Number: Core Concepts and Applications

Application of Prime Factorisation, HCF, and LCM

Solving Puzzles with Number DNA 🧬

1. 1
   First, we need to find the 'DNA' of our numbers by breaking them down into their prime factors. This is the key to unlocking both the HCF and LCM. Let's use factor trees.

   $108 = 2 \times 54 = 2 \times 2 \times 27 = 2^2 \times 3^3$ $72 = 2 \times 36 = 2 \times 2 \times 18 = 2^3 \times 3^2$
2. 2
   For part (a), the greatest number of identical bags means we need the Highest Common Factor (HCF). We find this by multiplying the common prime factors raised to their lowest powers.

   Common factors are 2 and 3. $\text{Lowest power of 2 is } 2^2.$ $\text{Lowest power of 3 is } 3^2.$ $HCF = 2^2 \times 3^2 = 4 \times 9 = 36$ So, you can make 36 identical goody bags.
3. 3
   Just to be thorough, we can figure out what's in each bag. Divide the total items by the number of bags (the HCF).

   $\text{Glow sticks per bag: } 108 \div 36 = 3$ $\text{Sweets per bag: } 72 \div 36 = 2$
4. 4
   For part (b), we need to find the next time the lights flash together, which means we're looking for the Lowest Common Multiple (LCM). We find this by multiplying all the prime factors from both numbers, raised to their highest powers.

   $\text{Highest power of 2 is } 2^3.$ $\text{Highest power of 3 is } 3^3.$ $LCM = 2^3 \times 3^3 = 8 \times 27 = 216$ They will flash together after 216 seconds.
5. 5
   The question asks for the answer in minutes, so we need to do a final conversion. Don't get caught out!

   $216 \text{ seconds} \div 60 \text{ s/min} = 3.6 \text{ minutes}$ (This is 3 minutes and 0.6 x 60 = 36 seconds).

Worked Example: Reverse Percentages

That 'Sale Price' Is Hiding Something... 😉

1. 1
   First, figure out what percentage of the original price you're actually paying. A 15% discount means you're paying 100% - 15%.

   $100\% - 15\% = 85\%$
2. 2
   So, the sale price of $\$102$ is equal to 85% of the original price. Let's call the original price 'P'. We can set this up as an equation, converting the percentage to a decimal.

   $85\% \times P = \$102 \implies 0.85 \times P = 102$
3. 3
   Now we just need to do some algebra to find 'P'. To get 'P' by itself, we need to divide both sides of the equation by 0.85. This is the 'reverse' part of the calculation.

   $P = \frac{102}{0.85}$
4. 4
   Whip out your calculator and solve it. This gives you the full, original price before that tempting discount was applied.

   $P = \$120$
5. 5
   Quick sanity check! Does a 15% discount on $\$120$ give you $\$102$? Let's see: 15% of $\$120$ is $0.15 \times 120 = 18$. So, $\$120 - 18 = 102$. Perfect! The original price was $\$120$.

   $\text{Check: } \$120 - (0.15 \times 120) = \$102$

Worked Example: Calculations with Standard Form

Let's Crunch Some Cosmic Numbers ✨

1. 1
   First, we need to figure out what the question is asking. 'How many times more massive' means we need to divide the Sun's mass by the Earth's mass. Let's set up the division.

   $\text{Ratio} = \frac{\text{Mass of Sun}}{\text{Mass of Earth}} = \frac{1.989 \times 10^{30}}{5.972 \times 10^{24}}$
2. 2
   Now, let's split the calculation into two parts: we'll divide the decimal numbers ('A' parts) and then divide the powers of 10. It makes the whole thing way less intimidating.

   $\text{Ratio} = \left( \frac{1.989}{5.972} \right) \times \left( \frac{10^{30}}{10^{24}} \right)$
3. 3
   Let's grab our calculator for the first part. Remember the question asks for 3 significant figures in the final answer, so let's keep a few extra decimal places for now to be safe.

   $\frac{1.989}{5.972} \approx 0.333054...$
4. 4
   For the powers of 10, we use our index laws. When you divide powers with the same base, you subtract the exponents. So, it's $30 - 24$.

   $\frac{10^{30}}{10^{24}} = 10^{30-24} = 10^6$
5. 5
   Now we combine our two results. We get $0.333054... \times 10^6$. But wait! This isn't proper standard form because the 'A' part (0.333...) is not between 1 and 10. We need to adjust it.

