1 Irrational and Rational Numbers Question 1 Circle the rational numbers. Numbers: 9, π, 0.1111 β¦ , 4, 12, 1, 3 Step 1: A rational number can be written as a fraction π πwhere π, πare integers and π β 0. Step 2: β’ 9 = 9 1rational β’ 0.1111 β¦recurring decimal, can be written as a fraction, rational β’ 4 = 4 1rational β’ 12 = 12 1 rational β’ 1 = 1 1rational β’ 3 = 3 1rational Answer: 9, 0.1111 β¦ , 4, 12, 1, 3are rational. Question 2 1. Integers: Whole numbers , no decimals, no fractions. β’ Examples: β365, 0, 1, 17, 989 β’ Not integers: 0.5, 2/3, β7, 13ΒΎ, β1000.1, 66.66, Ο Key idea: Integers = whole numbers. 2. Rational Numbers: Numbers that CAN be written as a fraction. They come in 3 forms: β’ Integers β 4 = 4/1 β’ Fractions β 1/4, β1/2, 7/4 β’ Terminating_or_recurring_decimals o 0.125 (stops) o 0.333333β¦ (repeats) Key idea: Rational = neat, tidy, can be written as p/q. 3. Irrational Numbers: Numbers that CANNOT be written as a fraction. β’ Never-ending β’ Never repeating β’ Messy decimals Examples: β2, β7, Ο Key idea: Irrational = messy decimals that never repeat. 4. Surds:A surd is an expression with an irrational square root. Examples: β2, β5, 3β7 Key idea: Surds = irrational roots kept in root form. Unit 1: Numbers
2 Katie says 0.6666 β¦is irrational because it is a recurring decimal. Is Katie correct? Step 1: Recurring decimals can be written as fractions. Step 2: 0.6666 β¦ = 2 3 This is a fraction of integers, so it is rational. Answer: Katie is not correct. 0.6666 β¦is rational. Key Think: Any recurring decimal is rational. Question 3 Write down an irrational number. Step 1: Irrational numbers cannot be written as π π. Step 2: They are non-terminating, non-recurring decimals. Examples: β7, π Answer: β7or π. Key Think: Square roots of non-square integers and πare irrational. Question 4 Write down an irrational number. Same idea as Q3. Answer: β2or β5. Key Think: If the number inside the square root is not a perfect square, the result is irrational. Question 5 π₯is an irrational number between 7 and 10. Find a value for π₯. Step 1: We need an irrational number whose value is between 7 and 10. Step 2: Try β50. β49 = 7, β64 = 8 So β50 β 7.07 This is between 7 and 10 and is irrational. Answer: π₯ = β50. Key Think: Use square roots between two perfect squares. Question 6 π¦is an irrational number between 3 and 4. Find a value for π¦. Step 1: We need an irrational number between 3 and 4. Step 2: Try β10. β9 = 3, β16 = 4 So β10 β 3.16 This is between 3 and 4 and is irrational. Answer: π¦ = β10. Key Think: Again, use square roots between perfect squares. Question 7 π§is a rational number between 1105 and 1135. Find a value for π§.
3 Step 1: Rational numbers include integers. Step 2: Any whole number between 1105 and 1135 works. Answer: π§ = 1121(or any integer in that range). Key Think: All integers are rational. Question 8 Which equation has a rational solution? Equations: 1. π₯2 = 39 2. π₯2 = 144 3. π₯2 = 350 Step 1: Solve each: β’ Equation 1: π₯ = β39(not a whole number, irrational) β’ Equation 2: π₯ = β144 = 12(whole number, rational) β’ Equation 3: π₯ = β350(not a whole number, irrational) Answer: Equation 2 has a rational solution. Key Think: If the number under the square root is a perfect square, the solution is rational. Question 9 Radius of a circle is 10 cm. Is the circumference rational or irrational? radius = 10 cm diameter = 20 cm Step 1: Diameter π = 2π = 2 Γ 10 = 20cm. Step 2: Circumference formula: πΆ = ππ So πΆ = π Γ 20 This includes π, which is irrational. Answer: The circumference is irrational. Key Think: Any exact expression with πis irrational. Question 10 5π₯2 = π (a) Write a value for πwhich gives rational solutions. (b) Write a value for πwhich gives irrational solutions. Step 1 (a): We want π₯2to be a perfect square. If π = 20: 5π₯2 = 20 β π₯2 = 20 5 = 4 β π₯ = Β±2 So π₯is rational. Answer (a): π = 20(also π = 5works: π₯2 = 1). Step 2 (b): We want π₯2not to be a perfect square. If π = 8: 5π₯2 = 8 β π₯2 = 8 5 β π₯ = Β±β8 5 This is irrational. Answer (b): π = 8. Key Think: Check if π/5is a perfect square.
4 Question 11 Right angled triangle ABC. Is length BC rational or irrational? Triangle (white background): AC = 15 cm, CB = 4 cm, AB = BC (hypotenuse). Use Pythagoras: π΄πΆ2 + πΆπ΅2 = π΄π΅2 But here AC and CB are the legs, BC is the hypotenuse. Step 1: 42 + π΅πΆ2 = 152 16 + π΅πΆ2 = 225 Step 2: π΅πΆ2 = 225 β 16 = 209 π΅πΆ = β209 209 is not a perfect square. Answer: BC is irrational. Key Think: If the square of the length is not a perfect square, the length is irrational. Question 12 Show (5 β 12)(5 + 12)is rational. Step 1: Use difference of squares: (π β π)(π + π) = π2 β π2 Here π = 5, π = 12. Step 2: (5 β 12)(5 + 12) = 52 β 122 = 25 β 144 = β119 Answer: β119is rational. Key Think: Product of conjugates is always rational. Question 13 Circle the rational numbers. Numbers: 18, 54 2 , β8, 12, π, 2, 1 15 , 2, 55 Step 1: β’ 18 = 18 1 rational β’ 54 2 = 27rational β’ β8not a whole number, irrational β’ 12rational β’ 2rational β’ 1 15rational β’ 55rational Answer: 18, 54 2 , 12, 2, 1 15 , 2, 55are rational.
5 Key Think: Square roots of non-square numbers are irrational. Question 14 Show 7Γ112 21Γ3 is rational. Step 1: 7 Γ 112 = 784 21 Γ 3 = 63 So 7 Γ 112 21 Γ 3 = 784 63 Step 2: Simplify the fraction: 784 63 = 784 Γ· 7 63 Γ· 7 = 112 9 This is a fraction of integers, so rational. Answer: 112 9 is rational. Key Think: Any fraction of integers is rational. Question 15 Find two different surds that multiply together and give a rational number. Step 1: A surd is an irrational root, like β3. Step 2: β3 Γ β3 = 3 3 is rational. Answer: β3 Γ β3 = 3. Key Think: βπ Γ βπ = π, which is rational if πis an integer. Question 16 A number is defined as π₯ = β18 β β8 Is π₯rational or irrational? Step 1 Write each square root in simplest surd form. β18 = 3β2 β8 = 2β2 Step 2 Subtract them. π₯ = 3β2 β 2β2 = β2 Step 3 β2is irrational. Answer π₯is irrational. Key Think If a surd simplifies to another surd, it stays irrational. Question 17 The number
6 π¦ = β45 3 Is π¦rational or irrational? Step 1 Simplify the square root. β45 = 3β5 Step 2 Substitute. π¦ = 3β5 3 Step 3 Cancel the 3. π¦ = β5 Step 4 β5is irrational. Answer π¦is irrational. Key Think Dividing a surd by a whole number does not make it rational. Question 18 A number is defined as π§ = β12 Γ β3 Is π§rational or irrational? Step 1 Multiply the square roots. π§ = β12 Γ 3 = β36 Step 2 β36 = 6 Step 3: 6 is rational. Answer π§is rational. Key Think Sometimes multiplying two surds gives a perfect square. Question 19 A number is defined as π€ = β50 β 5 Is π€ rational or irrational? Step 1: Simplify the surd. β50 = 5β2 Step 2 Substitute. π€ = 5β2 β 5 Step 3: 5 is rational. 5β2is irrational. Step 4: Irrational minus rational = irrational. Answer π€is irrational. Key Think Irrational Β± rational always stays irrational.
7 Surds 1. What is a Surd? A surd is an irrational number written using a root sign. It is a way of keeping an answer exact instead of turning it into a long decimal. Example of surds: β2, β5, β11 These are not surds: β9 = 3, β16 = 4, β25 = 5 Reason: The roots give whole numbers. General Note: A surd stays a surd unless the root becomes a whole number. 2. How to Know if a Number is a Surd Step 1: Check if the number is written with a root sign. Step 2: Check if the number inside the root is a perfect square. Step 3: If the root gives a whole number, it is not a surd. Step 4: If the root does not give a whole number, it is a surd. Example: β50 50 is not a perfect square β surd. Example: β49 = 7 49 is a perfect square β not a surd. Key Think: Surds are irrational roots. 3. Simplest Form of a Surd A surd is in simplest form when the number inside the root has no perfect square factors greater than 1. Step 1: Find square factors of the radicand. Step 2: Use the highest square factor. Step 3: Rewrite the surd. Step 4: Simplify.
8 Example: β18 Step 1: Square factors of 18 are 1 and 9. Step 2: Highest square factor is 9. Step 3: Rewrite: β18 = β9 Γ 2 Step 4: Simplify: β18 = 3β2 4. Like Surds Surds with the same radicand are like surds. Examples: 2β3, 5β3, 11β3 These can be added or subtracted. Not like surds: β2, β3, β5 These cannot be added. Key Think: Like surds behave like like terms in algebra. 5. Adding Surds You can only add surds with the same radicand. Example: Adding Like Surds 2β3 + 5β3 Step 1: Check the radicand: both are β3. Step 2: Add the coefficients: 2 + 5 = 7 Step 3: Keep the surd the same:
9 7β3 Final Answer: 2β3 + 5β3 = 7β3 2β3 + 5β3 + 11β3 = 18β3 Example: Adding Like Surds After Simplifying β12 + 3β3 Step 1: Simplify β12: β12 = β4 Γ 3 = 2β3 Step 2: Rewrite the expression: 2β3 + 3β3 Step 3: Add the coefficients: 2 + 3 = 5 Step 4: Keep the surd the same: 5β3 Final Answer: β12 + 3β3 = 5β3 General Notes o Like surds have the same radicand. o Add or subtract the coefficients only. o Keep the surd part the same. o If surds are not like surds, simplify first to check if they become like surds. o Surds with different radicands cannot be added..
