Introduction
Yo! Welcome to the ultimate guide for IGCSE Number. I know, I know, 'maths' can sometimes feel like a final boss battle on the hardest difficulty. But trust me, once you get the cheat codes, it's actually super satisfying. We're going to break down everything from percentages to fractions like you're learning a new TikTok dance – one step at a time. By the end of this, you'll be crunching numbers faster than you can scroll through your feed. Let's get this bread! 🍞
Understanding the Order of Operations (BODMAS/PEMDAS)C1.5, C1.6
Okay, so first things first. You can't just solve a math problem from left to right like you're reading a text. There's an official order, a set of rules that everyone follows. Think of it like the unspoken rules of Snapchat – you just know what to do. In maths, we use BODMAS (or PEMDAS, they're the same vibe).<br><br>It stands for:<br>*B*rackets<br>*O*rders (like powers/indices $x^2$ and square roots $\sqrt{x}$)<br>*D*ivision & Multiplication (these are partners, you do them left-to-right)<br>*A*ddition & Subtraction (also partners, done left-to-right)<br><br>Getting this order wrong is like putting your shoes on before your socks. It just doesn't work. Stick to the rules, and you'll always get the right answer. 🎯<br>
Worked Example: Applying the Order of OperationsC1.5, C1.6
Problem Your friend sends you this calculation to solve: $10 + (3 \times 2^2) \div 6 - 4$. Let's break it down.
Step 1 First up, Brackets. We solve what's inside the $(...)$ first. Inside the brackets, we have an Order ($2^2$), so we do that. $$2^2 = 4$$
Step 2 Now we finish the calculation inside the brackets: $3 \times 4$. $$10 + (12) \div 6 - 4$$
Step 3 Brackets are done. Next is Division and Multiplication. We have a division: $12 \div 6$. $$10 + 2 - 4$$
Step 4 Last up is Addition and Subtraction. We work from left to right. First, $10 + 2$. $$12 - 4$$
Step 5 And the final step... $$12 - 4 = 8$$
Calculations with Mixed Numbers, Fractions, and DecimalsC1.4, C1.13
Dealing with fractions, mixed numbers, and decimals in one problem can feel like trying to get your friend group to agree on what to watch. The secret? Convert them to the same format! Most of the time, it's easiest to turn everything into a decimal or an improper fraction. For a mixed number like $2\frac{1}{4}$, you can think of it as $2 + \frac{1}{4}$. To convert it to an improper fraction, you do $(2 \times 4) + 1$ for the top number, keeping the bottom number: $\frac{9}{4}$. To convert to a decimal, just do $1 \div 4 = 0.25$ and add it to the whole number: $2.25$. Easy!
Worked Example: Unifying Number Formats in CalculationsC1.4, C1.13
Problem You're working a shift at a café and need to calculate the total litres of milk used. You used $3\frac{1}{2}$ litres in the morning and $1.75$ litres in the afternoon. How much did you use in total?
Step 1 Let's convert the mixed number to a decimal. We know that $\frac{1}{2}$ is the same as $0.5$. $$3\frac{1}{2} = 3 + 0.5 = 3.5$$
Step 2 Now that both numbers are decimals, it's a simple addition problem. $$3.5 + 1.75 = 5.25$$
Step 3 So, you used $5.25$ litres of milk. You could also write this as $5\frac{1}{4}$ litres. $$Total = 5.25 \text{ litres}$$
Comparing and Ordering Numbers in Different FormatsC1.5, C1.6
Imagine you're comparing stats in a video game. One is a fraction, one is a percentage, one is a decimal. Who's winning? To order numbers, you need to see them in the same format. The easiest way is to convert everything to a decimal. To convert a percentage to a decimal, just divide by 100 (e.g., $75\% = 75 \div 100 = 0.75$). To convert a fraction to a decimal, divide the top number (numerator) by the bottom number (denominator) (e.g., $\frac{3}{5} = 3 \div 5 = 0.6$). Once they're all decimals, lining them up from smallest to largest is a piece of cake. 
Worked Example: Ranking Different Number FormatsC1.4, C1.13
Problem You're comparing phone battery levels with your friends. Your phone is at $\frac{4}{5}$, Sarah's is at $85\%$, and Ben's is at $0.82$. Who has the most battery? Order them from largest to smallest.
Step 1 Let's convert everyone's battery level to a decimal so we can compare them easily. $$\text{You: } \frac{4}{5} = 4 \div 5 = 0.80$$
Step 2 Next, convert Sarah's percentage to a decimal by dividing by 100. $$\text{Sarah: } 85\% = 85 \div 100 = 0.85$$
Step 3 Ben's is already a decimal, so we don't need to change it. Now we compare the three decimals: $0.80$, $0.85$, and $0.82$. $$0.85 > 0.82 > 0.80$$
Step 4 So, Sarah has the most battery. The order from largest to smallest is Sarah ($85\%$), then Ben ($0.82$), then you ($\frac{4}{5}$). $$85\% > 0.82 > \frac{4}{5}$$
Calculating a Percentage of a QuantityC1.4, C1.13
This is one of the most useful math skills you'll ever learn, for real. It's how you figure out discounts, tips, and more. The key is to remember that 'percent' means 'out of 100'. So $20\%$ is just $\frac{20}{100}$. The word 'of' in maths almost always means multiply. So, to find a percentage of a quantity, you turn the percentage into a decimal or fraction and multiply it by the amount. A pro-gamer move is to use a decimal multiplier: just convert the percentage to a decimal ($25\% \rightarrow 0.25$) and multiply. It's way faster on a calculator.
Worked Example: Calculating a Percentage DiscountC1.4, C1.13
Problem You've been saving up for a pair of sneakers that cost $£120$. You find a store that's offering a $30\%$ discount. How much money do you save?
