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Express 6.455 correct to 1 d.p, 2 s.f, and nearest whole number. Express 0.52 as a common fraction. Express 102 as a product of its prime factors. Find HCF and LCM of 30 and 24.

Mathematics

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1. (a) Express 6.455 correct to:

i) 1 d.p

ii) 2 s.f

iii) nearest whole number

i) 1 d.p (1 decimal place)
The first decimal place is 4. The digit after it is 5, so we round up.

6.455≈6.56.455 \approx 6.5

ii) 2 s.f (2 significant figures)
The first significant figure is 6, the second is 4. The digit after it is 5, so we round up.

6.455≈6.56.455 \approx 6.5

iii) nearest whole number
The whole number is 6. The digit after it is 4, so we round down.

6.455≈66.455 \approx 6

1. (b) Express 0.52 as a common fraction in its simplest form.

  1. Write the decimal as a fraction.

    0.52=521000.52 = \frac{52}{100}
  2. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.

    52÷4100÷4=1325\frac{52 \div 4}{100 \div 4} = \frac{13}{25}

    The fraction is already in its simplest form.

    1325\boxed{\frac{13}{25}}

    2. (a) Express 102 as a product of its prime factors in index form.

  3. Find the prime factors of 102.

    102÷2=51102 \div 2 = 51 51÷3=1751 \div 3 = 17 17÷17=117 \div 17 = 1

    The prime factors are 2, 3, and 17.

  4. Write the prime factors in index form.

    102=21×31×171102 = 2^1 \times 3^1 \times 17^1 2×3×17\boxed{2 \times 3 \times 17}

    2. (b) Find the HCF and LCM of 30 and 24.

  5. Find the prime factors of each number.

    30=2×3×530 = 2 \times 3 \times 5 24=2×2×2×3=23×324 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3
  6. Find the HCF (Highest Common Factor).

    The common prime factors are 2 and 3. Take the lowest power of each common factor.

    HCF=21×31=2×3=6HCF = 2^1 \times 3^1 = 2 \times 3 = 6
  7. Find the LCM (Lowest Common Multiple).

    Take the highest power of all prime factors present in either number.

    LCM=23×31×51=8×3×5=120LCM = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120 HCF=6,LCM=120\boxed{HCF = 6, LCM = 120}

    3. (a) If 2n462n4_6 is a number in base 6. State the largest possible value of n.
    In base 6, the digits used must be less than 6.
    The digits in 2n462n4_6 are 2, n, and 4.
    For n to be a valid digit in base 6, it must satisfy 0≤n<60 \le n < 6.
    The largest possible integer value for n is 5.

    5\boxed{5}

    3. (b) i) Convert 3334333_4 to base 10.

  8. Write the number in expanded form using powers of the base.

    3334=3×42+3×41+3×40333_4 = 3 \times 4^2 + 3 \times 4^1 + 3 \times 4^0
  9. Calculate the value.

    3×16+3×4+3×13 \times 16 + 3 \times 4 + 3 \times 1 48+12+3=6348 + 12 + 3 = 63 6310\boxed{63_{10}}

    3. (b) ii) Convert 5016501_6 to base 5.

  10. Convert 5016501_6 to base 10.

    5016=5×62+0×61+1×60501_6 = 5 \times 6^2 + 0 \times 6^1 + 1 \times 6^0 5×36+0×6+1×15 \times 36 + 0 \times 6 + 1 \times 1 180+0+1=18110180 + 0 + 1 = 181_{10}
  11. Convert 18110181_{10} to base 5.

    Divide 181 by 5 and record the remainders.

    181÷5=36 remainder 1181 \div 5 = 36 \text{ remainder } 1 36÷5=7 remainder 136 \div 5 = 7 \text{ remainder } 1 7÷5=1 remainder 27 \div 5 = 1 \text{ remainder } 2 1÷5=0 remainder 11 \div 5 = 0 \text{ remainder } 1

    Read the remainders from bottom to top.

    12115\boxed{1211_5}

    4. Given that E={1−10}\mathcal{E} = \{1-10\}, A={1,4,5,7,8}A = \{1,4,5,7,8\}, B={2,6,8,10}B = \{2,6,8,10\}

    a) Represent the sets on a Venn diagram.
    The universal set E={1,2,3,4,5,6,7,8,9,10}\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}.
    A={1,4,5,7,8}A = \{1,4,5,7,8\}
    B={2,6,8,10}B = \{2,6,8,10\}
    First, find the intersection A∩BA \cap B.
    A∩B={8}A \cap B = \{8\}
    Next, find the elements unique to A: A∖B={1,4,5,7}A \setminus B = \{1,4,5,7\}
    Next, find the elements unique to B: B∖A={2,6,10}B \setminus A = \{2,6,10\}
    Finally, find the elements outside A and B (in E\mathcal{E} but not in A or B):
    E∖(A∪B)={1,2,3,4,5,6,7,8,9,10}∖{1,2,4,5,6,7,8,10}={3,9}\mathcal{E} \setminus (A \cup B) = \{1,2,3,4,5,6,7,8,9,10\} \setminus \{1,2,4,5,6,7,8,10\} = \{3,9\}

    graph TD
        subgraph E
            A(A)
            B(B)
            A -- 1,4,5,7 --> A
            B -- 2,6,10 --> B
            A -- 8 --> B
            E -- 3,9 --> E
        end
    