10 6. Multiplying Surds Multiply the radicands. Example: β3 Γ β12 Step 1: Put both numbers under the same root: β3 Γ 12 Step 2: Multiply the radicands: 3 Γ 12 = 36 Step 3: Root the result: β36 = 6 Final Answer: β3 Γ β12 = 6 Multiply Surds With Coefficients 3β5 Γ 2β3 Step 1: Multiply the coefficients: 3 Γ 2 = 6 Step 2: Put radicands under the same root: β5 Γ 3 Step 3: Multiply radicands: 5 Γ 3 = 15 Step 4: Combine the results: 6β15 Final Answer: 3β5 Γ 2β3 = 6β15 General Note: Multiplying surds can sometimes give whole numbers.
11 7. Dividing Surds Divide the radicands. Example: β50 β2 Step 1: Divide radicands: 50 Γ· 2 = 25 Step 2: Root the result: β25 = 5 8. Rationalising Denominators This section now covers every type of denominator: o A single surd o A number β surd o A number + surd 8A. Single Surd in the Denominator Example: 5 β3 Step 1: Multiply top and bottom by β3. 5 Γ β3 β3 Γ β3 Step 2: Simplify denominator. β3 Γ β3 = 3 Step 3: Final answer. 5β3 3 General Note: A denominator must not contain a surd. 8B. Denominator of the Form (a β βb)
12 Example: 7 4 β β3 To remove the surd, use the conjugate. Conjugate of 4 β β3 is 4 + β3. Step 1: Multiply top and bottom by the conjugate. 7 Γ (4 + β3) (4 β β3)(4 + β3) Step 2: Use the difference of two squares. (4 β β3)(4 + β3) = 42 β (β3)2 Step 3: Simplify. 42 = 16, (β3)2 = 3 16 β 3 = 13 Step 4: Final answer. 7(4 + β3) 13 8C. Denominator of the Form (a + βb) Example: 2 5 + β2 Conjugate of 5 + β2 is 5 β β2. Step 1: Multiply top and bottom by the conjugate. 2 Γ (5 β β2) (5 + β2)(5 β β2) Step 2: Use the difference of two squares. (5 + β2)(5 β β2) = 52 β (β2)2 Step 3: Simplify. 52 = 25, (β2)2 = 2
13 25 β 2 = 23 Step 4: Final answer. 2(5 β β2) 23 General Notes: β’ Use the conjugate when the denominator has two terms. β’ Conjugates remove surds using the identity: (π β π)(π + π) = π2 β π2 β’ A denominator must always be rational. β’ Rationalising makes future steps easier. 9. Comparing Surds β Step 1: Simplify each surd. β Step 2: Compare coefficients. β Step 3: Decide which is bigger. Example: β45 and β20 Step 1: β45 = 3β5 β20 = 2β5 Step 2: Compare 3β5 and 2β5. Step 3: 3β5 is greater. 10. Estimating the Value of a Surd (Without a Calculator) Using Square Numbers to Estimate β87 To estimate a square root, find two perfect squares that the number lies between. Step 1: Find two perfect squares We need perfect squares around 87. 9 Γ 9 = 81 10 Γ 10 = 100
14 So: Write this as a line: 81 ββ 87 ββ 100 100 β 81 = 19 87 β 81 = 6. This is the small gap from 81 up to 87. Compare the gaps We compare the small gap to the whole gap: 6 19 β 0.3 Lower root = 9 Add 0.3: 9 + 0.3 = 9.3 Estimated Answer; β87 β 9.3 β50 Step 1: Find nearest square numbers. 49 < 50 < 64 Step 2: Use their roots. β49 = 7, β64 = 8 Step 3: 50 is very close to 49 β estimate slightly above 7. Estimated value: β50 β 7.1 Example: 3β5 Step 1: Estimate β5. β4 = 2, β9 = 3 5 is close to 4 β estimate slightly above 2. β5 β 2.2 Step 2: Multiply.
15 3 Γ 2.2 = 6.6 Estimated value: 3β5 β 6.6 SURDS Q1. Simplify fully: β12 + β75 β12 = β4 Γ 3 = 2β3β75 = β25 Γ 3 = 5β3 So: 2β3 + 5β3 = 7β3 Q2. Rationalise the denominator: 7 β8 Answer + Explanation Multiply top & bottom by β8: 7β8 8 Simplify β8 = 2β2: 7Γ2β2 8 = 14β2 8 = 7β2 4 Q3. Expand and simplify: (β5 β 3)2 Use (π β π)2 = π2 β 2ππ + π2: (β5)2 β 2 Γ β5 Γ 3 + 95 β 6β5 + 9 = 14 β 6β5 Q4. Simplify fully: β18 β β50 + β8 β18 = 3β2β50 = 5β2β8 = 2β2 So: 3β2 β 5β2 + 2β2 = 0 Q5. Write in the form πβπ: β27 + 2β12 β27 = 3β3β12 = 2β3 So: 3β3 + 2 Γ 2β3 = 3β3 + 4β3 = 7β3
16 Q6. Rationalise the denominator: 5 2 β β3 Multiply by conjugate 2 + β3: 5(2+β3) (2ββ3)(2+β3) Denominator: 4 β 3 = 1 So answer: 5(2 + β3) = 10 + 5β3 Q7. Simplify: β45 β β5 β45 = 3β5 So: 3β5 β β5 = 2β5 Q8. Expand and simplify: (4 + β2)(3 β β2) Multiply term-by-term: 4 Γ 3 = 124 Γ (ββ2) = β4β2β2 Γ 3 = 3β2β2 Γ (ββ2) = β2 Combine: 12 β 2 β 4β2 + 3β2 = 10 β β2 Q9. Simplify fully: β200 β β32 β200 = 10β2β32 = 4β2 So: 10β2 β 4β2 = 6β2 Q10. Write in the form π + πβπ: (β3 + 5)(β3 β 2) Multiply: β3 Γ β3 = 3β3 Γ (β2) = β2β35 Γ β3 = 5β35 Γ (β2) = β10 Combine: 3 β 10 + 3β3 = β7 + 3β3 Q11. Rationalise the denominator: 9 β7 + 1
17 Multiply by conjugate: 9(β7β1) 7β1 = 9(β7β1) 6 Simplify: 3 2 (β7 β 1) Q12. Simplify fully: β72 + β18 β β8 β72 = 6β2β18 = 3β2β8 = 2β2 So: 6β2 + 3β2 β 2β2 = 7β2 Q13. Expand and simplify: (2β5 β 3)2 (2β5)2 = 20 β 2 Γ 2β5 Γ 3 = β12β532 = 9 So: 29 β 12β5 Q14. Simplify: β48 β 2β3 β48 = 4β3 So: 4β3 β 2β3 = 2β3 Q15. Rationalise the denominator: 4 3 β β2 Multiply by conjugate: 4(3+β2) 9β2 = 4(3+β2) 7 Q16. Write in the form πβπ: β24 + β150 Answer + Explanation β24 = 2β6β150 = 5β6 So: 7β6 Q17. Expand and simplify: (β7 + 4)(β7 β 6)
18 β7 Γ β7 = 7β7 Γ (β6) = β6β74 Γ β7 = 4β74 Γ (β6) = β24 Combine: 7 β 24 β 2β7 = β17 β 2β7 Q18. Simplify fully: β300 β β75 β300 = 10β3β75 = 5β3 So: 5β3 Q19. Rationalise the denominator: 11 β5 β 2 Multiply by conjugate: 11(β5+2) 5β4 = 11(β5 + 2) Q20. Expand and simplify: (3 β β11)2 9 β 6β11 + 11 = 20 β 6β11 Q21. Simplify fully: β162 + β18 β162 = 9β2β18 = 3β2 So: 12β2 Q22. Write in the form π + πβπ: (5 + β2)(3 + β2) Multiply: 15 + 5β2 + 3β2 + 2 = 17 + 8β2 Q23. Rationalise the denominator: 6 β3 + β2 Multiply by conjugate: 6(β3ββ2) 3β2 = 6(β3 β β2) Q24. Simplify fully: β500 β 5β2
19 β500 = 10β5Oops β different radicals β cannot combine. Final answer: 10β5 β 5β2 Q25. Expand and simplify: (2 + β3)(2 β β3) Difference of squares: 4 β 3 = 1 Q26. Simplify fully: β27 + β12 β β3 β27 = 3β3β12 = 2β3 So: 3β3 + 2β3 β β3 = 4β3 Q27. Rationalise the denominator: 8 4 β β6 Multiply by conjugate: 8(4+β6) 16β6 = 8(4+β6) 10 Simplify: 4 5 (4 + β6) Q28. Expand and simplify: (β10 + 3)(β10 β 5) β10 Γ β10 = 10β10 Γ (β5) = β5β103 Γ β10 = 3β103 Γ (β5) = β15 Combine: β5 β 2β10 Q29. Simplify fully: β250 + β40 β250 = 5β10β40 = 2β10 So: 7β10 Q30. Write in the form π + πβπ: (6 β β7)(4 + β7) Multiply: 24 + 6β7 β 4β7 β 717 + 2β7 31. Simplify fully
20 4β72 β 3β32 + β18 Solution First simplify each surd: β72 = β36 Γ 2 = 6β2 β32 = β16 Γ 2 = 4β2 β18 = β9 Γ 2 = 3β2 Now substitute: 4(6β2) β 3(4β2) + 3β2 24β2 β 12β2 + 3β2 15β2 32. Rationalise the denominator 9 3 β β5 Solution Multiply by the conjugate: 9 3 β β5 Γ 3 + β5 3 + β5 Denominator: (3 β β5)(3 + β5) = 9 β 5 = 4 Numerator: 9(3 + β5) = 27 + 9β5
21 Final: 27 + 9β5 4 Explanation: Conjugates remove the surd because the middle terms cancel. 33. Expand and simplify (β11 + 4)(β11 β 2) Solution Multiply term by term: β11 Γ β11 = 11 β11 Γ (β2) = β2β11 4 Γ β11 = 4β11 4 Γ (β2) = β8 Combine: 11 β 2β11 + 4β11 β 8 3 + 2β11 Explanation:Expand like normal algebra, then combine surd terms. 34. Write in the form πβπ 2β27 + 5β12 β 3β75 Solution
22 Simplify each: β27 = 3β3 β12 = 2β3 β75 = 5β3 Substitute: 2(3β3) + 5(2β3) β 3(5β3) 6β3 + 10β3 β 15β3 1β3 Explanation: All surds simplify to multiples of β3, so they combine easily. 35. Hard rationalising 6 β7 + β2 Solution Multiply by conjugate: 6 β7 + β2 Γ β7 β β2 β7 β β2 Denominator: (β7)2 β (β2)2 = 7 β 2 = 5 Numerator: 6(β7 β β2) = 6β7 β 6β2