   $0.333054... \times 10^6 = (3.33054... \times 10^{-1}) \times 10^6$
6. 6
   Finally, we combine the powers of 10 and round our 'A' part to 3 significant figures as requested. This gives us our final, properly formatted answer.

   $3.33054... \times 10^{-1+6} = 3.33054... \times 10^5 \approx 3.33 \times 10^5$

Worked Example: Estimating a Calculation

Solving it without breaking a sweat 💪

1. 1
   First up, we follow the golden rule: round every number in the expression to 1 significant figure. This makes the calculation way easier.

   $41.3 \approx 40$ $\sqrt{98.5} \approx \sqrt{100} = 10$ $8.09 \approx 8$ $2.78 \approx 3$
2. 2
   Now, substitute these new, friendly numbers back into the original expression. It's like swapping out difficult opponents in a video game for easier ones.

   $\approx \frac{40 \times 10}{8 - 3}$
3. 3
   Solve the numerator (top part) and the denominator (bottom part) of the fraction separately. This keeps our working clean and easy to follow.

   $\frac{400}{5}$
4. 4
   Finally, perform the last division to get our estimated answer. See? No calculator needed for this boss battle!

   $\frac{400}{5} = 80$

Worked Example: Dividing a Quantity in a Given Ratio

The Part-Time Job Payout Problem 💸

1. 1
   First, let's write down the ratio of the hours worked. This represents how we're going to 'share' the money.

   $\text{Your hours : Alex's hours : Ben's hours} = 8:5:3$
2. 2
   Next, we need to find the total number of 'parts' or 'shares' in our ratio. We just add the numbers together.

   $\text{Total parts} = 8 + 5 + 3 = 16 \text{ parts}$
3. 3
   Now, we figure out how much money one single 'part' is worth. We do this by dividing the total earnings by the total number of parts.

   $\text{Value of one part} = \frac{£480}{16} = £30$
4. 4
   Time to distribute the cash! We multiply the value of one part (£30) by each person's share in the ratio.

   $\text{Your share} = 8 \times £30 = £240 \\ \text{Alex's share} = 5 \times £30 = £150 \\ \text{Ben's share} = 3 \times £30 = £90$
5. 5
   Always do a final check to make sure you haven't messed up! Add your answers together to see if they equal the original total.

   $£240 + £150 + £90 = £480$

Worked Example: Average Speed Calculation

That Tricky Time Conversion Problem 🚴‍♀️

1. 1
   First, let's identify what we've been given. We have the distance ($D$) and the total time ($T$). The goal is to find the average speed ($S$).
   Distance = 21 km
   Time = 1 hour 15 minutes

   $D = 21 \text{ km} \quad , \quad T = 1 \text{ hr } 15 \text{ min}$
2. 2
   Hold up! The speed needs to be in km/h, but our time is in hours and minutes. We need to convert the entire time into just hours. There are 60 minutes in an hour, so we can convert 15 minutes to hours by dividing by 60.

   $15 \text{ minutes} = \frac{15}{60} \text{ hours} = 0.25 \text{ hours}$
3. 3
   Now we can find the total time in hours by adding the hour part and the decimal part we just calculated.

   $\text{Total Time} = 1 \text{ hour} + 0.25 \text{ hours} = 1.25 \text{ hours}$
4. 4
   Great, we're ready to use the average speed formula. Remember, it's Speed = Distance / Time.

   $\text{Average Speed} = \frac{\text{Distance}}{\text{Time}}$
5. 5
   Finally, substitute our values into the formula and solve. Don't forget to include the correct units in your final answer!

   $\text{Average Speed} = \frac{21 \text{ km}}{1.25 \text{ h}} = 16.8 \text{ km/h}$

Worked Example: Rationalising a Binomial Denominator

Kicking that Surd Out of the Basement 👇

1. 1
   First up, we need to get rid of the surd in the denominator, $5 - \sqrt{2}$. We do this by multiplying by its conjugate. The conjugate is the same expression but with the sign flipped. So, the conjugate of $5 - \sqrt{2}$ is $5 + \sqrt{2}$.

   $Conjugate = 5 + \sqrt{2}$
2. 2
   Now, we multiply the entire fraction by a 'fancy version of 1', which is the conjugate over itself. This keeps the value of the fraction the same while changing how it looks.

