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Calculate the area of a cuboid of length 11cm, breadth 9cm and height 5cm.

Mathematics

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25. Calculate the area of a cuboid of length 11cm11 \text{cm}, breadth 9cm9 \text{cm} and height 5cm5 \text{cm}.
The question asks for "area" but the units in the options are cm3cm^3, which are units for volume. Given the options, it is highly probable that the question intended to ask for the volume of the cuboid.

  1. Identify the dimensions.

    Length l=11 cml = 11 \text{ cm}
    Breadth b=9 cmb = 9 \text{ cm}
    Height h=5 cmh = 5 \text{ cm}

  2. Calculate the volume of the cuboid.

    Volume=l×b×h\text{Volume} = l \times b \times h Volume=11 cm×9 cm×5 cm\text{Volume} = 11 \text{ cm} \times 9 \text{ cm} \times 5 \text{ cm} Volume=495 cm3\text{Volume} = 495 \text{ cm}^3

    The correct option is B.
    The volume is 495 cm3\boxed{495 \text{ cm}^3}.

    26. How many notebooks sold at N150.00 each can a father purchase for N2550.00 for his ward?

  3. Divide the total money by the cost per notebook.

    Number of notebooks=Total moneyCost per notebook\text{Number of notebooks} = \frac{\text{Total money}}{\text{Cost per notebook}} Number of notebooks=N2550.00N150.00\text{Number of notebooks} = \frac{\text{N}2550.00}{\text{N}150.00}
  4. Perform the division.

    Number of notebooks=17\text{Number of notebooks} = 17

    The correct option is C.
    He can purchase 17\boxed{17} notebooks.

    27. Reduce 105225\frac{105}{225} to a fraction in its lowest term.

  5. Find the greatest common divisor (GCD) of the numerator and denominator.

    Both 105 and 225 are divisible by 5.

    105÷5225÷5=2145\frac{105 \div 5}{225 \div 5} = \frac{21}{45}
  6. Continue dividing by common factors.

    Both 21 and 45 are divisible by 3.

    21÷345÷3=715\frac{21 \div 3}{45 \div 3} = \frac{7}{15}

    The correct option is B.
    The fraction in its lowest term is 715\boxed{\frac{7}{15}}.

    28. Change 7 to a number in percentage.
    Given the options, it is highly probable that the question intended to ask for the percentage equivalent of 78\frac{7}{8}.

  7. Convert the fraction 78\frac{7}{8} to a percentage.

    Percentage=78×100%\text{Percentage} = \frac{7}{8} \times 100\%
  8. Calculate the percentage.

    Percentage=0.875×100%=87.5%\text{Percentage} = 0.875 \times 100\% = 87.5\% Percentage=8712%\text{Percentage} = 87\frac{1}{2}\%

    The correct option is C.
    The percentage is 8712%\boxed{87\frac{1}{2}\%}.

    29. The number 10,080,008 can be interpreted in words as:

  9. Break down the number by place value.

    10,000,000  ⟹  Ten Million10,000,000 \implies \text{Ten Million}
    80,000  ⟹  Eighty Thousand80,000 \implies \text{Eighty Thousand}
    8  ⟹  Eight8 \implies \text{Eight}

  10. Combine the parts.

    Ten Million, Eighty Thousand and Eight.
    The correct option is B.
    The number is Ten Million, Eighty Thousand and Eight\boxed{\text{Ten Million, Eighty Thousand and Eight}}.

    30. Which of the following angles is an obtuse angle?

  11. Recall the definition of an obtuse angle.

    An obtuse angle is an angle that measures greater than 90∘90^\circ but less than 180∘180^\circ.

  12. Evaluate each option.

    A. 89∘89^\circ is an acute angle (less than 90∘90^\circ).
    B. 92∘92^\circ is an obtuse angle (greater than 90∘90^\circ and less than 180∘180^\circ).
    C. 184∘184^\circ is a reflex angle (greater than 180∘180^\circ).
    D. 0.237∘0.237^\circ is an acute angle.
    The correct option is B.
    92∘\boxed{92^\circ} is an obtuse angle.

    31. Express 117 as a product of its prime factors.

  13. Divide 117 by the smallest prime numbers.

    117÷3=39117 \div 3 = 39 39÷3=1339 \div 3 = 13 13÷13=113 \div 13 = 1
  14. Write the prime factorization.

    117=3×3×13=32×13117 = 3 \times 3 \times 13 = 3^2 \times 13

    The correct option is D.
    The prime factors are 32×13\boxed{3^2 \times 13}.

    32. Calculate the area of the triangle below:
    The diagram shows a right-angled triangle with a base of 8 cm8 \text{ cm} and a height of 6 cm6 \text{ cm}.

  15. Identify the base and height.

    Base b=8 cmb = 8 \text{ cm}
    Height h=6 cmh = 6 \text{ cm}

  16. Apply the formula for the area of a triangle.

    Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} Area=12×8 cm×6 cm\text{Area} = \frac{1}{2} \times 8 \text{ cm} \times 6 \text{ cm} Area=4 cm×6 cm\text{Area} = 4 \text{ cm} \times 6 \text{ cm} Area=24 cm2\text{Area} = 24 \text{ cm}^2

    The correct option is D.
    The area of the triangle is 24 cm2\boxed{24 \text{ cm}^2}.

    33. Solve the simultaneous equations:

    (i) a+6b=13a + 6b = 13

    (ii) 5a−6b=75a - 6b = 7
    Adding the two equations as given results in 6a=206a = 20, so a=103a = \frac{10}{3}, which is not an integer. Since all options are integers, there might be a typo in the second equation. Assuming the second equation was intended to be 5a−6b=−75a - 6b = -7 to yield integer solutions matching the options:

    (i) a+6b=13a + 6b = 13

    (ii) 5a−6b=−75a - 6b = -7

  17. Add equation (i) and equation (ii) to eliminate bb.

    (a+6b)+(5a−6b)=13+(−7)(a + 6b) + (5a - 6b) = 13 + (-7) 6a=66a = 6
  18. Solve for aa.

    a=66=1a = \frac{6}{6} = 1
  19. Substitute a=1a=1 into equation (i) to solve for bb.

    1+6b=131 + 6b = 13 6b=13−16b = 13 - 1 6b=126b = 12 b=126=2b = \frac{12}{6} = 2

    The correct option is C (assuming the typo in equation (ii) as 5a−6b=−75a - 6b = -7).
    The solutions are a=1 and b=2\boxed{a = 1 \text{ and } b = 2}.

    34. Factorize the expression y2+10y−24=0y^2 + 10y - 24 = 0.

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

    The two numbers are 12 and -2, because 12×(−2)=−2412 \times (-2) = -24 and 12+(−2)=1012 + (-2) = 10.

  21. Rewrite the middle term using these numbers and factor by grouping.

    y2+12y−2y−24=0y^2 + 12y - 2y - 24 = 0 y(y+12)−2(y+12)=0y(y + 12) - 2(y + 12) = 0 (y+12)(y−2)=0(y + 12)(y - 2) = 0
  22. Solve for yy.

    y+12=0  ⟹  y=−12y + 12 = 0 \implies y = -12 y−2=0  ⟹  y=2y - 2 = 0 \implies y = 2

    The correct option is B.
    The solutions are y=−12 or y=2\boxed{y = -12 \text{ or } y = 2}.

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