23 Final: 6β7 β 6β2 5 36. Geometry Surd (image) A triangle has sides: β50, β18, π₯ The triangle is right-angled. Find π₯. Solution Hypotenuse is the largest: β50. Use Pythagoras: (β50)2 = (β18)2 + π₯2 50 = 18 + π₯2 π₯2 = 32 π₯ = β32 = 4β2 37. Hard expansion (β3 + β2)3 Solution Use binomial expansion: (π + π)3 = π3 + 3π2π + 3ππ2 + π3 Let π = β3, π = β2:
24 (β3)3 = 3β3 3(β3)2 (β2) = 3(3)(β2) = 9β2 3(β3)(β2)2 = 3(β3)(2) = 6β3 (β2)3 = 2β2 Combine: 3β3 + 6β3 + 9β2 + 2β2 9β3 + 11β2 Explanation: Cube surds carefully using the binomial formula. 38. Very hard simplification β63 β β28 β7 Solution Simplify numerator: β63 = 3β7 β28 = 2β7 So numerator becomes: 3β7 β 2β7 = β7 Now divide: β7 β7 = 1
25 39. Write in the form πβπ β π 4β18 3 + β2 Solution Step 1: Simplify numerator β18 = 3β2 4β18 = 12β2 So expression becomes: 12β2 3 + β2 Step 2: Rationalise using conjugate 12β2 3 + β2 Γ 3 β β2 3 β β2 Denominator: (3 + β2)(3 β β2) = 9 β 2 = 7 Numerator: 12β2(3 β β2) Expand: 12β2 Γ 3 = 36β2 12β2 Γ (ββ2) = β12 Γ 2 = β24
26 So numerator is: 36β2 β 24 Final: 36β2 β 24 7 Write in required form: 36 7 β2 β 24 7 Explanation:You simplify the surd first, then rationalise using the conjugate. The final form separates the surd term and the rational term. 40. Expand and simplify (β10 β β6)2 Solution (β10)2 β 2β10β6 + (β6)2 10 β 2β60 + 6 Simplify β60: β60 = β4 Γ 15 = 2β15 So: 10 + 6 β 2(2β15) 16 β 4β15
27 Homework: Surds 1. Simplify fully: πβππ β πβππ + πβππ 2. Rationalise the denominator: π πββπ 3. Expand and simplify: (βπ + π)(βπ β π) 4. Write in the form πβπ: πβππ + πβπ β βπππ 5. Simplify: βππββππ βπ 6. Rationalise the denominator: ππβπ π+βπ 7. Expand and simplify: (βππ β βπ)π 8. Given that π = πβπ β βππ, simplify πfully. 9. Write in the form π + πβπ: π+πβπ πββπ 10. Expand and simplify: (βπ + βπ)(βπ β πβπ) Prime Numbers PRIME NUMBERS A prime number is a whole number greater than 1 that has exactly two factors: β’ 1 β’ itself Examples: 2, 3, 5, 7, 11, 13, 17, 19β¦ How do we know a number is prime? To check if a number is prime: 1. Try dividing it by 2, 3, 5, 7, 11β¦ (all primes up to βn). 2. If none divide exactly β the number is prime. 3. If any divide exactly β the number is not prime. Why stop at βn? If a number has a factor bigger than βn, the matching factor must be smaller than βn. So checking up to βn is enough. Example: Check if 47 is prime. β47 β 6.8 β test primes β€ 6 β 2, 3, 5. β’ 47 Γ· 2 = not whole β’ 47 Γ· 3 = not whole β’ 47 Γ· 5 = not whole No exact division β 47 is prime. Q1: Show whether 89 is prime Step 1: Find β89 92 = 81 and 102 = 100, so β89is between 9 and 10. So we only need to test 2, 3, 5, 7. Step 2: Test small primes 2: 89 Γ· 2 = 44.5(not whole) 3: 89 Γ· 3 = 29.6 β¦(not whole)
28 5: last digit is 9, so not a multiple of 5 7: 89 Γ· 7 = 12.7 β¦(not whole) Step 3: Conclusion No factor found up to β89. So 89 is prime. Q2: Is 121 prime? Justify your answer Step 1: Find β121 112 = 121, so β121 = 11. Step 2: Test 11 121 Γ· 11 = 11(whole number) Step 3: Conclusion 121 has factors 1, 11, 121, and also 11 Γ 11. So 121 is not prime. Q3: Determine if 143 is prime. Show all factor checks Step 1: Estimate β143 112 = 121, 122 = 144, so β143is just under 12. Step 2: Test small primes 2: 143 Γ· 2 = 71.5(not whole) 3: 143 Γ· 3 = 47.6 β¦(not whole) 5: last digit is 3, so not a multiple of 5 7: 143 Γ· 7 = 20.4 β¦(not whole) 11: 143 Γ· 11 = 13(whole number) Step 3: Conclusion 143 = 11 Γ 13, so it has more factors than 1 and itself. 143 is not prime. Q4: Prove whether 157 is prime Step 1: Estimate β157 122 = 144, 132 = 169, so β157is between 12 and 13. We test 2, 3, 5, 7, 11. Step 2: Test small primes 2: 157 Γ· 2 = 78.5(not whole) 3: 157 Γ· 3 = 52.3 β¦(not whole) 5: last digit is 7, so not a multiple of 5 7: 157 Γ· 7 = 22.4 β¦(not whole) 11: 157 Γ· 11 = 14.27 β¦(not whole) Step 3: Conclusion No factor found up to β157. So 157 is prime. Q5: Is 221 prime? Explain fully Step 1: Estimate β221 142 = 196, 152 = 225, so β221is between 14 and 15. We test 2, 3, 5, 7, 11, 13. Step 2: Test small primes 2: 221 Γ· 2 = 110.5(not whole) 3: 221 Γ· 3 = 73.6 β¦(not whole) 5: last digit is 1, so not a multiple of 5 7: 221 Γ· 7 = 31.57 β¦(not whole) 11: 221 Γ· 11 = 20.09 β¦(not whole) 13: 221 Γ· 13 = 17(whole number) Step 3: Conclusion 221 = 13 Γ 17, so it has more factors than 1 and itself. 221 is not prime.
29 Homework Prime Numbers: Question 1: Decide if 167 is prime. Show all checks. Question 2: Decide if 169 is prime. Show all checks. Question 3: Decide if 187 is prime. Show all checks. Question 4: Decide if 211 is prime. Show all checks. Question 5: Decide if 247 is prime. Show all checks. Question 6: Decide if 257 is prime. Show all checks. Question 7: Decide if 289 is prime. Show all checks. Question 8: Decide if 299 is prime. Show all checks. Question 9: Decide if 301 is prime. Show all checks. Question 10: Decide if 323 is prime. Show all checks. Factors vs Multiples Students often confuse these Multiples Multiples are the times table of a number. Example: Multiples of 24 β 24, 48, 72, 96, 120, β¦ Factors Factors are numbers that divide exactly into the number. Example: Factors of 24 β 1, 2, 3, 4, 6, 8, 12, 24 How to find factors Step-by-step method 1. Start with 1 Γ number 2. Try 2 Γ, 3 Γ, 4 Γ, etc. 3. If the multiplication does not give the number β cross it out. 4. Stop when the pair repeats (e.g., 6 Γ 4 then 4 Γ 6). 5. All uncrossed rows are the factors. Example: Factors of 24 Try pairs: β’ 1 Γ 24 β β’ 2 Γ 12 β β’ 3 Γ 8 β β’ 4 Γ 6 β β’ 5 Γ (no exact division) β β’ Stop (next pair would repeat) Factors = 1, 2, 3, 4, 6, 8, 12, 24 QUESTIONS: Factors & Multiples Q1 :List the first 6 multiples of 17. Answer: 17, 34, 51, 68, 85, 102 (Just the times table.)
30 Q2: Find all factors of 36 using the βpair methodβ. Answer: 1Γ36 2Γ18 3Γ12 4Γ9 6Γ6 Stop Factors = 1, 2, 3, 4, 6, 9, 12, 18, 36 Q3:Find all factors of 84. Answer: 1Γ84 2Γ42 3Γ28 4Γ21 6Γ14 7Γ12 Stop Factors = 1,2,3,4,6,7,12,14,21,28,42,84 Homework: Factors & Multiples 1. List the first 10 multiples of 42. 2. A number has factors 1, 2, 4, 8, 16, 32, 64. Explain whether the number is prime, composite or neither. 3. Find all the factors of 144. Use the method: 1 Γ 144 2 Γ β¦ 3 Γ β¦ Continue until the factor pairs repeat. 4. Write the first 12 multiples of 18. Explain why multiples increase in a straight pattern. 5. A number has exactly 4 factors. Explain what type of number it must be. Give one example and justify your reasoning. 6. Find all the factors of 210 using the systematic method. Show each factor pair clearly. 7. A number ends with 000. Explain why this guarantees the number has factors 10, 100 and 1000. Then list three other factors the number must have. 8. Find all the factors of 252. Use the pair method starting from 1 Γ 252. 9. A number has factors 1, 5, 25, 125, 625. Explain why this number is a power of 5. Write the number in index form.