   $\frac{12}{5 - \sqrt{2}} \times \frac{5 + \sqrt{2}}{5 + \sqrt{2}}$
3. 3
   Let's handle the numerators and denominators separately. For the numerator, we multiply 12 by the whole bracket $(5 + \sqrt{2})$. For the denominator, we use the difference of two squares trick: $(a-b)(a+b) = a^2 - b^2$.

   $\frac{12(5 + \sqrt{2})}{(5 - \sqrt{2})(5 + \sqrt{2})} = \frac{60 + 12\sqrt{2}}{5^2 - (\sqrt{2})^2}$
4. 4
   Time to simplify the denominator. Squaring 5 gives 25, and squaring $\sqrt{2}$ just gives 2. The surd is gone! Mission accomplished.

   $Denominator = 25 - 2 = 23$
5. 5
   Finally, put the simplified numerator and denominator back together. We check if the fraction can be simplified further, but since 23 is a prime number and doesn't go into 60 or 12, this is our final answer. Looking sharp!

   $\frac{60 + 12\sqrt{2}}{23}$

Worked Example: Problem Solving with a Three-Set Venn Diagram

The Ultimate Social Media Showdown 📱

1. 1
   First, draw a Venn diagram with three overlapping circles for Instagram, Snapchat, and TikTok. The golden rule is to always start from the inside and work your way out. The very middle represents the students who use all three apps.

   $n(I \cap S \cap T) = 4$
2. 2
   Now, let's fill in the regions where two sets intersect. Remember to subtract the middle part! For example, 12 students use I and S, but 4 of those also use T. So, the number who use only I and S is...

   $n(I \cap S \text{ only}) = 12 - 4 = 8 <br> n(S \cap T \text{ only}) = 15 - 4 = 11 <br> n(I \cap T \text{ only}) = 10 - 4 = 6$
3. 3
   Next, we find the number of students who use only one app. For each circle, take the total for that app and subtract the three intersection parts we've already filled in.

   $n(I \text{ only}) = 30 - (8 + 4 + 6) = 12 <br> n(S \text{ only}) = 35 - (8 + 4 + 11) = 12 <br> n(T \text{ only}) = 42 - (6 + 4 + 11) = 21$
4. 4
   Let's find the total number of students who use at least one of the apps. We just add up every number we've placed inside the circles.

   $n(I \cup S \cup T) = 12 + 12 + 21 + 8 + 11 + 6 + 4 = 74$
5. 5
   The final step! The question asks for the number of students who use none of the apps. We take the total number of students in the survey (our universal set, $E$) and subtract the number of students who use at least one app.

   $\text{None} = n(E) - n(I \cup S \cup T) = 80 - 74 = 6$

Worked Example: Combining Powers and Roots

Let's Solve This Thing 🧠

1. 1
   First, let's break down the problem. We have three separate parts to calculate before we can do the addition and subtraction: a power ($4^3$), a square root ($\sqrt{121}$), and a cube root ($\sqrt[3]{125}$). We need to deal with these first, just like following the order of operations (BIDMAS/PEMDAS).

   $4^3 + \sqrt{121} - \sqrt[3]{125}$
2. 2
   Let's calculate the power. $4^3$ means $4 \times 4 \times 4$. This is one of the cubes you should have memorized!

   $4^3 = 4 \times 4 \times 4 = 64$
3. 3
   Next up, the roots. These are straight from our 'must-know' list. The square root of 121 is 11, and the cube root of 125 is 5.

   $\sqrt{121} = 11 \quad \text{and} \quad \sqrt[3]{125} = 5$
4. 4
   Now we substitute these values back into our original expression. It's looking way simpler now, right?

   $64 + 11 - 5$
5. 5
   Finally, we just work from left to right with the addition and subtraction to get our final answer. You got this!

   $64 + 11 = 75 \\ 75 - 5 = 70$

Worked Example: Bacteria Population and Car Depreciation

1. 1
   For (a), grab the ingredients: $V_0 = 500$, rate $r = +8\%$, periods $n = 6$. Multiplier per hour is $1 + \tfrac{8}{100} = 1.08$.

   $V = 500 \times (1.08)^6$
2. 2
   Type it into the calculator in one go — don't round mid-way, that's how you lose marks.

   $V = 500 \times 1.586874... = 793.437...$
3. 3
   Round to the nearest whole cell because you can't have half a bacterium.