Step 1 We need to calculate $30\%$ of $£120$. First, convert the percentage to a decimal. This is our multiplier. $$30\% = 30 \div 100 = 0.30$$
Step 2 Now, multiply this decimal by the total cost of the sneakers. $$0.30 \times £120 = £36$$
Step 3 Boom! You save $£36$. The final price would be $£120 - £36 = £84$. Score! $$\text{Discount} = £36$$
Calculating Percentage Increase and DecreaseC1.4, C1.13
Life is full of ups and downs, and so are numbers! When you want to increase an amount by a percentage (like a pay rise), you start with $100\%$ (the original amount) and add the percentage increase. To decrease an amount (like a price drop), you start with $100\%$ and subtract the percentage decrease. Then, you use this new percentage as a decimal multiplier. For example, a $15\%$ increase means you find $115\%$ of the original ($1.15$ multiplier). A $10\%$ decrease means you find $90\%$ of the original ($0.90$ multiplier). It's a super quick way to find the new amount directly.
Worked Example: Calculating a Price IncreaseC1.4, C1.13
Problem Your music streaming subscription is currently $£9.99$ per month. They announce a $10\%$ price increase. What will the new monthly cost be?
Step 1 This is a percentage increase. We start with the original $100\%$ and add the $10\%$ increase. $$100\% + 10\% = 110\%$$
Step 2 Convert this new percentage into a decimal multiplier. $$110\% = 110 \div 100 = 1.10$$
Step 3 Multiply the original price by this multiplier to find the new price. Remember to round your answer to two decimal places for money. $$£9.99 \times 1.10 = £10.989$$
Step 4 Rounding to the nearest penny, the new price is $£10.99$. $$\text{New Price} = £10.99$$
Finding the Percentage Change Between Two ValuesC1.4, C1.13
Ever wonder by what percentage your followers on Instagram grew? Or how much your score in a game improved? That's percentage change! It tells you how much something has changed in relation to its starting value. The formula is your best friend here, so memorize it:<br><br>$$ \text{Percentage Change} = \frac{\text{Change}}{\text{Original}} \times 100 $$<br><br>The 'Change' is just the difference between the new value and the original value. This one formula works for both increases and decreases!
Worked Example: Calculating Percentage GrowthC1.4, C1.13
Problem Last week you had 450 followers on your TikTok account. This week, you have 540. What is the percentage increase in your followers?
Step 1 First, let's find the 'Change' in followers. This is the new amount minus the original amount. $$\text{Change} = 540 - 450 = 90$$
Step 2 Next, we use the percentage change formula. We divide the change (90) by the original number of followers (450). $$\frac{90}{450} = 0.2$$
Step 3 Finally, multiply by 100 to turn the decimal into a percentage. $$0.2 \times 100 = 20\%$$
Step 4 You had a $20\%$ increase in followers! Time to go viral. ✨ $$\text{Percentage Increase} = 20\%$$
Introduction to Reverse PercentagesC1.4, C1.13
This one sounds tricky, but it's just working backwards. It's when you know the final price after a discount or increase, but you need to find the original price. For example, a jacket is $£60$ in a '20% off' sale. That $£60$ isn't $100\%$ of the price, it's the price after the 20% was removed, so it's $100\% - 20\% = 80\%$ of the original price. You can set up an equation: $80\% \times \text{Original Price} = £60$. Then you just solve for the original price! 
Worked Example: Finding the Original ValueC1.4, C1.13
Problem A concert ticket costs $£45$ after a $10\%$ booking fee was added. What was the original price of the ticket before the fee?
Step 1 The final price includes the original $100\%$ plus a $10\%$ fee. So, the $£45$ represents $110\%$ of the original price. $$100\% + 10\% = 110\%$$
Step 2 We can set up an equation. Let 'P' be the original price. We know that $110\%$ of P is $£45$. Let's write $110\%$ as a decimal. $$1.10 \times P = £45$$
Step 3 To find P, we need to do the reverse of multiplying. We divide both sides by 1.10. $$P = \frac{£45}{1.10} = £40.90909...$$
Step 4 Rounding to two decimal places for money, the original ticket price was $£40.91$. $$\text{Original Price} = £40.91$$
Calculating Successive Percentage ChangesC1.4, C1.13
What happens when a shop has a 25% off sale, and you have a student discount card for an extra 10% off? Warning: it's NOT a 35% discount! You have to apply the changes one after the other. It's a combo move. First, you calculate the price after the first discount. Then, you apply the second discount to that new price. Using decimal multipliers makes this super fast. Just multiply the original amount by the first multiplier, then by the second multiplier.
Worked Example: Applying Successive DiscountsC1.4, C1.13
Problem A graphics card costs $£500$. It's in a $20\%$ off sale. You also have a coupon for an additional $10\%$ off the sale price. What is the final price you pay?
Step 1 Let's find the decimal multipliers for both discounts. A $20\%$ discount means you pay $80\%$ ($100\% - 20\%$). A $10\%$ discount means you pay $90\%$ ($100\% - 10\%$). $$\text{Multiplier 1} = 0.80 \quad \text{and} \quad \text{Multiplier 2} = 0.90$$
Step 2 Now, apply the first discount to the original price. $$£500 \times 0.80 = £400$$
Step 3 This $£400$ is the sale price. Now apply the second discount to this new price. $$£400 \times 0.90 = £360$$
Step 4 The final price is $£360$. Alternatively, you could multiply the multipliers together first ($0.8 \times 0.9 = 0.72$) and then multiply by the original price ($£500 \times 0.72 = £360$). Same result, fewer steps! $$\text{Final Price} = £360$$
Introduction to Simple InterestC1.15, C1.16
Okay, let's talk money. 💰 When you put money in a savings account, the bank pays you extra for keeping it there. This is called interest. Simple interest is the most basic kind. You get a fixed percentage of your original amount (the 'principal') every year. The formula is another one to lock in your memory:<br><br>$$ I = P \times R \times T $$<br><br>Where:<br>*I* = Interest earned<br>*P* = Principal (the starting amount)<br>*R* = Rate (the percentage as a decimal)<br>*T* = Time (in years)<br><br>It's called 'simple' because the interest you earn doesn't start earning its own interest. It's always calculated on that starting amount.