31 LCM LCM (Least Common Multiple) Step 1: write each number using prime factors Step 2: include every prime factor that appears in either number Step 3: if a prime appears more times in one number, include it that many times Step 4: multiply all the primes together Example: HCM of 18 and 30 18 2 3 3 30 2 3 5 Take only one copy of each prime factor 2 3 3 3 LCM = 2 Γ 3 Γ 3 Γ 5 (multiply all selected primes) The LCM (Least Common Multiple) is used when two or more processes repeat, and you want to know when they match again. It is also used when you need two quantities to reach the same total. You use LCM when you see words like: β’ repeat β’ cycle β’ interval β’ same total β’ meet again β’ match again β’ next time together LCM is for repeating events LCM : Questions Example 1 Find the LCM of 18 and 24. Step 1 : write prime factors 18 2 3 3 24 2 3 2 2 Take only one copy of each prime factor 2 3 3 2 2 Step 2 : take one copy of each prime factor using the highest repeats 2 , 2 , 2 , 3 , 3 Step 3 : multiply LCM = 2 Γ 2 Γ 2 Γ 3 Γ 3 LCM = 72 Example 2 Find the LCM of 15 and 28. 15 3 5 28 7 2 2 Take only one copy of each prime factor 3 5 7 2 2 Multiply LCM = 2 Γ 2 Γ 3 Γ 5 Γ 7 LCM = 420
32 Example 3 Find the LCM of 12 , 18 and 30. Step 1 : prime factors 12 = 2 Γ 2 Γ 3 18 = 2 Γ 3 Γ 3 30 = 2 Γ 3 Γ 5 Step 2 : take one copy of each prime factor using the highest repeats 2 , 2 , 3 , 3 , 5 Step 3 : multiply LCM = 2 Γ 2 Γ 3 Γ 3 Γ 5 LCM = 180 LCM: (Wordy Questions) Example 4 Two alarms ring at regular intervals. Alarm A rings every 18 minutes. Alarm B rings every 24 minutes. They ring together at 7:00. Find the next time they ring together. Step 1 : prime factors 18 = 2 Γ 3 Γ 3 24 = 2 Γ 2 Γ 2 Γ 3 Step 2 : take one copy of each prime factor using the highest repeats 2 , 2 , 2 , 3 , 3 Step 3 : multiply LCM = 72 Meaning: They ring together every 72 minutes. 7:00 + 72 minutes = 8:12 Example 5 A factory packs items in boxes of 36 and boxes of 48. The manager wants the same number of items from both types of boxes. Find the smallest number of items. Step 1 : prime factors 36 = 2 Γ 2 Γ 3 Γ 3 48 = 2 Γ 2 Γ 2 Γ 2 Γ 3 Step 2 : take one copy of each prime factor using the highest repeats 2 , 2 , 2 , 2 , 3 , 3 Step 3 : multiply LCM = 144 Meaning: The smallest number of items they can match is 144. Example 6 Two runners start at the same point. Runner A completes a lap every 45 seconds. Runner B completes a lap every 60 seconds. Find how long it takes until they meet again at the starting point. Step 1 : prime factors 45 = 3 Γ 3 Γ 5 60 = 2 Γ 2 Γ 3 Γ 5 Step 2 : take one copy of each prime factor using the highest repeats 2 , 2 , 3 , 3 , 5 Step 3 : multiply LCM = 180 Meaning: They meet again after 180 seconds.
33 Homework: LCM Note: When a question says smallest, it means: Find the first number that both processes reach. This is the LCM. LCM is used when two things repeat and you want the first time they match. 1. Find the LCM of 18 and 24 using prime factors. 2. Find the LCM of 15 and 28 using prime factors. 3. Find the LCM of 12 , 18 and 30 using prime factors. 4. Three buses leave a station at the same time. One returns every 15 minutes, one every 20 minutes, and one every 30 minutes. After how many minutes will all three be back together at the station? 5. A school bell rings every 12 minutes and a clock chimes every 18 minutes. If they both ring together at 9:00, at what time will they next ring together? 6.Two buses leave a station at the same time. Bus A returns every 20 minutes. Bus B returns every 30 minutes. Find the next time they return together. 7.A school orders packs of 28 pencils and packs of 36 pens. The school wants the same number of pencils and pens. Find the smallest number of pencils the school can order. 8. A machine produces a beep every 45 seconds. Another machine produces a beep every 75 seconds. They beep together at 9:00. Find the next time they beep together. 9.A gardener waters two plants. Plant A needs watering every 12 days. Plant B needs watering every 18 days. They are watered together today. Find the next day they will be watered together. 10.Three lights flash at intervals of 15 seconds, 20 seconds and 24 seconds. Find how long it takes until all three flash together. 11. A cyclist completes one lap of a track every 45 seconds. A runner completes one lap every 60 seconds. They start together. Find how long it takes until they meet again at the starting point. 12. A builder has two lengths of timber: 84 cm and 126 cm. He wants to cut both lengths into equal pieces with no waste. Find the greatest possible length of each piece. 13.A builder has two lengths of timber: 84 cm and 126 cm. He wants to cut both lengths into equal pieces with no waste. Find the greatest possible length of each piece.
34 HCF HCF (Highest Common Factor) Step 1: write each number using prime factors Step 2: choose only the primes that appear in both numbers Step 3 : multiply them together Example: 180 and 84 180 2 2 3 3 5 84 2 2 3 7 only the primes that appear in both numbers 2 2 3 HCF = 2 Γ 2 Γ 3 What HCF is used for HCF is used when: β’ you want equal pieces β’ you want no waste β’ you want the largest possible size β’ you want the maximum length β’ you want the biggest group size β’ you want the largest number that divides both. HCF is for cutting or grouping Example Find the HCF of 84 and 126. Step 1: prime factors 84 = 2 Γ 2 Γ 3 Γ 7 126 = 2 Γ 3 Γ 3 Γ 7 Step 2: take common primes 2 , 3 , 7 Step 3: multiply HCF = 2 Γ 3 Γ 7 = 42 Final Answer: 42 Example Find the HCF of 90 and 150. Step 1: prime factors 90 = 2 Γ 3 Γ 3 Γ 5 150 = 2 Γ 3 Γ 5 Γ 5 Step 2: take common primes 2 , 3 , 5 Step 3: multiply HCF = 2 Γ 3 Γ 5 = 30 Final Answer: 30 Example Find the HCF of 96 and 120. Step 1: prime factors 96 = 2 Γ 2 Γ 2 Γ 2 Γ 2 Γ 3 120 = 2 Γ 2 Γ 2 Γ 3 Γ 5 Step 2: take common primes 2 , 2 , 2 , 3 Step 3: multiply HCF = 2 Γ 2 Γ 2 Γ 3 = 24 Final Answer: 24
35 Example Find the HCF of 180 and 270. Step 1: prime factors 180 = 2 Γ 2 Γ 3 Γ 3 Γ 5 270 = 2 Γ 3 Γ 3 Γ 3 Γ 5 Step 2: take common primes 2 , 3 , 3 , 5 Step 3: multiply HCF = 2 Γ 3 Γ 3 Γ 5 = 90 Final Answer: 90 Example A builder has two lengths of timber: 84 cm and 126 cm. He wants to cut both lengths into equal pieces with no waste. Find the greatest possible length of each piece. Step 1: prime factors 84 = 2 Γ 2 Γ 3 Γ 7 126 = 2 Γ 3 Γ 3 Γ 7 Step 2: take common primes 2 , 3 , 7 Step 3: multiply HCF = 2 Γ 3 Γ 7 = 42 Final Answer: 42 cm Example A teacher has 32 students and 40 students in two classes. She wants to split both classes into equal groups with no student left out. Find the greatest group size. Step 1: prime factors 32 = 2 Γ 2 Γ 2 Γ 2 Γ 2 40 = 2 Γ 2 Γ 2 Γ 5 Step 2: take common primes 2 , 2 , 2 Step 3: multiply HCF = 2 Γ 2 Γ 2 = 8 Final Answer: 8 students per group Example A baker has 60 biscuits and 84 biscuits. He wants to pack them into bags with equal numbers and no waste. Find the greatest number of biscuits per bag. Step 1: prime factors 60 = 2 Γ 2 Γ 3 Γ 5 84 = 2 Γ 2 Γ 3 Γ 7 Step 2: take common primes 2 , 2 , 3 Step 3: multiply HCF = 2 Γ 2 Γ 3 = 12 Final Answer: 12 biscuits per bag Example Two ropes are 96 cm and 120 cm long. Find the greatest length of equal pieces with no waste. Step 1: prime factors 96 = 2 Γ 2 Γ 2 Γ 2 Γ 2 Γ 3 120 = 2 Γ 2 Γ 2 Γ 3 Γ 5 Step 2: take common primes 2 , 2 , 2 , 3 Step 3: multiply HCF = 2 Γ 2 Γ 2 Γ 3 = 24 Final Answer: 24 cm Example A gardener has two plant beds: 180 cm and 210 cm long. He wants to divide both into equal sections. Find the greatest possible section length. Step 1: prime factors 180 = 2 Γ 2 Γ 3 Γ 3 Γ 5 210 = 2 Γ 3 Γ 5 Γ 7 Step 2: take common primes 2 , 3 , 5 Step 3: multiply HCF = 2 Γ 3 Γ 5 = 30 Final Answer: 30 cm
36 Homework: HCF 1. Express 84 as a product of prime factors. Use a factor tree. 2. Find the LCM of 24 and 36 using prime factors. 3. Find the HCF of 90 and 126 using prime factors. 4. A baker has 48 muffins and 60 muffins. He wants to pack them into equal boxes with no waste. Find the greatest number of muffins per box. 5. Two metal rods are 144 cm and 180 cm long. Find the greatest length of equal pieces. 6. A teacher wants to split 36 students and 54 students into equal groups. Find the greatest group size. Fractions 1. Cancelling Down: Cancelling down means making a fraction simpler by dividing the top and bottom by the same number. Example Simplify 18 24 Step 1 : divide top and bottom by 6 18Γ·6 24Γ·6 = 3 4 Final Answer: 3 4 General Note: You can cancel in small steps or one big step. Always divide top and bottom by the same number. 2. Mixed Numbers: A mixed number has a whole number and a fraction. Example Write π π πas an improper fraction Step 1 : multiply the whole number by the denominator. 4 Γ 3 = 12 Step 2 : add the numerator.12 + 2 = 14 Step 3 : write the answer over the denominator 14 3 Final Answer: 14 3 4 +2 Γ 3
37 Second Example Write 31 4 as a mixed number. Step 1 : divide 31 Γ· 4 = 7 remainder 3 Step 2 : write the mixed number 7 3 4 3. Multiplying Fractions To multiply fractions, multiply the top numbers and multiply the bottom numbers. Cancel first if possible. Example 8 15 Γ 5 12 Step 1 : cancel 8 Γ· 4 = 2 12 Γ· 4 = 3 5 Γ· 5 = 1 15 Γ· 5 = 3 Step 2 : multiply 2 3 Γ 1 3 = 2 9 Final Answer: 2 9 4. Dividing Fractions: To divide fractions, turn the second fraction upside down and multiply. Example 2 1 3 Γ· 3 1 2 Step 1 : convert to improper fractions 2 1 3 = 7 3 3 1 2 = 7 2 Step 2 : flip the second fraction 7 3 Γ 2 7 Step 3 : multiply 2 3 Final Answer: 2 3 5. Common Denominators: To compare or add fractions, make the denominators the same. How to find a common denominator β’ Step 1 : list the denominators β’ Step 2 : find the LCM β’ Step 3 : rewrite each fraction using the LCM β’ Step 4 : compare or add Example Put in ascending order: 8 3 , 5 4 , 12 5 Step 1 : denominators β 3, 4, 5 Step 2 : LCM = 60 Step 3 : rewrite 8 3 = 160 60 5 4 = 75 60 12 5 = 144 60 Step 4 : order 75 60 , 144 60 , 160 60 Final Answer: 5 4 , 12 5 , 8 3 7 ______ 4 β 31 28 ββ 3 ascending = small β big descending = big β small
38 6. Adding and Subtracting Fractions Sort the denominators first. Example Calculate 2 1 5 β 1 1 2 Step 1 : convert 2 1 5 = 11 5 1 1 2 = 3 2 Step 2 : common denominator = 10 11 5 = 22 10 3 2 = 15 10 Step 3 : subtract 22β15 10 = 7 10 Final Answer: 7 10 7. Fractions of a Quantity Example 1 9 20of Β£360 ( of is x) 9 20 x Β£360 = Β£162 Final Answer: Β£162 Example 2 3 8of 64 3 8 Γ64= 24 8. Expressing One Number as a Fraction of Another Example 1 Write 180 as a fraction of 80. Step 1 : write 180 80 Step 2 : cancel 9 4 Final Answer: 9 4 Example 2 Write 45 as a fraction of 60. Step 1 : write 45 60 Step 2 : cancel 3 4 Final Answer: 3 4 Homework: Fractions 1 Simplify 42 56. 2 Write 3 4 7as an improper fraction. 3 Write 53 6 as a mixed number. 4 Calculate 9 14 Γ 7 18. 5 Calculate 4 1 3 Γ· 2 2 5. 6 Put in ascending order: 7 3 , 9 5 , 11 4 . 7 Calculate 3 2 5 + 1 3 10. 8 Find 7 12of 96.
