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In construction, all the lines and arcs that lead to the final results should be left visible, not to be erased. (A) Erased (B) Seen (C) Drawn (D) None of the above

Mathematics

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here are the solutions to the questions:

28. In construction, all the lines and arcs that lead to the final results should be left visible, not to

  • (A) Erased In geometric constructions, construction lines and arcs are typically left visible to show the steps and method used.

29. Convert 29 in base ten to binary

  1. Divide 29 by 2 and record the remainder.

    29÷2=14 remainder 129 \div 2 = 14 \text{ remainder } 1
  2. Continue dividing the quotient by 2 until the quotient is 0.

    14÷2=7 remainder 014 \div 2 = 7 \text{ remainder } 0 7÷2=3 remainder 17 \div 2 = 3 \text{ remainder } 1 3÷2=1 remainder 13 \div 2 = 1 \text{ remainder } 1 1÷2=0 remainder 11 \div 2 = 0 \text{ remainder } 1
  3. Read the remainders from bottom to top.

    The binary representation of 29 is 11101211101_2.

    • (A) 11101\boxed{11101}

    30. How many liters of water does a 5m×4m×3m5\text{m} \times 4\text{m} \times 3\text{m} tank hold?

  4. Calculate the volume of the tank in cubic meters.

    Volume=length×width×height\text{Volume} = \text{length} \times \text{width} \times \text{height} Volume=5 m×4 m×3 m=60 m3\text{Volume} = 5 \text{ m} \times 4 \text{ m} \times 3 \text{ m} = 60 \text{ m}^3
  5. Convert the volume from cubic meters to liters.

    Recall that 1 m3=1000 liters1 \text{ m}^3 = 1000 \text{ liters}.

    Volume in liters=60 m3×1000litersm3=60000 liters\text{Volume in liters} = 60 \text{ m}^3 \times 1000 \frac{\text{liters}}{\text{m}^3} = 60000 \text{ liters}
    • (A) 60000 liters\boxed{60000 \text{ liters}}

    31. When data is collected without being processed, it is known as:

    • (C) Raw Data Raw data refers to information that has been collected but not yet processed, analyzed, or organized.

    32. A rectangle has a length of 15 cm and a width of 6 cm. What is its area?

  6. Recall the formula for the area of a rectangle.

    Area=length×width\text{Area} = \text{length} \times \text{width}
  7. Substitute the given values.

    Area=15 cm×6 cm=90 cm2\text{Area} = 15 \text{ cm} \times 6 \text{ cm} = 90 \text{ cm}^2
    • (A) 90 cm2\boxed{90 \text{ cm}^2}

    33. A rectangular piece of land measure 57m by 42m. What is the perimeter of the land?

  8. Recall the formula for the perimeter of a rectangle.

    Perimeter=2×(length+width)\text{Perimeter} = 2 \times (\text{length} + \text{width})
  9. Substitute the given values.

    Perimeter=2×(57 m+42 m)\text{Perimeter} = 2 \times (57 \text{ m} + 42 \text{ m}) Perimeter=2×(99 m)\text{Perimeter} = 2 \times (99 \text{ m}) Perimeter=198 m\text{Perimeter} = 198 \text{ m}
    • (A) 198m\boxed{198 \text{m}}

    34. Calculate the perimeter of a square with a side length of 9 cm.

  10. Recall the formula for the perimeter of a square.

    Perimeter=4×side\text{Perimeter} = 4 \times \text{side}
  11. Substitute the given side length.

    Perimeter=4×9 cm=36 cm\text{Perimeter} = 4 \times 9 \text{ cm} = 36 \text{ cm}
    • (C) 36cm\boxed{36 \text{cm}}

    35. Which is not among the reasons for collecting of data?

    • (D) For exercise Data is collected for various purposes like planning, decision-making, and prediction, but not for physical exercise.

    36. If the volume of a container is 200cm3200 \text{cm}^3, while its area is 50cm250 \text{cm}^2, what is height of the container?

  12. Recall the formula for the volume of a prism or container.

    Volume=Base Area×Height\text{Volume} = \text{Base Area} \times \text{Height}
  13. Substitute the given values and solve for the height.

    200 cm3=50 cm2×Height200 \text{ cm}^3 = 50 \text{ cm}^2 \times \text{Height} Height=200 cm350 cm2\text{Height} = \frac{200 \text{ cm}^3}{50 \text{ cm}^2} Height=4 cm\text{Height} = 4 \text{ cm}
    • (A) 4cm\boxed{4 \text{cm}}

    37. A triangle whose angles are less than 90∘90^\circ is called...

    • (D) Acute angle triangle An acute angle triangle is a triangle where all three interior angles are less than 90∘90^\circ.

    38. What instrument is commonly used to construct perpendicular and parallel lines?

    • (D) All of the above A ruler (straightedge) and compass are fundamental for classical geometric constructions of perpendicular and parallel lines. A protractor can also be used to draw a 90∘90^\circ angle for a perpendicular line.

    39. What is the perimeter of a square with side length 6 cm?

  14. Recall the formula for the perimeter of a square.

    Perimeter=4×side\text{Perimeter} = 4 \times \text{side}
  15. Substitute the given side length.

