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Find the area of the figure below: 60cm 2 B. 90cm 2 C. 72cm 2 D. 105cm 2 E. 135cm 2

Mathematics

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D

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31. Find the product of 10 and 6, subtract 24 and divide the result by 4.

  1. Find the product of 10 and 6.

    10×6=6010 \times 6 = 60
  2. Subtract 24 from the result.

    60−24=3660 - 24 = 36
  3. Divide the new result by 4.

    36÷4=936 \div 4 = 9

    The correct option is D.

    32. What is the value of two right angles?

  4. Recall the value of a right angle.

    A right angle is 90∘90^\circ.

  5. Calculate the value of two right angles.

    2×90∘=180∘2 \times 90^\circ = 180^\circ

    The correct option is B.

    33. How many lines of symmetry has a rectangle?

  6. Consider the properties of a rectangle.

    A rectangle has two lines of symmetry: one horizontal line passing through the midpoints of its vertical sides, and one vertical line passing through the midpoints of its horizontal sides.
    The correct option is C.

    34. This year written in Roman numeral is A. AMXV B. MMVC C. MMXV D. N E. MCMXV

  7. Interpret "This year" in the context of common exam questions.

    Often, "this year" refers to a recent year like 2015 in past papers. Let's assume "this year" refers to 2015.

  8. Convert 2015 to Roman numerals.

    M = 1000
    MM = 2000
    X = 10
    V = 5
    So, 2015=2000+10+5=MMXV2015 = 2000 + 10 + 5 = \text{MMXV}.
    The correct option is C.

    35. If 500 oranges were shared among Bill, Jill and Pearl in the ratio 3:3:4 respectively, what is Pearl's share?

  9. Calculate the total number of ratio parts.

    3+3+4=10 parts3 + 3 + 4 = 10 \text{ parts}
  10. Determine Pearl's share of the ratio.

    Pearl's share is 4 parts out of 10.

  11. Calculate Pearl's share of the oranges.

    Pearl’s share=410×500 oranges\text{Pearl's share} = \frac{4}{10} \times 500 \text{ oranges} Pearl’s share=4×50=200 oranges\text{Pearl's share} = 4 \times 50 = 200 \text{ oranges}

    The correct option is D.

    36. Which of the following pairs of angles is supplementary?

  12. Understand the definition of supplementary angles.

    Supplementary angles are two angles whose sum is 180∘180^\circ.

  13. Check each option.

    A. 30∘,60∘  ⟹  30∘+60∘=90∘30^\circ, 60^\circ \implies 30^\circ + 60^\circ = 90^\circ (complementary)
    B. 30∘,60∘  ⟹  30∘+60∘=90∘30^\circ, 60^\circ \implies 30^\circ + 60^\circ = 90^\circ (complementary)
    C. 80∘,100∘  ⟹  80∘+100∘=180∘80^\circ, 100^\circ \implies 80^\circ + 100^\circ = 180^\circ (supplementary)
    D. 190∘,80∘  ⟹  190∘+80∘=270∘190^\circ, 80^\circ \implies 190^\circ + 80^\circ = 270^\circ
    E. 30∘,150∘  ⟹  30∘+150∘=180∘30^\circ, 150^\circ \implies 30^\circ + 150^\circ = 180^\circ (supplementary)
    Both C and E are correct. Assuming only one answer is expected, we select the first one.
    The correct option is C.

    37. A bicycle wheel has a diameter of 42cm. Find its circumference (take π=227\pi = \frac{22}{7}).

  14. Recall the formula for the circumference of a circle.

    C=πdC = \pi d

    where CC is the circumference, π\pi is pi, and dd is the diameter.

  15. Substitute the given values into the formula.

    C=227×42 cmC = \frac{22}{7} \times 42 \text{ cm}
  16. Calculate the circumference.

    C=22×427 cmC = 22 \times \frac{42}{7} \text{ cm} C=22×6 cmC = 22 \times 6 \text{ cm} C=132 cmC = 132 \text{ cm}

    The correct option is B.