39 9 Write 144 as a fraction of 90 and simplify. 10 A recipe uses 3 8of a bag of flour. How much flour is used from a 320 g bag? Converting Between Fractions, Decimals and Percentages Common conversions you should know: Fraction β Decimal β Percentage To convert a fraction into a decimal: β divide the top by the bottom β multiply the decimal by 100 to get a percentage Example 7 20 7 Γ· 20 = 0.35 0.35 Γ 100 = 35% Final Answer: Decimal = 0.35 Percentage = 35% Percentage β Decimal β Fraction To convert a percentage into a decimal: Γ· 100 To convert a decimal into a fraction: write the digits after the decimal over a power of 10 Example 35% β Step 1: decimal = 35 Γ· 100 = 0.35 β Step 2: fraction = 35 100 β Step 3: cancel β 7 20 Final Answer: 7 20 Terminating Decimals β Fractions Write the digits after the decimal over a power of 10. 1 2 = 0.5 = 50% 1 4 = 0.25 = 25% 3 4 = 0.75 = 75% 1 3 = 0.333 β¦ = 33 1 3 % 2 3 = 0.666 β¦ = 66 2 3 % 0.35 _______ 20 β 7.00 0 -- 70 60 -- 100 100 --- 0 A terminating decimal ends.
40 Example 0.78 Step 1: write 78 100 Step 2: cancel 39 50 Final Answer: 39 50 Recurring Decimals β Fractions A. The decimal is 0.abc: The number before the decimal point is zero. CASE 1: Recurring digits start immediately after the decimal Ex (0.3βΎ) Count how many digits repeat. The denominator is: β’ 1 repeating digit β 9 0.3- 0.3333333β¦. β’ 2 repeating digits β 99 0.3-4- 0.343434343β¦β¦.. β’ 3 repeating digits β 999 0.3-4- 5- 0.345345345345β¦β¦. The numerator is the repeating digits. Example 1 One repeating digit The decimal is: 0.3βΎ The repeating part starts straight away. The repeated digit is 3. This means there is one repeating digit, so the denominator is 9. 0.3βΎ π = 3 9 Final Answer:1 3 Example 2: Two repeating digits The decimal is: 0.45βΎ The repeating part starts straight away. The repeated digits are 45. This means there are two repeating digits, so the denominator is 99. Working A recurring decimal repeats forever. Working: 10 r= 3.3333β¦.. -r =0.333β¦.. ------------ 9r= 3.000000 π = 3 9
41 0.4βΎ5βΎ means 0.45454545 Let π = 0.4βΎ5βΎ Then π = 45 99 Final Answer 5 11 Example 3: Three repeating digits The decimal is: 0.128βΎ The repeating part starts straight away. The repeated digits are 128. This means there are three repeating digits, so the denominator is 999. Working Let π = 0.128βΎ Then π = 128 999 Final Answer 128 999 (Already simplified) Case2: Recurring digits do NOT start immediately ex 0.34- Golden Rule β Denominator Count repeating digits β write that many 9s Count non-repeating digits β write that many 0s after the 9s β Numerator The non-repeating part - one full repeating block Example 0.123βΎ Non-recurring part = 12 Recurring part = 3 Step 1: numerator Write digits up to one full repeat: 123 Subtract the non-recurring part: 123 β 12 = 111 Step 2 : denominator Non-recurring digits = 2 β write 2 zeros Recurring digits = 1 β write 1 nine Denominator = 900 Final Answer 111 900 B. When the number before the decimal point is NOT zero Working: 100 r= 45.454545 β¦ -r =0.45454545 β¦ ------------ 99r= 45.000000 π = 45 99 Working: 1000 r= 123.123123 β¦ -r =0.123123123 β¦ ------------ 999r= 123.000 π = 128 999 There is a non-recurring part first, then the repeating part.
42 The method is exactly the same as when the number before the decimal point is 0. The only difference: The whole number becomes part of the numerator. Everything else stays the same: β’ count repeating digits β’ denominator = 9, 99, 999 β’ numerator = whole number + one complete repeating block Case 1: The recurring digits start immediately after the decimal point This is the easy case. Golden Rule Denominator β’ Count the repeating digits. β’ Write the same number of 9s. Numerator β’ Write the whole number and one complete repeating block (ignore the decimal point). Example 1 The decimal is: 1.3βΎ Whole number = 1 Repeating digit = 3 There is 1 repeating digit β denominator = 9 Working Let π = 1.3βΎ 10π = 13.3βΎ β π = 1.3βΎ β β β β β β β β β β β β β β β β 9π = 12 So: π = 12 9 = 4 3 Example 2 The decimal is: 1.25βΎ Whole number = 1 Repeating digits = 25 There are 2 repeating digits β denominator = 99 Working Let π = 1.25βΎ 100π = 125.25βΎ β π = 1.25βΎ β β β β β β β β β β β β β β β β 99π = 124 So: π = 124 99 Final Answer: 124 99 Case 2: There is a non-recurring part before the repeating digits Example: 2.41βΎ6 Whole number = 2
43 Non-recurring digits = 41 Recurring digit = 6 Golden Rule Denominator β’ One 9 for each repeating digit β’ One 0 for each non-recurring decimal digit Numerator 1. Write all digits up to one complete repeating block 2. Subtract the digits before the repeating part Example π = 1.24 Only 4 is repeated digit Decimal: 1.24444β¦ Whole number = 1 Non-recurring digit = 2 ( 00) Repeating digit = 4 ( 9) So tha denominator is 990 1000π = 1244.4βΎ β 10π = 12.4βΎ β β β β β β β β β β β β β β β β β β β β β 990π = 1232 So: π = 1232 990 Now simplify: 1232 990 = 616 495 = 56 45
44
45 Questions: Recurring Decimals β Fractions Q1. Convert 0.4βΎ to a fraction Let π = 0.4βΎ 10π = 4.4βΎ β π = 0.4βΎ β β β β β β β β β β β β β β 9π = 4 So: π = 4 9 Q2. Convert 1.3βΎ to a fraction Let π = 1.3βΎ 10π = 13.3βΎ β π = 1.3βΎ β β β β β β β β β β β β β β 9π = 12 So: π = 12 9 = 4 3 Q3. Convert 0.12βΎ3 to a fraction Non-recurring = 12 Recurring = 3 Numerator: 123 β 12 = 111 Denominator: 1 repeating β 9 2 non-recurring β 00 = 900 111 900 = 37 300 Q4. Convert 1.2βΎ4 to a fraction (only 4 repeats) Let π = 1.2βΎ4 1000π = 1244.4βΎ β 10π = 12.4βΎ β β β β β β β β β β β β β β β β β β β β β 990π = 1232 So: π = 1232 990 = 56 45 HARD QUESTIONS (with full answers) Q5. Convert 2.41βΎ6 to a fraction (only 6 repeats) Let π = 2.41βΎ6 1000π = 2416.6βΎ β π = 2.41βΎ6 β β β β β β β β β β β β β β β β β β β β β 999π = 2414 So: π = 2414 999 Shortcut fraction: Numerator: 2416 β 241 = 2175 Denominator: 900 2175 900 = 29 12 Q6. Convert 3.45βΎ67 to a fraction
46 (two digits repeat) Whole number = 3 Non-recurring = 45 Recurring = 67 Numerator: 34567 β 345 = 34222 Denominator: 2 repeating β 99 2 non-recurring β 00 = 9900 34222 9900 = 17111 4950 Q7. Convert 4.1βΎ23 to a fraction (23 repeats) Whole number = 4 Non-recurring = 1 Recurring = 23 Numerator: 4123 β 41 = 4082 Denominator: 2 repeating β 99 1 non-recurring β 0 = 990 4082 990 = 2041 495 Q8. Convert 5.678βΎ9 to a fraction (only 9 repeats) Whole number = 5 Non-recurring = 678 Recurring = 9 Numerator: 56789 β 5678 = 51111 Denominator: 1 repeating β 9 3 non-recurring β 000 = 9000 51111 9000 Q9. Convert 0.βΎ128 to a fraction (three digits repeat) Let π = 0.βΎ128 1000π = 128.βΎ128 β π = 0.βΎ128 β β β β β β β β β β β β β β β β β β β β β 999π = 128 So: π = 128 999 Q10. Convert 7.βΎ345 to a fraction (three digits repeat) Let π = 7.βΎ345 1000π = 7345.βΎ345 β π = 7.βΎ345 β β β β β β β β β β β β β β β β β β β β β 999π = 7338 So: π = 7338 999