Worked Example: Calculating Simple Interest Over TimeC1.15, C1.16
Problem You saved $£800$ from your part-time job and put it in a savings account that pays $3\%$ simple interest per year. How much interest will you have earned after 4 years?
Step 1 Let's identify our P, R, and T from the problem. $$P = £800, \quad R = 3\% = 0.03, \quad T = 4 \text{ years}$$
Step 2 Now, we plug these values into the simple interest formula: $I = P \times R \times T$. $$I = 800 \times 0.03 \times 4$$
Step 3 Calculate the result. $$I = £96$$
Step 4 After 4 years, you will have earned $£96$ in interest. The total amount in your account would be $£800 + £96 = £896$. Not bad for doing nothing! $$\text{Interest Earned} = £96$$
Number Types and PropertiesC1.1
Alright, let's talk numbers! It might seem basic, but trust me, understanding the different 'squads' of numbers is a total game-changer for IGCSE Maths. Think of it like character classes in a video game – each has its own unique skills and properties.<br><br>First up, we have Natural Numbers (1, 2, 3...). These are your basic counting numbers, like when you're counting your followers. Add zero and all the negatives, and you get Integers (..., -2, -1, 0, 1, 2,...). Your bank account after a shopping spree? Definitely in the negative integers zone. 😬<br><br>Now for the VIPs: Prime Numbers. These are numbers greater than 1 that can only be divided by 1 and themselves (like 2, 3, 5, 7, 11...). They're the building blocks of all other numbers. Speaking of building, we have Square Numbers ($1, 4, 9, 16...$ which are $1^2, 2^2, 3^2, 4^2...$) and Cube Numbers ($1, 8, 27, 64...$ which are $1^3, 2^3, 3^3, 4^3...$). Then there are Factors (numbers that divide into another number perfectly) and Multiples (basically a number's times table). A number's factors are like its core friend group, while its multiples are like its ever-expanding list of followers.<br><br>This leads us to two super important ideas: the Highest Common Factor (HCF) and the Lowest Common Multiple (LCM). The HCF is the biggest factor that two numbers share – useful for simplifying fractions or splitting things into equal groups. The LCM is the smallest number that is a multiple of both numbers – perfect for figuring out when two things will happen at the same time. To find these, we often use Prime Factorisation, which is breaking a number down into a product of its prime factors. 
.<br><br>Finally, we have Rational Numbers (any number that can be written as a fraction, like $\frac{1}{2}$, 7, or 0.25) and Irrational Numbers (numbers that can't, like $\pi$ or $\sqrt{2}$ – their decimals go on forever without repeating!). And don't forget the Reciprocal, which is just 1 divided by the number. The reciprocal of 5 is $\frac{1}{5}$. It's like flipping the number upside down. Mastering these types will give you a massive power-up for the rest of the course! 💪
Worked Example: Prime Factorisation, HCF, and LCMC1.1
Problem You're organising a gaming tournament party. You have 72 mini sausage rolls and 90 spring rolls. You want to create identical snack plates for your friends with the maximum number of plates possible and no leftovers.<br>a) What is the greatest number of snack plates you can make?<br>b) If two separate prize draws happen every 72 seconds and 90 seconds, after how many seconds will they first happen at the same time?
Step 1 First, let's break down our numbers into their prime factors. This is like finding the basic DNA of 72 and 90. We'll use factor trees to make it visual. $$72 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9 = 2^3 \times 3^2$$<br>$$90 = 2 \times 45 = 2 \times 3 \times 15 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$$
Step 2 To find the greatest number of snack plates (part a), we need the Highest Common Factor (HCF). We look at the prime factors and multiply the lowest power of each factor that they both share. $$\text{Common factors are } 2 \text{ and } 3^2.<br>\text{HCF} = 2^1 \times 3^2 = 2 \times 9 = 18$$<br>So, you can make a maximum of 18 identical snack plates!
Step 3 Now for the prize draws (part b), we need to find when they'll sync up. This means finding the Lowest Common Multiple (LCM). To find the LCM, we multiply the highest power of all the prime factors from both numbers. $$\text{Highest power of 2 is } 2^3.<br>\text{Highest power of 3 is } 3^2.<br>\text{Highest power of 5 is } 5^1.<br>\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$$
Step 4 Let's finalise the answers. The HCF gives us the answer for the snack plates, and the LCM gives us the answer for the prize draws. a) The greatest number of snack plates is 18. Each plate will have $72 \div 18 = 4$ sausage rolls and $90 \div 18 = 5$ spring rolls.<br>b) The prize draws will happen at the same time after 360 seconds (or 6 minutes).
Powers, Roots, and IndicesC1.3, C1.7
Alright, let's talk about Powers and Roots, or as they're sometimes called, Indices. Think of a power like a 'level-up' in a game or a multiplier for your streak on Snapchat. The little number at the top, the index, tells you how many times to multiply the main number (the base) by itself. So, $7^2$ isn't $7 \times 2$, it's $7 \times 7$, which is 49. This is called '7 squared'. Similarly, $2^3$ (or '2 cubed') is $2 \times 2 \times 2$, which equals 8.<br><br>For your IGCSE, you basically need to have a playlist of key squares and cubes on repeat in your head. You need to know the squares of all the numbers from 1 to 15 (like $13^2 = 169$) and the cubes of 1, 2, 3, 4, 5, and 10 (like $4^3 = 64$). 