    Perimeter=4×6 cm=24 cm\text{Perimeter} = 4 \times 6 \text{ cm} = 24 \text{ cm}
    • (C) 24cm\boxed{24 \text{cm}}

    40. The following are the instrument found in mathematical set except

    • (A) Razor blade A standard mathematical set (geometry box) typically includes a set square, protractor, and compass, but not a razor blade.

    SECTION B: ESSAY

    1(A) Write 3025 in Roman numerals.

  16. Break down the number into thousands, hundreds, tens, and ones.

    3025=3000+20+53025 = 3000 + 20 + 5
  17. Convert each part to Roman numerals.

    3000=MMM3000 = MMM 20=XX20 = XX 5=V5 = V
  18. Combine the Roman numerals.

    3025=MMMXXV3025 = \boxed{MMMXXV}

    1(B) What are adjacent angles?
    Adjacent angles are two angles that share a common vertex and a common side, but do not overlap.

    1(C) What is the value of 8 in the figure 6.0657?
    The digit 8 is not present in the figure 6.0657.

    2(A) Express 3450cm to metres.

  19. Recall the conversion factor between centimeters and meters.

    1 meter=100 centimeters1 \text{ meter} = 100 \text{ centimeters}
  20. Divide the given length in centimeters by 100 to convert to meters.

    3450 cm=3450100 m3450 \text{ cm} = \frac{3450}{100} \text{ m} 3450 cm=34.5 m3450 \text{ cm} = \boxed{34.5 \text{ m}}

    2(B) If the volume of a cuboid is 150cm3150 \text{cm}^3, calculate the height given that the length is 10cm and the breadth is 5cm.

  21. Recall the formula for the volume of a cuboid.

    Volume=length×breadth×height\text{Volume} = \text{length} \times \text{breadth} \times \text{height}
  22. Substitute the given values into the formula.

    150 cm3=10 cm×5 cm×height150 \text{ cm}^3 = 10 \text{ cm} \times 5 \text{ cm} \times \text{height}
  23. Simplify and solve for the height.

    150 cm3=50 cm2×height150 \text{ cm}^3 = 50 \text{ cm}^2 \times \text{height} height=150 cm350 cm2\text{height} = \frac{150 \text{ cm}^3}{50 \text{ cm}^2} height=3 cm\text{height} = \boxed{3 \text{ cm}}

    3(A) Construct a parallelogram ABCD with AB = 3cm, BC = 4cm and angle B = 70 degrees.
    To construct the parallelogram:

    1. Draw a line segment AB of length 3 cm.
    2. At point B, use a protractor to draw a ray such that the angle with AB is 70∘70^\circ.
    3. Along this ray, measure and mark point C such that BC is 4 cm.
    4. From point A, draw an arc with a radius of 4 cm (equal to BC).
    5. From point C, draw an arc with a radius of 3 cm (equal to AB).
    6. The intersection of these two arcs is point D.
    7. Join AD and CD to complete the parallelogram ABCD.

    3(B) Express 504 as a product of its prime factors.

  24. Divide 504 by the smallest prime number, 2, repeatedly until the result is odd.

    504÷2=252504 \div 2 = 252 252÷2=126252 \div 2 = 126 126÷2=63126 \div 2 = 63
  25. Divide 63 by the next smallest prime number, 3, repeatedly.

    63÷3=2163 \div 3 = 21 21÷3=721 \div 3 = 7
  26. Divide 7 by the next smallest prime number, 7.

    7÷7=17 \div 7 = 1
  27. Write 504 as the product of its prime factors.

    504=2×2×2×3×3×7=23×32×7504 = 2 \times 2 \times 2 \times 3 \times 3 \times 7 = \boxed{2^3 \times 3^2 \times 7}

    4(A) Add 15231 and 18162.

  28. Align the numbers by place value and add.

    1523115231 +18162+ 18162 33393‾\overline{33393}

    The sum is 33393\boxed{33393}.

    4(B) Multiply 23.81 by 8.24.

  29. Multiply the numbers as if they were whole numbers.

    23.81×8.24=196.284423.81 \times 8.24 = 196.2844

    The product is 196.2844\boxed{196.2844}.

    5(A) If 6 students bought a pen and a book for N50 and N300, respectively, find the total cost of the items.

  30. Calculate the cost of one pen and one book.

    Cost per student=Cost of pen+Cost of book\text{Cost per student} = \text{Cost of pen} + \text{Cost of book} Cost per student=N50+N300=N350\text{Cost per student} = \text{N}50 + \text{N}300 = \text{N}350
  31. Calculate the total cost for 6 students.

    Total cost=6×Cost per student\text{Total cost} = 6 \times \text{Cost per student} Total cost=6×N350=N2100\text{Total cost} = 6 \times \text{N}350 = \text{N}2100

    The total cost is N2100\boxed{\text{N}2100}.

    5(B) Solve, n - 1 = 8.

  32. Isolate the variable nn by adding 1 to both sides of the equation.

    n−1=8n - 1 = 8 n−1+1=8+1n - 1 + 1 = 8 + 1 n=9n = \boxed{9}

    5(C) Convert 0.075 to fraction in it's lowest form.

  33. Write the decimal as a fraction with a denominator that is a power of 10.

    0.075=7510000.075 = \frac{75}{1000}
  34. Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor.

    Both 75 and 1000 are divisible by 25.

    75÷251000÷25=340\frac{75 \div 25}{1000 \div 25} = \frac{3}{40}

    The fraction in its lowest form is 340\boxed{\frac{3}{40}}.

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