47 Homework: Recurring Decimals β Fractions Q1. Write 0.6βΎ (repeated digit: 6) as a fraction in its simplest form. Q2. Write 1.7βΎ (repeated digit: 7) as a fraction in its simplest form. Q3. Write 2.45βΎ (repeated digits: 45) as a fraction in its simplest form. Q4. Write 0.128βΎ (repeated digits: 128) as a fraction in its simplest form. Q5. Write 3.4βΎ (repeated digit: 4) as a fraction in its simplest form. Q6. Write 5.23βΎ (repeated digits: 23) as a fraction in its simplest form. Q7. Write 7.128βΎ (repeated digits: 128) as a fraction in its simplest form. Q8. Write 0.12βΎ3 (repeated digit: 3) as a fraction in its simplest form. Q9. Write 1.45βΎ6 (repeated digit: 6) as a fraction in its simplest form. Q10. Write 2.378βΎ9 (repeated digit: 9) as a fraction in its simplest form. Q11. Write 1.2βΎ4 (repeated digit: 4) as a fraction in its simplest form. Q12. Write 3.57βΎ8 (repeated digit: 8) as a fraction in its simplest form. Q13. Write 4.903βΎ1 (repeated digit: 1) as a fraction in its simplest form. Q14. Write 0.βΎ27 (repeated digits: 27) as a fraction in its simplest form. Q15. Write 2.3βΎ45 (repeated digits: 45) as a fraction in its simplest form. Q16. Write 5.12βΎ67 (repeated digits: 67) as a fraction in its simplest form. Q17. Write 0.βΎ345 (repeated digits: 345) as a fraction in its simplest form. Q18. Write 6.βΎ128 (repeated digits: 128) as a fraction in its simplest form. Q19. Write 3.45βΎ678 (repeated digits: 678) as a fraction in its simplest form. Q20. Write 1.034βΎ7 (repeated digit: 7) as a fraction in its simplest form. Q21. Write 2.4βΎ56 (repeated digits: 56) as a fraction in its simplest form. Q22. Write 7.89βΎ3 (repeated digit: 3) as a fraction in its simplest form. Q23. Write 4.1βΎ23 (repeated digits: 23) as a fraction in its simplest form. Q24. Write 0.45βΎ678 (repeated digits: 678) as a fraction in its simplest form. Q25. Write 3.204βΎ9 (repeated digit: 9) as a fraction in its simplest form.
48 Bounds (Upper & Lower Bounds) When a number is rounded, the actual value is not exact. It lies between a lower bound and an upper bound. If a value is rounded to the nearest: β’ 1 β’ 0.1 β’ 0.01 β’ 0.001 Then the error half-unit is: β’ nearest 1 β Β± 0.5 β’ nearest 0.1 β Β± 0.05 β’ nearest 0.01 β Β± 0.005 β’ nearest 0.001 β Β± 0.0005 Lower Bound Rounded value β half unit Upper Bound Rounded value + half unit Error Interval: lower bound β€x<upper bound Error Interval β the inequality that shows the full range between those bounds. Truncation: Lower and Upper Bounds Truncated means the number has been cut off at a certain decimal place. Nothing is rounded. You simply stop at that decimal place and throw away everything after it. Example: Truncated to 1 d.p. 2.4 β€ x < 2.5 Bounds in Calculations 1. Multiplying A rectangle has a length of 12 cm and a width of 8 cm, both measured to the nearest centimetre. Find the minimum and maximum possible values of the area. Bounds: 11.5 β€ length < 12.5 7.5 β€ width < 8.5 Minimum area = 11.5 Γ 7.5 = 86.25 cmΒ² Maximum area = 12.5 Γ 8.5 = 106.25 cmΒ² Error interval for area: 86.25 cmΒ² β€ A < 106.25 cmΒ² 2. Adding Two friends run a race. Samβs time is 42 minutes, correct to the nearest minute. Aliβs time is 38 minutes, correct to the nearest minute. Multiplying Small Γ small gives the smallest area. Big Γ big gives the biggest area.
49 Find the minimum and maximum possible total time. Bounds: 41.5 β€ Sam < 42.5 37.5 β€ Ali < 38.5 Minimum total = 41.5 + 37.5 = 79 minutes Maximum total = 42.5 + 38.5 = 81 minutes Error interval for total time: 79 β€ T < 81 3. Subtracting A plank of wood is 150 cm long to the nearest centimetre. A piece cut from it is 47 cm long to the nearest centimetre. Find the minimum and maximum possible remaining length. Bounds: 149.5 β€ plank < 150.5 46.5 β€ cut < 47.5 Minimum remaining length = 149.5 β 47.5 = 102 cm Maximum remaining length = 150.5 β 46.5 = 104 cm Error interval for remaining length: 102 β€ R < 104 Reason: To make subtraction small, subtract the biggest number. To make subtraction big, subtract the smallest number. 4. Dividing A cyclist travels 20 km to the nearest kilometre. Her time is 45 minutes to the nearest minute. Find the minimum and maximum possible values of her speed in km/min. Bounds: 19.5 β€ distance < 20.5 44.5 β€ time < 45.5 Minimum speed = min distance Γ· max time = 19.5 Γ· 45.5 β 0.428 km/min Maximum speed = max distance Γ· min time = 20.5 Γ· 44.5 β 0.460 km/min Error interval for speed: 0.428β¦ < s < 0.460β¦ Adding Small + small gives the smallest total. Big + big gives the biggest total. Subtracting To make subtraction small, subtract the biggest number. To make subtraction big, subtract the smallest number. Dividing Minimum value = (min numerator Γ· max denominator) Maximum value = (max numerator Γ· min denominator) Example The mass of a cake is given as 2.4 kg, rounded to the nearest 0.1 kg. Nearest 0.1 β half unit = 0.05 Lower bound = 2.4 β 0.05 = 2.35 kg Upper bound = 2.4 + 0.05 = 2.45 kg Error Interval:2.35 β€ π < 2.45 Example A length is 12 cm, correct to the nearest centimetre. Nearest 1 β error = Β±0.5
50 Lower bound: 12 β 0.5 = 11.5 Upper bound: 12 + 0.5 = 12.5 Error interval: 11.5 β€ length < 12.5 Example A mass is 6.8 kg, correct to the nearest 0.1 kg. Nearest 0.1 β error = Β±0.05 Lower bound: 6.8 β 0.05 = 6.75 Upper bound: 6.8 + 0.05 = 6.85 Lower bound = 6.75 kg Upper bound = 6.85 kg Example A temperature is 18.42Β°C, correct to the nearest 0.01Β°C. Nearest 0.01 β error = Β±0.005 Lower bound: 18.42 β 0.005 = 18.415 Upper bound: 18.42 + 0.005 = 18.425 Error interval: 18.415 β€ T < 18.425 Example A car travels 250 km, correct to the nearest 10 km. Nearest 10 β error = Β±5 Greatest possible distance: 250 + 5 = 255 km Example A pinboard is measured as being 0.89 m wide and 1.23 m long, to the nearest cm. a) Calculate the minimum and maximum possible values for the area of the pinboard. Find the bounds for the width and length: 0.885 m β€ width < 0.895 m 1.225 m β€ length < 1.235 m Find the minimum area by multiplying the lower bounds, and the maximum by multiplying the upper bounds: minimum possible area = 0.885 Γ 1.225 maximum possible area = 0.895 Γ 1.235 Example Mass = 2.4 kg (nearest 0.1 kg) Half unit = 0.05 Lower bound = 2.4 β 0.05 = 2.35 Upper bound = 2.4 + 0.05 = 2.45 Interval: 2.35 β€ m < 2.45 Example (Truncated) Mass = 2.4 truncated to 1 d.p. Lower bound = 2.4 Upper bound = 2.5 Interval: 2.4 β€ x < 2.5 Example Width = 0.89 m (nearest cm) Length = 1.23 m (nearest cm) Half unit = 0.005 m
51 Width: 0.885 β€ w < 0.895 Length: 1.225 β€ l < 1.235 Minimum area = 0.885 Γ 1.225 Maximum area = 0.895 Γ 1.235 Example The mass of a letter is 58 grams to the nearest gram. Complete the error interval for the mass of the letter. Solution: Nearest 1 g β half = 0.5 g Lower bound: 58 β 0.5 = 57.5 g Upper bound: 58 + 0.5 = 58.5 g Error interval: 57.5 g β€ mass < 58.5 g Example The length of a mobile phone is 142 mm to the nearest millimetre. Complete the error interval for the length of the mobile phone. Solution: Nearest 1 mm β half = 0.5 mm Lower bound: 142 β 0.5 = 141.5 mm Upper bound: 142 + 0.5 = 142.5 mm Error interval: 141.5 mm β€ length < 142.5 mm Example The distance between two towns is 300 miles to the nearest 100 miles. Complete the error interval for distance. Solution: Nearest 100 miles β half = 50 miles Lower bound: 300 β 50 = 250 miles Upper bound: 300 + 50 = 350 miles Error interval: 250 miles β€ distance < 350 miles Example Frank rounds a number, y, to the nearest ten. His result is 80 Write down the error interval for y. Solution: Nearest 10 β half = 5 Lower bound: 80 β 5 = 75 Upper bound: 80 + 5 = 85 Error interval: 75 β€ y < 85 Example Freya rounds a number, y, to one decimal place. Her result is 6.4 Write down the error interval for y. Solution: Nearest 0.1 β half = 0.05 Lower bound: 6.4 β 0.05 = 6.35 Upper bound: 6.4 + 0.05 = 6.45 Error interval: 6.35 β€ y < 6.45 Example A number, p, is rounded to 2 decimal places to give 10.68 Using inequalities, write down the error interval for p. Solution: Nearest 0.01 β half = 0.005 Lower bound: 10.68 β 0.005 = 10.675 Upper bound: 10.68 + 0.005 = 10.685 Error interval: 10.675 β€ p < 10.685 Example