    38. A, B and C are three houses in this order along a street in Lagos. The distance from A to B is 120m and the distance from B to C is 2/3 of the distance from A to B. What is the distance between A and C?

  17. Identify the given distances.

    Distance from A to B (AB) = 120 m.

  18. Calculate the distance from B to C (BC).

    BC=23×AB\text{BC} = \frac{2}{3} \times \text{AB} BC=23×120 m\text{BC} = \frac{2}{3} \times 120 \text{ m} BC=2×40 m=80 m\text{BC} = 2 \times 40 \text{ m} = 80 \text{ m}
  19. Calculate the total distance between A and C (AC).

    AC=AB+BC\text{AC} = \text{AB} + \text{BC} AC=120 m+80 m=200 m\text{AC} = 120 \text{ m} + 80 \text{ m} = 200 \text{ m}

    The correct option is E.

    39. The difference between two numbers is 457. If the smaller number is 610, what is the larger number?

  20. Let the larger number be LL and the smaller number be SS.

    The difference is given by L−S=457L - S = 457.

  21. Substitute the value of the smaller number.

    L−610=457L - 610 = 457
  22. Solve for the larger number LL.

    L=457+610L = 457 + 610 L=1067L = 1067

    The correct option is C.

    40. Find the area of the figure below:

  23. Decompose the figure into a rectangle and a right-angled triangle.

    Let the rectangle have width ww and height hh.
    Let the triangle have base bb and height hh.
    From the visual, it appears that the base of the triangle is equal to the width of the rectangle (b=wb=w), and their heights are also equal.

  24. Formulate the total area.

    Area of rectangle = w×hw \times h
    Area of triangle = 12×b×h=12×w×h\frac{1}{2} \times b \times h = \frac{1}{2} \times w \times h
    Total Area = Area of rectangle + Area of triangle

    Total Area=wh+12wh=32wh\text{Total Area} = wh + \frac{1}{2}wh = \frac{3}{2}wh
  25. Test the options by assuming dimensions that fit the visual and the formula.

    If we assume the rectangle has dimensions 10 cm×6 cm10 \text{ cm} \times 6 \text{ cm} (so w=10,h=6w=10, h=6):
    Area of rectangle = 10 cm×6 cm=60 cm210 \text{ cm} \times 6 \text{ cm} = 60 \text{ cm}^2.
    Area of triangle = 12×10 cm×6 cm=30 cm2\frac{1}{2} \times 10 \text{ cm} \times 6 \text{ cm} = 30 \text{ cm}^2.
    Total Area = 60 cm2+30 cm2=90 cm260 \text{ cm}^2 + 30 \text{ cm}^2 = 90 \text{ cm}^2.
    This matches option B.
    The correct option is B.

    A woman went to the market to buy some articles. She bought 3 dozen cups and 1121\frac{1}{2} gross of plates. How many articles did she buy altogether?

  26. Convert dozens to individual articles.

    1 dozen = 12 articles.
    3 dozen cups = 3×12=363 \times 12 = 36 cups.

  27. Convert gross to individual articles.

    1 gross = 12 dozen = 12×12=14412 \times 12 = 144 articles.
    1121\frac{1}{2} gross of plates = 1.5×1441.5 \times 144 plates.

    1.5×144=32×144=3×72=216 plates1.5 \times 144 = \frac{3}{2} \times 144 = 3 \times 72 = 216 \text{ plates}
  28. Calculate the total number of articles.

    Total articles = cups + plates

    Total articles=36+216=252\text{Total articles} = 36 + 216 = 252

    The correct option is E.

    What is the HCF of 21, 63, 91 and 105?

  29. Find the prime factorization of each number.

    21=3×721 = 3 \times 7 63=32×763 = 3^2 \times 7 91=7×1391 = 7 \times 13 105=3×5×7105 = 3 \times 5 \times 7
  30. Identify the common prime factors and their lowest powers.

    The only prime factor common to all four numbers is 7. The lowest power of 7 is 717^1.

  31. The Highest Common Factor (HCF) is the product of these common prime factors.

    HCF=7\text{HCF} = 7

    The correct option is A.

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