<br><br>Now, what about roots? Roots are the reverse operation. They're like hitting 'undo' or finding out the original number before it was powered up. The square root ($\sqrt{...}$) asks: 'What number, when multiplied by itself, gives me this number?' So, $\sqrt{144}$ is 12. A cube root ($\sqrt[3]{...}$) asks: 'What number, multiplied by itself twice, gives me this?' So, $\sqrt[3]{27}$ is 3, because $3 \times 3 \times 3 = 27$. The exam will often mix these up in a single question, asking you to calculate something like $5^2 \times \sqrt[3]{8}$. Just stay calm, solve each part separately ($5^2 = 25$ and $\sqrt[3]{8} = 2$), and then do the final operation ($25 \times 2 = 50$). You've got this!
Worked Example: Calculating with Powers and RootsC1.3, C1.7
Problem Without using a calculator, find the value of $12^2 + \sqrt[3]{125} - \sqrt{49}$.
Step 1 Okay, let's break this expression down. It looks complicated, but it's just three mini-problems in one. We need to handle the square ($12^2$), the cube root ($\sqrt[3]{125}$), and the square root ($\sqrt{49}$) separately before we combine them. $$ (12^2) + (\sqrt[3]{125}) - (\sqrt{49}) $$
Step 2 First up, the square. $12^2$ means $12 \times 12$. This is one of the key squares you need to memorise for the exam! $$ 12^2 = 12 \times 12 = 144 $$
Step 3 Next, the cube root. We're looking for the number that, when multiplied by itself three times, gives 125. Since it ends in a 5, the answer is probably 5. Let's check: $5 \times 5 \times 5 = 25 \times 5 = 125$. Perfect. This is another one to lock in your memory. $$ \sqrt[3]{125} = 5 $$
Step 4 Last of the mini-problems: the square root. What number times itself equals 49? That's a classic from your times tables. $$ \sqrt{49} = 7 $$
Step 5 Now we're on the final level. We just substitute our answers back into the original expression and solve it like a normal calculation. Follow the order of operations. $$ 144 + 5 - 7 = 149 - 7 = 142 $$
Equivalence and Conversion of Fractions, Decimals, and PercentagesC1.4, C1.13
Alright, let's break down fractions, decimals, and percentages. Think of them as three different ways to say the same thing, like posting a pic on Insta, sending it on Snap, and using it as your Discord profile pic. It's the same content, just a different format. A fraction is all about parts of a whole, like $\frac{3}{4}$ of a pizza. We have proper fractions where the top number (numerator) is smaller than the bottom (denominator), like $\frac{1}{2}$. Then there are improper fractions where the top is bigger, like $\frac{5}{2}$, which is more than one whole. You can write that as a mixed number, $2\frac{1}{2}$. Your calculator will probably show it as a decimal, which is a number with a decimal point, like $2.5$. Finally, a percentage is just a special fraction out of 100, like $50\%$.<br><br>The real skill is being able to switch between these forms seamlessly. It's like a quick-change in a video game. 
To go from a fraction to a decimal, just divide the top by the bottom: $\frac{3}{4}$ becomes $3 \div 4 = 0.75$. To turn a decimal into a percentage, just multiply by 100 (or slide the decimal point two places to the right): $0.75$ becomes $75\%$. Going backwards? To get a percentage to a decimal, divide by 100 (slide the decimal two places left): $75\%$ becomes $0.75$. And for percentage to a fraction, stick the number over 100 and simplify. $75\% = \frac{75}{100}$. Both 75 and 100 can be divided by 25, so we simplify it to $\frac{3}{4}$. Always simplify your fractions—it’s like closing unused apps to make your phone run faster. It’s just cleaner!
Worked Example: Multi-Step Percentage DecreaseC1.4, C1.13
Problem You want to buy a new headset that originally costs $80. A store is having a 15% off sale. Because you have a loyalty card, you get an additional 10% off the sale price. How much do you actually pay for the headset?
Step 1 First, let's figure out the initial 15% discount. The easiest way is to convert the percentage to a decimal and multiply it by the original price. $$15\% = 0.15 \\ \text{Discount} = 0.15 \times \$80 = \$12$$
Step 2 Now, subtract that discount from the original price to find the first sale price. This is the price before your extra loyalty discount. $$\text{Sale Price} = \$80 - \$12 = \$68$$
Step 3 Super important: The next 10% discount is taken from the new sale price ($68), not the original $80. Let's calculate this second discount. $$10\% = 0.10 \\ \text{Second Discount} = 0.10 \times \$68 = \$6.80$$
Step 4 Finally, subtract the second discount from the sale price to get the final price you'll pay at the checkout. GG! 🎮 $$\text{Final Price} = \$68 - \$6.80 = \$61.20$$
Ordering Quantities by MagnitudeC1.5, C1.6
Alright, let's talk about ordering numbers. Honestly, it's something you do all the time without even thinking about it. When you're checking a gaming leaderboard, scrolling through TikTok to see which video has more views, or figuring out if you have enough money for a snack, you're ordering quantities. In math, we just have a specific set of symbols to make it official. Let's break them down:<br><br>First, the basics: $=$ means 'is equal to' and $≠$ means 'is not equal to'. Simple enough.<br><br>Now for the cool kids, the inequalities. Think of the symbols $>$ (greater than) and $<$ (less than) as a hungry alligator that always wants to eat the bigger number. So, in $10 > 5$, the alligator's mouth is open towards the 10. Easy peasy. The symbols $≥$ (greater than or equal to) and $≤$ (less than or equal to) are just a little extra. For example, to get your driver's license, your age must be $≥ 17$. You can be 17 or older. To get a student discount, you might need to be $≤ 18$, meaning 18 or younger. 