    (Note: The Venn diagram cannot be fully rendered in text. The description above details the placement of elements.)

    b) Find i) n(A∩B)n(A \cap B) ii) n(B′)n(B') iii) List the elements in (A∩B)′(A \cap B)'

    i) n(A∩B)n(A \cap B)
    A∩B={8}A \cap B = \{8\}
    The number of elements in A∩BA \cap B is 1.

    n(A∩B)=1\boxed{n(A \cap B) = 1}

    ii) n(B′)n(B')
    B′B' is the complement of B, meaning all elements in the universal set E\mathcal{E} that are not in B.
    E={1,2,3,4,5,6,7,8,9,10}\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}
    B={2,6,8,10}B = \{2, 6, 8, 10\}
    B′={1,3,4,5,7,9}B' = \{1, 3, 4, 5, 7, 9\}
    The number of elements in B′B' is 6.

    n(B′)=6\boxed{n(B') = 6}

    iii) List the elements in (A∩B)′(A \cap B)'
    (A∩B)′(A \cap B)' is the complement of A∩BA \cap B.
    A∩B={8}A \cap B = \{8\}
    (A∩B)′=E∖{8}={1,2,3,4,5,6,7,9,10}(A \cap B)' = \mathcal{E} \setminus \{8\} = \{1, 2, 3, 4, 5, 6, 7, 9, 10\}

    {1,2,3,4,5,6,7,9,10}\boxed{\{1, 2, 3, 4, 5, 6, 7, 9, 10\}}

    5. (a) Find the radius of a circle whose area is 440 cm2440 \text{ cm}^2.

  12. Use the formula for the area of a circle, A=πr2A = \pi r^2.

    Given A=440 cm2A = 440 \text{ cm}^2. Use π≈227\pi \approx \frac{22}{7}.

    440=227r2440 = \frac{22}{7} r^2
  13. Solve for r2r^2.

    r2=440×722r^2 = \frac{440 \times 7}{22} r2=20×7r^2 = 20 \times 7 r2=140r^2 = 140
  14. Solve for rr.

    r=140r = \sqrt{140} r≈11.832 cmr \approx 11.832 \text{ cm}

    Rounding to 3 significant figures:

    11.8 cm\boxed{11.8 \text{ cm}}

    5. (b) Find the volume of a cylinder with radius 14 cm and height 40 cm.

  15. Use the formula for the volume of a cylinder, V=πr2hV = \pi r^2 h.

    Given r=14 cmr = 14 \text{ cm} and h=40 cmh = 40 \text{ cm}. Use π≈227\pi \approx \frac{22}{7}.

    V=227×(14 cm)2×40 cmV = \frac{22}{7} \times (14 \text{ cm})^2 \times 40 \text{ cm} V=227×196 cm2×40 cmV = \frac{22}{7} \times 196 \text{ cm}^2 \times 40 \text{ cm}
  16. Calculate the volume.

    V=22×28 cm2×40 cmV = 22 \times 28 \text{ cm}^2 \times 40 \text{ cm} V=616 cm2×40 cmV = 616 \text{ cm}^2 \times 40 \text{ cm} V=24640 cm3V = 24640 \text{ cm}^3 24640 cm3\boxed{24640 \text{ cm}^3}

    5. (c) Express 3.5 m23.5 \text{ m}^2 to km2\text{km}^2.

  17. Recall the conversion factors.

    1 km=1000 m1 \text{ km} = 1000 \text{ m}
    1 km2=(1000 m)2=1000000 m2=106 m21 \text{ km}^2 = (1000 \text{ m})^2 = 1000000 \text{ m}^2 = 10^6 \text{ m}^2

  18. Convert 3.5 m23.5 \text{ m}^2 to km2\text{km}^2.

    3.5 m2×1 km2106 m23.5 \text{ m}^2 \times \frac{1 \text{ km}^2}{10^6 \text{ m}^2} 3.5×10−6 km23.5 \times 10^{-6} \text{ km}^2 0.0000035 km2\boxed{0.0000035 \text{ km}^2}

    6. (a) Expand and simplify (x−1)(x+2)(x-1)(x+2).

  19. Use the FOIL method (First, Outer, Inner, Last).

    (x−1)(x+2)=x×x+x×2−1×x−1×2(x-1)(x+2) = x \times x + x \times 2 - 1 \times x - 1 \times 2 =x2+2x−x−2= x^2 + 2x - x - 2
  20. Combine like terms.