52 Elliott weighs 71.8 kg. This mass, m, is to the nearest 100 g. Write the error interval for m. Solution: 71.8 kg = 71800 g Nearest 100 g β half = 50 g Lower bound: 71800 β 50 = 71750 g Upper bound: 71800 + 50 = 71850 g Error interval in grams: 71750 g β€ m < 71850 g Or in kg: 71.75 kg β€ m < 71.85 kg Example The length of each side of a rhombus is 5 cm to the nearest centimetre. (a) Write the error interval for the length of each side of the rhombus. (b) Write the error interval for the perimeter of the rhombus. Solution (a): Nearest 1 cm β half = 0.5 cm Lower bound: 5 β 0.5 = 4.5 cm Upper bound: 5 + 0.5 = 5.5 cm Error interval: 4.5 cm β€ side length < 5.5 cm Solution (b): Perimeter = 4 Γ side Minimum perimeter: 4 Γ 4.5 = 18 cm Maximum perimeter: 4 Γ 5.5 = 22 cm Error interval: 18 cm β€ P < 22 cm Example The length of each side of a regular heptagon is 2.8 cm to 1 decimal place. Write the error interval for the perimeter, P. Solution: Side length to 1 d.p. β nearest 0.1 β half = 0.05 Side bounds: 2.8 β 0.05 = 2.75 cm 2.8 + 0.05 = 2.85 cm Perimeter of heptagon = 7 Γ side Minimum perimeter: 7 Γ 2.75 = 19.25 cm Maximum perimeter: 7 Γ 2.85 = 19.95 cm Error interval: 19.25 cm β€ P < 19.95 cm Example The length of a rectangle is 20 cm. The width of the rectangle is 6 cm. Both measurements are correct to the nearest centimetre. Write the error interval for the area of the rectangle, A. Solution: Length bounds: 20 β 0.5 = 19.5 cm 20 + 0.5 = 20.5 cm Width bounds: 6 β 0.5 = 5.5 cm 6 + 0.5 = 6.5 cm Minimum area: 19.5 Γ 5.5 = 107.25 cmΒ² Maximum area: 20.5 Γ 6.5 = 133.25 cmΒ² Error interval: 107.25 cmΒ² β€ A < 133.25 cmΒ² Example A band writes two songs. The first song is 3 minutes long to the nearest minute. The second song is 5 minutes long to the nearest minute. Show that the total time for both songs could be 8 minutes 58 seconds. Solution: First song bounds: 3 β 0.5 = 2.5 min 3 + 0.5 = 3.5 min Second song bounds: 5 β 0.5 = 4.5 min 5 + 0.5 = 5.5 min Total time bounds: Minimum: 2.5 + 4.5 = 7 min Maximum: 3.5 + 5.5 = 9 min
53 So total time can be any value with: 7 min β€ T < 9 min 8 min 58 s = 8.966β¦ min, which is less than 9 min, so it is possible. Example x is rounded to 3 significant figures. The answer is 12.7 Write the error interval for x. Solution: 12.7 to 3 s.f. β nearest 0.1 β half = 0.05 Lower bound: 12.7 β 0.05 = 12.65 Upper bound: 12.7 + 0.05 = 12.75 Error interval: 12.65 β€ x < 12.75 Example A number, y, is 9200 when rounded to 3 significant figures. Write down the error interval. Solution: 9200 to 3 s.f. β nearest 100 β half = 50 Lower bound: 9200 β 50 = 9150 Upper bound: 9200 + 50 = 9250 Error interval: 9150 β€ y < 9250 Example The length of a line, L, was given as 2.6 cm, truncated to 1 decimal place. Complete the error interval for L. Solution: Truncated to 1 d.p. β cut at 0.1 Lower bound = 2.6 Upper bound = 2.7 Error interval: 2.6 cm β€ L < 2.7 cm Example A number, y, is 0.04 when truncated to 2 decimal places. Complete the error interval for y. Solution: Truncated to 2 d.p. Lower bound = 0.04 Upper bound = 0.05 Error interval: 0.04 β€ y < 0.05 Example A number, n, is truncated to 1 decimal place. The result is 39.1 Using inequalities, write down the error interval for n. Solution: Truncated to 1 d.p. Lower bound = 39.1 Upper bound = 39.2 Error interval: 39.1 β€ n < 39.2 Example Sahil solves an equation to find the value of x. His answer for x is 8.25 His teacher has realised that Sahil has written down the first three digits of x from his calculator display. (a) Write down the error interval for x. (b) Explain why Sahil should not have truncated his answer. Solution (a): First three digits β truncated to 2 decimal places. Lower bound = 8.25 Upper bound = 8.26 Error interval: 8.25 β€ x < 8.26
54 Solution (b): His answer might be very close to 8.26 (for example 8.2599). Truncating makes it look further from 8.26 than it really is. Rounding would give a more accurate final answer. Example When asked her age, Summer says that she rounds her age to the nearest year. When asked her age, Ciara says that she truncates her age to the nearest year. Who uses the most common approach? Solution: Most people round their age to the nearest year. So the most common approach is used by Summer. Example The perimeter of a regular pentagon is 18 cm to the nearest centimetre. Using inequalities, write down the error interval for the side length, x. Solution: Perimeter 18 to nearest 1 cm β half = 0.5 Perimeter bounds: 17.5 cm β€ P < 18.5 cm Side length = P Γ· 5 Minimum side: 17.5 Γ· 5 = 3.5 cm Maximum side: 18.5 Γ· 5 = 3.7 cm Error interval: 3.5 cm β€ x < 3.7 cm Example A woman runs 400 metres to the nearest 10 metres. It takes her 80 seconds to the nearest 10 seconds. Work out the error interval for her speed, s. Solution: Distance 400 m to nearest 10 m β half = 5 m Distance bounds: 395 m β€ d < 405 m Time 80 s to nearest 10 s β half = 5 s Time bounds: 75 s β€ t < 85 s Speed s = distance Γ· time Minimum speed = min distance Γ· max time = 395 Γ· 85 β 4.647 m/s Maximum speed = max distance Γ· min time = 405 Γ· 75 = 5.4 m/s Error interval: 4.647β¦ m/s < s < 5.4 m/s Example A number, y, is 100 when rounded to 1 significant figure. Circle the correct error interval for y. Options include (correct one): 95 β€ y < 150 Solution: 100 to 1 s.f. β nearest 100 β half = 50 Lower bound: 100 β 50 = 50 Upper bound: 100 + 50 = 150 But numbers from 50 to 94.9β¦ round to 50, not 100. To round to 100, y must be at least 95. So correct interval: 95 β€ y < 150 Example The area of a circle is 50 cmΒ² to one significant figure. Find the error interval for the circumference of the circle. Solution: Area 50 to 1 s.f. β nearest 10 β half = 5 Area bounds: 45 cmΒ² β€ A < 55 cmΒ² Use A = ΟrΒ²
55 For A = 55: rΒ² = 55 Γ· Ο β 17.507β¦ r β β17.507β¦ β 4.1841β¦ Diameter β 2r β 8.368β¦ Circumference β Ο Γ diameter β 26.2897β¦ cm For A = 45: rΒ² = 45 Γ· Ο β 14.3239β¦ r β β14.3239β¦ β 3.78469β¦ Diameter β 7.569β¦ Circumference β Ο Γ diameter β 23.7799β¦ cm Error interval: 23.7799β¦ cm β€ C < 26.2897β¦ cm Homework: Bounds Q1. A rectangular garden is measured as 4.8 m long and 3.2 m wide, both correct to the nearest 0.1 m. Write the error interval for the area of the garden. Q2. A runner completes a race in 52 minutes, correct to the nearest minute. Her friend completes the same race in 47 minutes, also correct to the nearest minute. Write the error interval for the total time taken by both runners. Q3.A plank of wood is 2.6 m long to the nearest 0.1 m. A piece cut from it is 0.9 m long to the nearest 0.1 m. Write the error interval for the remaining length of the plank. Q4.A cyclist travels 18 km to the nearest kilometre. Her time is 42 minutes to the nearest minute. Write the error interval for her speed in km/min. Q5.The side length of a regular hexagon is 7.4 cm, correct to the nearest 0.1 cm. Write the error interval for the perimeter of the hexagon. Q6.A number, x, is 0.53 when rounded to 2 decimal places. Write down the error interval for x. Q7.The mass of a bag of flour is 1.8 kg, correct to the nearest 0.1 kg. A recipe uses 0.6 kg, correct to the nearest 0.1 kg. Write the error interval for the remaining mass of flour. Q8.A rectangular screen has a width of 32 cm and a height of 18 cm, both measured to the nearest centimetre. Write the error interval for the area of the screen. Q9. A number, y, is 700 when rounded to 2 significant figures. Write down the error interval for y. Q10. A car travels 240 miles to the nearest 10 miles. The journey takes 4 hours to the nearest 0.5 hours. Write the error interval for the average speed of the car in miles per hour.