The trickiest part is with negative numbers – remember that $-10$ is smaller than $-2$ because it's further to the left on the number line. When you have to order a mix of fractions, decimals, and integers, the best strategy is to convert them all to decimals first. It makes comparing them way easier!
Worked Example: Comparing Financial DataC1.5, C1.6
Problem You and your friends are comparing your weekly finances. You have a balance of $£15.50$. Your friend Sam spent more than he had and has a balance of $-£5$. Your other friend, Chloe, has saved exactly half of the $£20$ she was given, and David has a balance of $-£5.25$. Put these four amounts in ascending order (from smallest to largest).
Step 1 First up, let's list all the financial amounts we need to compare. We have your balance, Sam's, Chloe's, and David's. One of them needs a quick calculation. Your Balance: $£15.50$<br>Sam's Balance: $-£5$<br>Chloe's Balance: $\frac{1}{2} \times £20$<br>David's Balance: $-£5.25$
Step 2 To make them easy to compare, let's make sure every value is in the same format. The only one that isn't a simple decimal or integer is Chloe's, so let's calculate that. $$ \text{Chloe's Balance} = \frac{1}{2} \times 20 = £10 $$
Step 3 Now we have a clean list of decimals and integers. This is much easier to work with! The numbers to order are: $15.50, -5, 10, -5.25$
Step 4 Let's think about a number line. The negative numbers are the smallest. The more negative a number is, the smaller it is. So, $-5.25$ is smaller than $-5$. Then we have the positive numbers, where $10$ is smaller than $15.50$. Smallest to Largest: $-5.25$, then $-5$, then $10$, then $15.50$
Step 5 Finally, let's write it out as one continuous inequality, just like the question asks. This shows the final, ranked order of everyone's money. $$ -5.25 < -5 < 10 < 15.50 $$
Standard Form and CalculationsC1.8, C1.9
Alright, let's talk about Standard Form. Think of it as a super-compressed file, but for numbers. You know how you might zip a huge folder of photos before sending it? Standard form does that for massive or microscopic numbers, making them way easier to handle. The official 'rule' is that we write any number in the form $A \times 10^n$, where A is a number between 1 and 10 (so $1 \le A < 10$) and n is just an integer (a whole number, positive or negative).<br><br>So, how do we get there? Imagine a number like the views on a viral TikTok video: 580,000,000. To put this in standard form, we need to move the decimal point (which is hiding at the end) until there's only one non-zero digit in front of it. We move it 8 spots to the left to get 5.8. Since we moved it 8 places, our power of 10 is 8. So, it becomes $5.8 \times 10^8$. Easy, right? The positive power tells you it's a big number. 
<br><br>What about tiny numbers, like the diameter of a red blood cell, say 0.0000075 metres? Same idea, but we move the decimal to the right to get 7.5. We moved it 6 spots, but because we started with a tiny decimal, the power is negative. So, it's $7.5 \times 10^{-6}$. The negative power is a dead giveaway that you're dealing with a number smaller than 1.<br><br>For your IGCSE Core Paper 3, you'll need to calculate with these. Don't panic! Your calculator is your best friend here. Look for a button that looks like `x10^x`, `EXP`, or `EE`. To multiply $(2 \times 10^5)$ by $(3 \times 10^2)$, you'd just type it in. The calculator handles the magic, which is basically multiplying the main numbers ($2 \times 3 = 6$) and adding the powers ($5 + 2 = 7$) to get $6 \times 10^7$. It's basically a cheat code for handling huge numbers in calculations. 😉
Worked Example: Application of Standard FormC1.8, C1.9
Problem A hit song on Spotify has been streamed $1.5 \times 10^9$ times. If the song is 200 seconds long, calculate the total time in seconds that the song has been played worldwide. Give your answer in standard form.
Step 1 First, let's get both our numbers ready for the calculation. The number of streams is already in standard form: $1.5 \times 10^9$. We need to convert the song length, 200 seconds, into standard form. To do this, we move the decimal point two places to the left. $$200 = 2.0 \times 10^2$$
Step 2 To find the total time, we need to multiply the number of streams by the length of one stream. Think of it like this: if you play a game for 30 minutes every day for 10 days, you'd multiply $30 \times 10$ to get the total time. $$Total \; Time = (Number \; of \; Streams) \times (Length \; of \; Song)$$ $$Total \; Time = (1.5 \times 10^9) \times (2 \times 10^2)$$
Step 3 Now for the calculation. We multiply the front numbers together ($1.5 \times 2$) and then we add the powers of 10 ($9 + 2$). On Paper 3, you can just plug this straight into your calculator using the `x10^x` or `EXP` button, but it's good to know what's happening behind the scenes! $$1.5 \times 2 = 3$$ $$10^9 \times 10^2 = 10^{9+2} = 10^{11}$$
Step 4 Let's put it all together. Our final answer is the product of the two parts. We check that the front number (3) is between 1 and 10, which it is, so our answer is in correct standard form. That's a lot of listening time! $$Total \; Time = 3 \times 10^{11} \; seconds$$
Limits of Accuracy and Error IntervalsC1.10
Okay, let's talk about 'Limits of Accuracy'. It sounds super technical, but I promise it's something you already get. Think about when you're telling your friends about a part-time job that pays £10 per hour. When you say you worked for 4 hours, you've probably rounded it. Was it exactly 4.0000 hours? Or was it maybe 3 hours and 58 minutes? This little bit of uncertainty is what limits of accuracy are all about. It's about finding the true range of values a number could have been before it was rounded. We call this range the 'error interval'.<br><br>
<br><br>Here’s the simple trick to find it: First, identify the degree of accuracy (e.g., rounded to the nearest 10, nearest whole number, nearest metre). Then, just halve that accuracy value. You subtract this result from your rounded number to find the Lower Bound (the absolute minimum it could have been) and you add it to find the Upper Bound (the value it gets up to, but doesn't quite include). So if your gaming session was 5 hours long (to the nearest hour), the accuracy is 1 hour. Half of that is 0.5 hours. The lower bound is $5 - 0.5 = 4.5$ hours, and the upper bound is $5 + 0.5 = 5.5$ hours. We write this as an error interval: $4.5 \le \text{time} < 5.5$. That 'less than' sign ($<$) for the upper bound is crucial – it's because if the time was exactly 5.5 hours, it would round up to 6! It's the ultimate limit you can't quite reach.