   $V \approx 793 \text{ cells}$
4. 4
   For (b), decay mode: $V_0 = 24\,000$, $r = -15\%$, $n = 4$. Multiplier is $1 - \tfrac{15}{100} = 0.85$. (Resist the urge to use $-0.15$!)

   $V = 24\,000 \times (0.85)^4$
5. 5
   Evaluate and round to the nearest dollar. That car lost $\$11\,472$ in four years. Brutal. 💸

   $V = 24\,000 \times 0.52200625 = 12\,528.15$ $V \approx \$12\,528$

Worked Example: Calculator Use with Fractions, Time, and Brackets

1. 1
   For (a), bracket both top and bottom so the calculator knows what to do. Watch the difference when you forget brackets — it's chaos.

   $\text{Type: } (5.2+3.8) \div (4.1-2.6)$ $= \frac{9.0}{1.5} = 6$
2. 2
   For (b), $3.75$ hours = $3$ h plus $0.75$ of an hour. The $0.75$ part is $0.75 \times 60 = 45$ min. Total in minutes is just the original $\times 60$.

   $3.75 \times 60 = 225 \text{ minutes}$
3. 3
   For (c), handle the fraction under the root first. Bracket it so the root applies to the whole thing: `√((24^2+10^2)/13)`.

   $\frac{24^2 + 10^2}{13} = \frac{576 + 100}{13} = \frac{676}{13} = 52$
4. 4
   Now take the square root and round to 3 sig fig. Don't round before — that 7.2111... is your working value. 🎯

   $\sqrt{52} = 7.2111...$ $\approx 7.21$

Worked Example: Ordering Mixed Numbers, Fractions, Decimals and Percentages

1. 1
   Convert everything to decimals (4 d.p. gives enough precision). Same language = easy comparison.

   $\frac{5}{8} = 0.6250$ $0.6 = 0.6000$ $62\% = 0.6200$ $\frac{3}{5} = 0.6000$
2. 2
   Line them up on a mental number line from smallest to biggest. Note $0.6 = \tfrac{3}{5}$ (same number, different clothes).

   $0.6000 = 0.6000 < 0.6200 < 0.6250$
3. 3
   Write the final ordering using the original forms — that's what the question asked for.

   $0.6 = \frac{3}{5} < 62\% < \frac{5}{8}$

Worked Example: Four Operations with Mixed Numbers, Decimals and Negatives

1. 1
   For (a), convert both mixed numbers to improper fractions first. This is not optional.

   $2\tfrac{1}{4} = \frac{9}{4}, \quad 1\tfrac{1}{2} = \frac{3}{2}$
2. 2
   Keep-flip-change: division becomes multiplication by the reciprocal. Then simplify.

   $\frac{9}{4} \div \frac{3}{2} = \frac{9}{4} \times \frac{2}{3} = \frac{18}{12} = \frac{3}{2} = 1\tfrac{1}{2}$
3. 3
   For (b), BODMAS says multiplication before subtraction. $(-3) \times 2 = -6$. Then $-7 - (-6) = -7 + 6$.

   $-7 - (-6) = -7 + 6 = -1$
4. 4
   For (c), $0.6$ has 1 d.p. and $0.04$ has 2 d.p., so the answer has 3 d.p. Multiply the digits: $6 \times 4 = 24$. Then place the decimal.

   $0.6 \times 0.04 = 0.024$

Worked Example: Time Calculations Across Timetable and Time Zones

1. 1
   Add the flight time to the GMT departure.

   $22\!:\!30 + 7\!:\!45 = 30\!:\!15 \text{ GMT}$
2. 2
   $30\!:\!15$ is past midnight. Subtract $24\!:\!00$ (one full day) and roll the day forward.

   $30\!:\!15 - 24\!:\!00 = 06\!:\!15 \text{ Tuesday (GMT)}$
3. 3
   Cairo is $+2$ h, so add $2$ hours for local time. That's the answer the question wants.

   $06\!:\!15 + 2\!:\!00 = 08\!:\!15 \text{ Tuesday (local)}$

Worked Example: Currency Conversion with Rounding

1. 1
   For (a), dollars → euros: multiply by the rate.

   $250 \times 0.86 = \text{€}215.00$
2. 2
   For (b), euros → dollars: divide by the rate.

   $\frac{129}{0.86} = 150.0000$
3. 3
   Round to 2 d.p. (the nearest cent) and slap on the dollar sign. Money presented like a pro. 💸

   $= \$150.00$