56 Rounding , Significant Figures, Estimating Calculations 1. Rounding to Decimal Places (d.p.) Steps 1. Identify the last digit you are keeping. 2. Look at the next digit to the right (the decider). 3. If the decider is 5 or more, round the last digit up. 4. If the decider is 4 or less, keep the last digit the same. 5. After rounding, no extra digits are allowed. Example Round 7.45839 to 2 d.p. Last digit = 5 Decider = 8 7.45839 β 7.46 2. Significant Figures (s.f.) Rules 1. The 1st significant figure is the first non-zero digit. 2. The 2nd, 3rd, 4thβ¦ follow immediately after, even if they are zeros. 3. After rounding, fill in zeros up to the decimal point, but not beyond. Number 3 s.f. 2 s.f. 1 s.f. 54.7651 54.8 55 50 0.0045902 0.00459 0.0046 0.005 30895.4 30900 31000 30000 3. Estimating Calculations When estimating, you do not follow rounding rules. You choose the nearest easy number to make the calculation simpler. Example 1: 127.8 20 ββ 27 ββ 30 (We only care about the β27β part of 127.8) 27 is closer to 30 than 20. So 127.8 becomes 130 for estimating. Example 2: 41.9 Look at the tens: 40 ββ 41.9 ββ 50 41.9 is closer to 40. So 41.9 becomes 40 for estimating. Example 3: 56.5 Look at the tens: 50 ββ 56.5 ββ 60 56.5 is closer to 60. So 56.5 becomes 60 for estimating. Estimating Is NOT Rounding
57 Example 1: By rounding each number to the nearest 10, estimate: 46 Γ 32 Step 1: Choose nearest easy numbers 46 β 50 (closer to 50 than 40) 32 β 30 (closer to 30 than 40) Step 2: Multiply 50 Γ 30 = 1500 Estimated answer: 1500 Example 2: Estimate: 3.4 Γ 509 Step 1: Choose nearest easy numbers 3.4 β 3 (closer to 3 than 4) 509 β 500 (closer to 500 than 600) Step 2: Multiply 3 Γ 500 = 1500 Estimated answer: 1500 Example 3 Work out an estimate for: 6.8 Γ 104 Step 1: Choose nearest easy numbers 6.8 β 7 (closer to 7 than 6) 104 β 100 (closer to 100 than 110) Step 2: Multiply 7 Γ 100 = 700 Estimated answer: 700 Homework: Rounding, Significant Figures, Estimating Calculations Q1. By rounding each number to the nearest 10, estimate the value of: (76 Γ 53) Γ· 214 Q2. By rounding each number to the nearest 10, estimate: 45 Γ 28 Q3. Estimate: 3.7 Γ 612 Q4. Work out an estimate for the value of: 8.4 Γ 109 Q5. A student uses her mobile phone while abroad. She received 4 minutes of calls at 92p per minute. She made 6 minutes of calls at 67p per minute. She sent 5 text messages at 48p per message. Estimate the total cost of her mobile phone use. Q6. A merchandise stall sells: 62 bubble wands for $31.40 each 175 badges for $8.85 each. Estimate the total money paid for the items. Q7. Estimate the value of: 5.18 Γ 12.7 Γ· 6.9 Q8. A shop sells: 37 notebooks at Β£2.89 each 84 pens at Β£1.22 each Estimate the total cost of all items. Q9. Estimate: 9.6 Γ 304 Q10. By rounding each number to the nearest 10, estimate: (93 Γ 41) Γ· 187
58 Standard form Standard form is a way to write very big or very small numbers in the form: A Γ 10βΏ Case 1: Whole Number β Standard Form Step 1: Put the decimal point at the end of the whole number. β’ Example: 45600 β 45600. Step 2: Move the decimal point left until the first number is between 1 and 10. β’ 45600. β 4.56 Step 3: Count how many places the decimal moved. β’ 45600. β 4.56 = 4 places Step 4: Write the answer as: 4.56Γ104 Case 2: Decimal Number β Standard Form Step 1: Move the decimal point right until the first number is between 1 and 10. The number of moves becomes the negative exponent. Example: 0.00456 β 4.56 Step 2: Count how many places the decimal moved. 0.00456 β 4.56 = 3 places Step 3: Write the answer as: 4.56Γ10β3 When the Answer Needs Modification Case 1: A is 10 or More: Move the decimal 1 place left and add 1 to the exponent. Example 45.6 Γ 103 Since 45.6 is greater than 10: β’ Move decimal left: 45.6 β 4.56 β’ Add 1 to the exponent: 3 β 4 Answer: 4.56 Γ 104 Case 2: A is Less Than 1 Rule Move the decimal 1 place right and subtract 1 from the exponent. Example 0.56 Γ 104 Whole numbers β Move LEFT β Exponent +1 Decimal numbers (< 1) β Move RIGHT β Exponent β1 A β₯ 10 β Move decimal LEFT β Exponent +1 A < 1 β Move decimal RIGHT β Exponent β1
59 Since 0.56 is less than 1: β’ Move decimal right: 0.56 β 5.6 β’ Subtract 1 from the exponent: 4 β 3 Answer: 5.6 Γ 103 Operations in Standard Form 1. Multiplying Exampl: (3.2 Γ 104) Γ (5 Γ 103) Step 1: Multiply front numbers 3.2 Γ 5 = 16 Step 2: Add powers 10β΄ Γ 10Β³ = 10β· So we have: 16 Γ 10β· Step 3: Fix the front number 16 is too big β move decimal left: 16 β 1.6 Power increases by 1: 1.6 Γ 108 2. Dividing: (7.5 Γ 106) Γ· (2.5 Γ 102) Step 1: Divide front numbers 7.5 Γ· 2.5 = 3 Step 2: Subtract powers 10βΆ Γ· 10Β² = 10β΄ So we have: 3 Γ 104 Step 3: Front number already between 1 and 10 No change needed. 3. Adding: (4.8 Γ 105) + (7.2 Γ 104) Step 1: Make powers the same Rewrite 7.2 Γ 10β΄ as 0.72 Γ 10β΅ Step 2: Add front numbers 4.8 + 0.72 = 5.52 So we have: 5.52 Γ 105 Step 3: Front number is between 1 and 10 No change needed. 4. Subtracting: (9.1 Γ 106) β (3.4 Γ 105) Step 1: Make powers the same Rewrite 3.4 Γ 10β΅ as 0.34 Γ 10βΆ Step 2: Subtract front numbers 9.1 β 0.34 = 8.76 So we have: 8.76 Γ 106 Step 3: Front number is between 1 and 10 No change needed.
60 Q1. Write 84 300 000 in standard form. 84 300 000 = 8.43 Γ 10β· Q2. Write 0.00000782 in standard form. 0.00000782 = 7.82 Γ 10β»βΆ Q3. Convert 19.4 Γ 10βΆ into correct standard form. 19.4 is too big β move decimal left: 19.4 β 1.94 Power increases by 1: 1.94 Γ 10β· Q4. Convert 0.045 Γ 10β»Β³ into correct standard form. 0.045 is too small β move decimal right: 0.045 β 4.5 Power decreases by 1: 4.5 Γ 10β»β΄ Q5. Calculate in standard form: (3 Γ 10β΄) Γ (7.2 Γ 10β΅) Front numbers: 3 Γ 7.2 = 21.6 Powers: 10β΄ Γ 10β΅ = 10βΉ 21.6 Γ 10βΉ β fix front number: 2.16 Γ 10ΒΉβ° Q6. Calculate in standard form: (8.1 Γ 10β·) Γ· (2.7 Γ 10Β³) Front numbers: 8.1 Γ· 2.7 = 3 Powers: 10β· Γ· 10Β³ = 10β΄ , 3 Γ 10β΄ Q7.Add in standard form: (6.4 Γ 10β΅) + (9.2 Γ 10β΄) Rewrite 9.2 Γ 10β΄ as 0.92 Γ 10β΅ Add fronts: 6.4 + 0.92 = 7.32 7.32 Γ 10β΅ Q8. Subtract in standard form: (4.9 Γ 10βΆ) β (7.3 Γ 10β΅) Rewrite 7.3 Γ 10β΅ as 0.73 Γ 10βΆ Subtract fronts: 4.9 β 0.73 = 4.17 4.17 Γ 10βΆ Q9.Write 0.000000456 in standard form. Move decimal to after 4 β 7 places: 4.56 Γ 10β»β· Q10. Rewrite 12.8 Γ 10β»Β² into correct standard form. 12.8 is too big β move decimal left: 12.8 β 1.28 Power increases by 1: 1.28 Γ 10β»ΒΉ
61 Homework: Standard Form Q1. A distant planet is approximately 5.2 Γ 10ΒΉβ΄ km from Earth. A probe travels 1.3 Γ 10βΉ km each year. How many years would it take the probe to reach the planet? Give your answer in standard form. Q2. A single virus particle has a mass of 4.6 Γ 10β»ΒΉΒ³ g. A sample contains 8.2 Γ 10βΈ virus particles. Calculate the total mass of the sample. Give your answer in standard form. Q3. A factory produces 3.1 Γ 10βΆ metal clips each week. The factory operates for 52 weeks in a year. Calculate the total number of clips produced in one year. Give your answer in standard form. Q4. A data centre stores 7.8 Γ 10βΉ bytes of information on each server. The centre has 24 identical servers. Calculate the total amount of data stored. Give your answer in standard form. Q5. A scientist has 9.2 Γ 10β»Β³ litres of a solution. Each test requires 3.7 Γ 10β»β΅ litres. How many tests can be carried out? Give your answer in standard form, where appropriate. Q6. The diameter of one cell is 1.4 Γ 10β»β· m. Another cell has a diameter of 6.2 Γ 10β»β· m. How much larger is the second cell? Give your answer in standard form. Q7. A country uses 4.9 Γ 10βΉ litres of water each day. Estimate the total amount of water used in 45 days. Give your answer in standard form. Q8. The Moon is approximately 3.8 Γ 10β΅ km from Earth. A signal travels at 2.4 Γ 10β΅ km/s. Calculate the time taken for the signal to reach the Moon. Give your answer in standard form and in seconds. Q9. A company manufactures 6.3 Γ 10β· light bulbs each year. During testing, 4.1 Γ 10β»Β³ of the bulbs are found to be faulty. Calculate the number of faulty bulbs produced each year. Give your answer in standard form. Q10. A warehouse contains 5.4 Γ 10βΆ crates. Each crate has a mass of 2.7 Γ 10ΒΉ kg. Calculate the total mass of all the crates. Give your answer in standard form.