Calculating Bounds for a Measured LengthC1.10
Problem A car journey is measured as 25 km, correct to the nearest kilometre. Find the upper and lower bounds for the distance and write down the error interval for the distance, $d$.
Step 1 First, let's pull out the key info. We have the measured value and how accurately it was measured. This is our starting point. Measured Value = 25 km<br>Accuracy = 1 km
Step 2 This is the most important move! To find the 'wiggle room' on either side of 25 km, we take the accuracy (1 km) and chop it in half. $$\frac{1}{2} = 0.5 \text{ km}$$
Step 3 To find the Lower Bound (the shortest the trip could possibly be), we subtract our 'wiggle room' value from the measured distance. $$25 - 0.5 = 24.5 \text{ km}$$
Step 4 Now for the Upper Bound. We add the 'wiggle room' value to the measured distance. This is the value the actual distance gets right up to, but can't be equal to (or it would round up!). $$25 + 0.5 = 25.5 \text{ km}$$
Step 5 Finally, we write this as an error interval using inequalities. We're saying the actual distance, $d$, is greater than or equal to the lower bound, but strictly less than the upper bound. Done! ✅ $$24.5 \le d < 25.5$$
Ratio, Proportion, and RatesC1.11, C1.12
Alright, let's break down ratios, proportions, and rates. Honestly, you use these all the time without even thinking about it. A ratio is just a way to compare quantities. Think about your favourite Spotify playlist: the ratio of pop to hip-hop might be 3:2, meaning for every 3 pop songs, there are 2 hip-hop tracks. Or in a game, the ideal team comp might have a ratio of 2 damage dealers : 1 tank : 1 healer. The first thing we often do is simplify ratios to make them easier to work with. If your part-time job earnings compared to your friend's are £20 to £30, the ratio is 20:30. To simplify, you find the biggest number that divides into both (the HCF), which is 10. Divide both sides by 10, and you get the simplest form: 2:3.<br><br>Now, let's get to the good part: dividing stuff up. Let's say you and your friend earn £100 on a joint project and you agree to split it in that 2:3 ratio. How do you do it? First, add the parts of the ratio: $2 + 3 = 5$ total parts. Then, divide the total amount by the number of parts to find what one part is worth: $£100 \div 5 = £20$. So, one part is £20. Your share is 2 parts ($2 \times £20 = £40$) and your friend's is 3 parts ($3 \times £20 = £60$). Easy, right? 
This same logic, called proportional reasoning, helps you adapt a recipe for more people or figure out the best value deal at the supermarket (is the giant bag of crisps actually cheaper per gram?).<br><br>Finally, let's talk rates. A rate compares two different types of units. The most common one you'll face is speed, which is a rate that compares distance and time. When you're learning to drive, you're constantly thinking about your speed in km/h or mph. The magic formula you need to tattoo on your brain is: $$Average\ Speed = \frac{Total\ Distance}{Total\ Time}$$ So, if you travel 150 km in 2 hours, your average speed is $150 \div 2 = 75$ km/h. It’s the same principle for things like download speeds (Megabits per second) or pay rates (Pounds per hour). It's all just about how much of one thing happens per unit of another thing. Master these, and you've got a massive advantage for your exams! 😉
Worked Example: Ratio and Average Speed in ContextC1.11, C1.12
Problem You and two friends, Liam and Sara, go on a road trip covering a total distance of 360 km. The total driving time was 8 hours. You agree to split the £80 petrol cost in the ratio of the hours you each drove, which was 3:2:5 (You:Liam:Sara).<br>(a) Calculate how much each person pays for petrol.<br>(b) Calculate the average speed of the car for the trip in km/h.
Step 1 First, let's tackle part (a), splitting the cost. We need to find the total number of parts in the ratio to figure out what one 'part' of the cost is worth. $$Total\ Parts = 3 + 2 + 5 = 10\ parts$$
Step 2 Now, we divide the total cost by the total number of parts to find the value of a single part. $$Value\ of\ 1\ part = \frac{Total\ Cost}{Total\ Parts} = \frac{£80}{10} = £8$$
Step 3 Finally for part (a), we multiply the value of one part by each person's number of parts in the ratio to find their individual shares. $$Your\ share (3\ parts): 3 \times £8 = £24<br>Liam's\ share (2\ parts): 2 \times £8 = £16<br>Sara's\ share (5\ parts): 5 \times £8 = £40$$
Step 4 On to part (b)! We need to calculate the average speed. Let's start by writing down the formula. $$Average\ Speed = \frac{Total\ Distance}{Total\ Time}$$
Step 5 Now we substitute the values given in the problem into our formula. The total distance is 360 km and the total time is 8 hours. $$Average\ Speed = \frac{360\ km}{8\ hours}$$
Step 6 Last step! We just do the division to find the final answer. Don't forget the units! $$Average\ Speed = 45\ km/h$$
Practical Calculations with Time and MoneyC1.15, C1.16
Okay, let's be real – some math feels like it was invented just for exams. But this stuff? You'll actually use it. We're talking about mastering time and money, which is basically a superpower for adulting. First up, time. Your phone probably uses the 24-hour clock (like 17:30 instead of 5:30 p.m.) because it's way clearer and avoids accidentally setting an alarm for the evening. When you're calculating how long that new Netflix series is, remember there are 60 seconds in a minute, and 60 minutes in an hour. So, a 150-minute movie is $150 \div 60 = 2.5$ hours, or 2 hours and 30 minutes.<br><br>This gets super important when reading timetables for a part-time job or a train to a festival. 