    =x2+x−2= x^2 + x - 2 x2+x−2\boxed{x^2 + x - 2}

    6. (b) Express 3x+53−x−22\frac{3x+5}{3} - \frac{x-2}{2} as a single fraction in its simplest form.

  21. Find a common denominator, which is 6.

    2(3x+5)2×3−3(x−2)3×2\frac{2(3x+5)}{2 \times 3} - \frac{3(x-2)}{3 \times 2} 2(3x+5)6−3(x−2)6\frac{2(3x+5)}{6} - \frac{3(x-2)}{6}
  22. Combine the numerators over the common denominator.

    2(3x+5)−3(x−2)6\frac{2(3x+5) - 3(x-2)}{6}
  23. Expand the numerators.

    6x+10−(3x−6)6\frac{6x + 10 - (3x - 6)}{6}
  24. Distribute the negative sign and combine like terms.

    6x+10−3x+66\frac{6x + 10 - 3x + 6}{6} 3x+166\frac{3x + 16}{6} 3x+166\boxed{\frac{3x+16}{6}}

    7. Factorise completely:

    a) am−an+bm−bnam - an + bm - bn

  25. Group terms with common factors.

    (am−an)+(bm−bn)(am - an) + (bm - bn)
  26. Factor out the common factor from each group.

    a(m−n)+b(m−n)a(m - n) + b(m - n)
  27. Factor out the common binomial factor (m−n)(m-n).

    (m−n)(a+b)(m - n)(a + b) (m−n)(a+b)\boxed{(m - n)(a + b)}

    b) 49−y249 - y^2
    This is a difference of squares, a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b).
    Here, a2=49  ⟹  a=7a^2 = 49 \implies a = 7 and b2=y2  ⟹  b=yb^2 = y^2 \implies b = y.

    (7−y)(7+y)(7 - y)(7 + y) (7−y)(7+y)\boxed{(7 - y)(7 + y)}

    c) x2−10x−24x^2 - 10x - 24

  28. Find two numbers that multiply to -24 and add to -10.

    The numbers are -12 and 2.

    −12×2=−24-12 \times 2 = -24 −12+2=−10-12 + 2 = -10
  29. Write the quadratic expression in factored form.

    (x−12)(x+2)(x - 12)(x + 2) (x−12)(x+2)\boxed{(x - 12)(x + 2)}

    8. $210 is shared in the ratio 1:2:4, find the difference between the largest and smallest share.

  30. Find the total number of parts in the ratio.

    1+2+4=7 parts1 + 2 + 4 = 7 \text{ parts}
  31. Calculate the value of one part.

    2107=30\frac{210}{7} = 30
  32. Calculate the smallest share (1 part) and the largest share (4 parts).

    Smallest share: 1×30=301 \times 30 = 30
    Largest share: 4×30=1204 \times 30 = 120

  33. Find the difference between the largest and smallest share.

    120−30=90120 - 30 = 90 90\boxed{90}

    9. A map is drawn to scale 1:35000

    i) Express the scale in the form 1 cm : n km

  34. Convert 35000 cm to km.

    1 km=1000 m1 \text{ km} = 1000 \text{ m}
    1 m=100 cm1 \text{ m} = 100 \text{ cm}
    So, 1 km=1000×100 cm=100000 cm1 \text{ km} = 1000 \times 100 \text{ cm} = 100000 \text{ cm}.

  35. Divide 35000 by 100000 to convert to km.

    35000 cm=35000100000 km=0.35 km35000 \text{ cm} = \frac{35000}{100000} \text{ km} = 0.35 \text{ km}

    So the scale is 1 cm:0.35 km1 \text{ cm} : 0.35 \text{ km}.

    1 cm:0.35 km\boxed{1 \text{ cm} : 0.35 \text{ km}}

    ii) Hence or otherwise find the actual area, in km2\text{km}^2, represented by 20 cm220 \text{ cm}^2 on the map.

  36. Determine the linear scale factor.

    From part (i), 1 cm1 \text{ cm} on the map represents 0.35 km0.35 \text{ km} in actual distance.

  37. Determine the area scale factor.

    If 1 cm1 \text{ cm} represents 0.35 km0.35 \text{ km}, then 1 cm21 \text{ cm}^2 represents (0.35 km)2(0.35 \text{ km})^2.

    (0.35 km)2=0.1225 km2(0.35 \text{ km})^2 = 0.1225 \text{ km}^2

    So, 1 cm21 \text{ cm}^2 on the map represents 0.1225 km20.1225 \text{ km}^2 in actual area.

  38. Calculate the actual area represented by 20 cm220 \text{ cm}^2 on the map.

    20 cm2×0.1225km2cm2=2.45 km220 \text{ cm}^2 \times 0.1225 \frac{\text{km}^2}{\text{cm}^2} = 2.45 \text{ km}^2 2.45 km2\boxed{2.45 \text{ km}^2}
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