832.13 seen, or 4000 + their (a), i.e. 4840 seen, compared with their (b); FT their (a) and their (b)
A1 states that Option A earns more (840 > 832.13, or 4840 > 4832.13), so Mai is not right for these options oe; dep on M1

(d) Mai chooses Option B and leaves all of the money in the account. Find the number of complete years until the value of her investment is first more than $5000. 3 marks

Mark scheme

M1 4000 × 1.032ⁿ evaluated for at least one whole number n ≥ 7, or 1.032ⁿ > 1.25 oe
M1 4986.75… (awrt 4987) for 7 years and 5146.33… (awrt 5146) for 8 years, both seen; or n = 7.08… from a correct logarithm or graphical method
A1 8 cao

Question 3.

(a) Work out $35 %$ of $260$ . 2 marks No calculator

Mark scheme

M1 30% = 78 and 5% = 13 seen (e.g. 3 × 26 + 13), or 35 ÷ 100 × 260 oe
A1 91 cao

(b) Show that an increase of 20% followed by a decrease of 20% gives an overall decrease of 4%. 2 marks No calculator

Mark scheme

M1 1.2 × 0.8 oe, or a correct worked example, e.g. 100 → 120 → 96
A1 0.96 (or 96) with a statement that 96% of the original is left, so the overall decrease is 4%; nfww

How this is examined

Finding the original amount when you are given the amount after a percentage change: A Calculate or Find part in a context: a sale price after a discount, a salary after a rise, or a selling price with its percentage profit, worth about 3 marks The given amount matched to the correct percentage (a sale price after a 15% discount is 85% of the original), then the division by the multiplier written down, e.g. $68 \div 0.85$ , then the original amount, $80$ . Finding the percentage of the given amount and adding it on is not a valid method.
Compound interest as a power of the multiplier, and the number of years to pass a target: A Calculate part for the value after $n$ years, followed by a Find part for the number of whole years, each worth 2 or 3 marks The calculation written with its numbers, e.g. $2500 \times {1.032}^{5}$ , then the value to the accuracy the question asks for. For the number of years: the values for the last year below the target and the first year above it, both written down, then the whole number of years.
Simple interest from a formula you have to remember: A 2-mark Calculate for the interest, or for the rate, the time or the principal when the interest is given $\frac{P \times R \times T}{100}$ with the numbers in it, e.g. $\frac{2500 \times 3.2 \times 5}{100}$ , then the value. Then the quantity the question asks for: the interest, or the total amount (principal + interest).
A percentage profit, loss or change, measured against the original amount: A 2-mark Calculate or Work out in a buying-and-selling or price-change context The change over the original, times 100, e.g. $\frac{572 - 500}{500} \times 100$ with the cost price on the bottom, then the percentage. A fraction with the selling price or the new value on the bottom is not a correct method.
Showing that two percentage changes combine into one overall change: A 2-mark Show (that) with the overall change stated in the question The two multipliers multiplied, e.g. $1.15 \times 0.85 = 0.9775$ , then a sentence that turns the result into the stated change ( $97.75 %$ is left, so a $2.25 %$ decrease). In a Show (that), writing the given result without the working that leads to it does not show it.
A percentage of an amount without a calculator: A 1- or 2-mark Work out on the non-calculator paper Each piece written down with what it is a percentage of, e.g. $10 %$ of $480 = 48$ and $5 % = 24$ , then the total.

Words to know

Percentage: A number out of 100. $35 %$ means $\frac{35}{100}$ , or $0.35$ .
Multiplier: The number you multiply by to make a percentage change: $1.15$ for a 15% increase, $0.85$ for a 15% decrease.
Principal: The amount of money first invested or borrowed.
Simple interest: Interest calculated on the original principal only, so it is the same every year: $I = \frac{P R T}{100}$ .
Compound interest: Interest added to the account each year, so later interest is also earned on earlier interest: $A = P {(1 + \frac{r}{100})}^{n}$ .
Cost price: The amount a seller paid for an item.
Selling price: The amount an item is sold for.
Profit and loss: Profit is selling price minus cost price. Loss is cost price minus selling price, when the item sells for less than it cost.
Discount: A reduction in a price, given as an amount of money or as a percentage of the original price.
Deposit: The part of a price paid at the start, with the rest paid later.
Reverse percentage: Finding the original amount from the amount after a percentage change, by dividing by the multiplier.

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