If your train leaves at 09:15 and arrives at 11:22, you can calculate the journey time by finding the difference. From 09:15 to 11:15 is exactly 2 hours. Then there are another 7 minutes to get to 11:22. Total journey: 2 hours and 7 minutes. Easy! It gets even cooler with time zones. Ever tried to squad up for a game with a friend in New York? If you're in London (GMT) and they're in New York (GMT-5), you're 5 hours ahead. So when it's 20:00 your time, it's only 15:00 for them.<br><br>Now for the money part. Let's say you want to buy some merch from a US artist's website. The price is in dollars (\$), but you have pounds (£). You'll need the exchange rate, like $£1 = \$1.25$. To find the price in pounds, you'd divide the dollar price by 1.25. Mastering these calculations means you can plan trips, shop online, and coordinate with anyone, anywhere, without getting your times or your money mixed up. It's practical math for a connected world.
Worked Example: International Travel and Currency ConversionC1.15, C1.16
Problem Yuki flies from Tokyo to London. Her flight departs from Tokyo at 11:30 on Tuesday. The flight duration is 14 hours and 15 minutes. Tokyo is 9 hours ahead of London (GMT+9). While in the airport, she buys a souvenir for ¥2500. The exchange rate is $£1 = ¥175$.<br><br>Calculate:<br>a) Her arrival time and date in London (local time).<br>b) The cost of the souvenir in pounds (£), to 2 decimal places.
Step 1 First, let's figure out the arrival time in Tokyo's time zone. We do this by adding the flight duration to the departure time. $$\text{Departure Time: } 11:30 \text{ (Tuesday)} \\ \text{Add flight duration: } + 14 \text{ hours } 15 \text{ minutes} \\ \text{Calculation: } 11:30 + 14h = 25:30. \text{ Then } 25:30 + 15m = 25:45 \\ \text{Since 24:00 is midnight, } 25:45 \text{ is } 01:45 \text{ the next day.} \\ \text{Arrival time (Tokyo time): } 01:45 \text{ on Wednesday}$$
Step 2 Now we need to adjust for the 9-hour time difference. London is 9 hours behind Tokyo, so we subtract 9 hours from the Tokyo arrival time. $$\text{Tokyo arrival time: } 01:45 \text{ (Wednesday)} \\ \text{Subtract time difference: } - 9 \text{ hours} \\ \text{Let's count back: } 01:45 \text{ back 1h 45m is } 00:00 \text{ (start of Wed). We still need to go back } 7h \text{ 15m.} \\ 24:00 \text{ (end of Tue) } - 7 \text{ hours } 15 \text{ minutes} = 16:45 \text{ (Tuesday)} \\ \text{Arrival time (London time): } 16:45 \text{ on Tuesday}$$
Step 3 For the souvenir, we need to convert Japanese Yen (¥) to British Pounds (£). The rate is $£1 = ¥175$. To find how many pounds we get, we divide the amount in yen by the exchange rate. $$\text{Cost in } £ = \frac{\text{Cost in } ¥}{\text{Exchange Rate}} \\ \text{Cost in } £ = \frac{2500}{175}$$
Step 4 Finally, we do the division and round the answer to 2 decimal places, because money always has two decimal places for cents or pence. $$2500 \div 175 = 14.285714... \\ \text{Rounding to 2 d.p. gives } £14.29 \\ \text{The souvenir cost } £14.29$$
Set Notation and Venn DiagramsC1.2
Alright, let's break down sets. Think of a set as any group of things, like your follower list on Instagram, your ultimate gaming squad, or a Spotify playlist. It's just a collection of distinct items. The big-picture group that contains everything we're talking about is called the universal set, which we note with a fancy E. For example, E could be 'all students in Year 11'.<br><br>To visualize how these groups relate, we use Venn diagrams. They're basically just overlapping circles inside a box. The box is our universal set E, and each circle is a specific set, like 'students who have a driver's license' (Set A) and 'students with a part-time job' (Set B). 
<br><br>Now for the lingo, which is actually super simple once you get it:<br>• $n(A)$: This just means 'the number of elements in set A'. So, if 12 of your friends have a driver's license, $n(A) = 12$.<br>• $A \cap B$: This is the intersection. It's the stuff that's in BOTH sets A AND B. Think of it as the overlap in the circles – the friends who both drive and have a job. The $\cap$ symbol looks like a bridge connecting them.<br>• $A \cup B$: This is the union. It's everyone who is in set A, OR set B, OR both. It's like merging two group chats – you get everyone from both groups combined.<br>• $A'$: This is the complement of A. It means everything in the universal set that is NOT in A. If A is people with a job, $A'$ is everyone without one.
Worked Example: Analyzing Student HobbiesC1.2
Problem In a group of 40 students, a survey was conducted on whether they use streaming services. Let E = {all students surveyed}. Let S = {students who use Spotify} and N = {students who use Netflix}. We find that $n(S) = 25$, $n(N) = 22$, and $n(S \cap N) = 10$. Represent this information in a Venn diagram and find the number of students who use neither service.
Step 1 First, let's draw the basic Venn diagram. We need a rectangle for our universal set, E, which has 40 students in total. Inside, we'll draw two overlapping circles for our sets, S (Spotify) and N (Netflix).
Step 2 The golden rule of Venn diagrams is to always start from the middle! The middle is the intersection, $S \cap N$. The problem tells us that 10 students use both services, so we put '10' in the overlapping section. $$ n(S \cap N) = 10 $$
Step 3 Now, let's figure out the 'Spotify only' part. We know the entire Spotify circle, $n(S)$, must have 25 students. Since 10 are already in the overlap, the number of students who only use Spotify is the total minus the overlap. $$ \text{Spotify only} = n(S) - n(S \cap N) = 25 - 10 = 15 $$
Step 4 We do the same thing for the 'Netflix only' part. The whole Netflix circle, $n(N)$, has 22 students. Subtract the 10 who are also in the Spotify group to find those who are exclusive to Netflix. $$ \text{Netflix only} = n(N) - n(S \cap N) = 22 - 10 = 12 $$
Step 5 Finally, we need to find how many students use neither service. This is the number that goes outside the circles but still inside the box. We add up everyone inside the circles and subtract that total from the universal set, E. $$ \text{Total in circles} = 15 (S \text{ only}) + 10 (Both) + 12 (N \text{ only}) = 37 <br> \text{Neither} = n(E) - 37 = 40 - 37 = 3 $$
Using a Calculator EffectivelyC1.5, C1.6
In Paper 3 you've got a calculator, and using it well is genuinely a marks machine. The #1 rule, the one that saves people every session: never round mid-calculation. Rounding early is like pausing a song halfway through — you miss the best part. Only round the final answer. 🎯<br><br>Stuff you actually need to know:<br>• Brackets around fractions. For $\dfrac{8+4}{3}$ type `(8+4)/3`, not `8+4/3`. Without the brackets the calculator does $8 + \tfrac{4}{3}$, which is a completely different number. Same vibe as autocorrect ruining your group chat.<br>• Use `Ans` to recall your last answer instead of retyping. Typos are the enemy.<br>• Time entry. $2$ hours $30$ min is $2.5$ on a calculator (because 30 min = half an hour). Or use the degree-minute-second button: `2° 30' 0"`. Never $2.30$ — that's $2$ h $18$ min, totally different.<br>• Money on the display. If the answer shows $4.8$, that's $\$4.80$. Always 2 d.p. for money. No exceptions. 💰<br>• Powers and roots. `^` for any power. Shifted root key for cube/nth roots. For $5^{-2}$ wrap the negative: `5^(-2)`.<br><br>
<br><br>Reading the display: a screen showing `1.234E3` or `1.234×10³` is just standard form (that's $1234$). Don't panic, don't copy the `E` — write it as $1.234 \times 10^{3}$ in your answer. Calculator talk ≠ maths talk. 📱
Worked Example: Calculator Use — Fractions, Time and BracketsC1.4, C1.13
Problem Using a calculator, find (a) $\dfrac{12.5 - 3.5}{1.5}$, (b) the total time in minutes between $08\!:\!45$ and $11\!:\!20$, and (c) $\sqrt{15^2 + 8^2}$. ⏱️
Step 1 For (a), bracket the numerator — otherwise the calculator thinks you meant $12.5 - \tfrac{3.5}{1.5}$. Different universe. $$\frac{12.5 - 3.5}{1.5} = \frac{9}{1.5} = 6$$
Step 2 For (b), subtract times in base 60. $20 - 45$ doesn't work, so borrow 1 hour (= 60 min): $11\!:\!20$ becomes $10\!:\!80$. $$10\!:\!80 - 8\!:\!45 = 2\!:\!35$$
Step 3 Now convert $2$ h $35$ min to total minutes. $$2 \times 60 + 35 = 120 + 35 = 155 \text{ minutes}$$
Step 4 For (c), bracket the expression under the root. Recognise this as Pythagoras — nice round answer incoming. 😎 $$\sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$$
Worked Example: Compound InterestC1.15, C1.16
Problem $\$2000$ is invested at a compound interest rate of $5\%$ per year. (a) Find the amount after $3$ years. (b) How much interest has been earned? Answers to the nearest cent. 💰
Step 1 Lock in the compound interest formula. $P$ = starting amount, $r$ = rate, $n$ = number of years. $$A = P \left(1 + \frac{r}{100}\right)^n$$
Step 2 Sub in $P = 2000$, $r = 5$, $n = 3$. Don't round yet — keep the full decimal. $$A = 2000 \times (1.05)^3 = 2000 \times 1.157625$$
Step 3 Evaluate and round to 2 d.p. (money = 2 d.p., always). $$A = \$2315.25$$
Step 4 For (b), interest is just the extra money — final amount minus what you started with. $$\text{Interest} = 2315.25 - 2000 = \$315.25$$
Key Points to Remember - The TL;DR
- BODMAS/PEMDAS is the law! Brackets, Orders, Division/Multiplication, Addition/Subtraction.
- To compare or calculate with different number types, convert them all to decimals or fractions first.
- Percentage of an amount: Turn the percentage into a decimal and multiply.
- Percentage Change formula: $$\frac{\text{Change}}{\text{Original}} \times 100$$
- Reverse Percentages: Find what percentage the final amount represents, then divide to find the original $100\%$.
- Successive Percentages: Apply changes one by one. Do not add the percentages together.
- Simple Interest formula: $$I = P \times R \times T$$