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Gather x terms on one side and constant terms on the other side of the inequality.

Mathematics

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C

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17. Find the truth set of 2x−4<6+3x2x - 4 < 6 + 3x.

  1. Gather xx terms on one side and constant terms on the other side of the inequality.

    Subtract 2x2x from both sides:

    −4<6+3x−2x-4 < 6 + 3x - 2x −4<6+x-4 < 6 + x

  2. Subtract 6 from both sides

    −4−6<x-4 - 6 < x −10<x-10 < x

  3. Rewrite the inequality with xx on the left side.

    x>−10x > -10

    The truth set is {x:x>−10}\{x: x > -10\}.
    The correct option is C.

    C. {x:x>−10}\boxed{\text{C. } \{x: x > -10\}}

    18. Evaluate 75+18−27\sqrt{75} + \sqrt{18} - \sqrt{27}.

  4. Simplify each square root by finding the largest perfect square factor.

    75=25×3=25×3=53\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3} 18=9×2=9×2=32\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2} 27=9×3=9×3=33\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}

  5. Substitute the simplified square roots back into the expression.

    53+32−335\sqrt{3} + 3\sqrt{2} - 3\sqrt{3}

  6. Combine like terms (terms with the same radical).

    (53−33)+32(5\sqrt{3} - 3\sqrt{3}) + 3\sqrt{2} (5−3)3+32(5-3)\sqrt{3} + 3\sqrt{2} 23+322\sqrt{3} + 3\sqrt{2}

    The correct option is B.

    B. 23+32\boxed{\text{B. } 2\sqrt{3} + 3\sqrt{2}}

    19. The locus of points equidistant from a fixed point is called a

    A locus of points equidistant from a fixed point is the definition of a circle. The fixed point is the center of the circle, and the constant distance is the radius.
    The correct option is B.

    B. circle\boxed{\text{B. circle}}

    20. Find the highest common factor of 54 and 72.

  7. Find the prime factorization of each number.

    54=2×27=2×3×9=2×3×3×3=2×3354 = 2 \times 27 = 2 \times 3 \times 9 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3 72=2×36=2×2×18=2×2×2×9=2×2×2×3×3=23×3272 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2

  8. Identify the common prime factors and their lowest powers.

    Common prime factors are 2 and 3.
    Lowest power of 2 is 212^1.
    Lowest power of 3 is 323^2.

  9. Multiply these common prime factors raised to their lowest powers.

    HCF=21×32=2×9=18\text{HCF} = 2^1 \times 3^2 = 2 \times 9 = 18

    The correct option is A.

    A. 18\boxed{\text{A. } 18}

    21. Arrange in ascending order the fractions 12\frac{1}{2}, 34\frac{3}{4} and 25\frac{2}{5}.

  10. Find the least common multiple (LCM) of the denominators (2, 4, 5).

    Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ...
    Multiples of 4: 4, 8, 12, 16, 20, ...
    Multiples of 5: 5, 10, 15, 20, ...
    The LCM is 20.

  11. Convert each fraction to an equivalent fraction with a denominator of 20.

    12=1×102×10=1020\frac{1}{2} = \frac{1 \times 10}{2 \times 10} = \frac{10}{20} 34=3×54×5=1520\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20} 25=2×45×4=820\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}

  12. Compare the numerators and arrange the fractions in ascending order.

    The numerators are 10, 15, and 8.
    In ascending order: 8, 10, 15.
    So, the fractions in ascending order are 820,1020,1520\frac{8}{20}, \frac{10}{20}, \frac{15}{20}.

  13. Write the original fractions in ascending order.

    25,12,34\frac{2}{5}, \frac{1}{2}, \frac{3}{4}

    The correct option is C.

    C. 25,12, and 34\boxed{\text{C. } \frac{2}{5}, \frac{1}{2}, \text{ and } \frac{3}{4}}

    22. A hawker is carrying a basket load of three types of fruits: oranges, mangoes and pears. If 14\frac{1}{4} of the fruits are oranges and 25\frac{2}{5} mangoes, what percentage of the fruits are pears?

  14. Calculate the total fraction of oranges and mangoes.

    Fraction of oranges =14= \frac{1}{4}
    Fraction of mangoes =25= \frac{2}{5}
    Total fraction of oranges and mangoes =14+25= \frac{1}{4} + \frac{2}{5}

  15. Find a common denominator for the fractions and add them.

    The LCM of 4 and 5 is 20.

    14=1×54×5=520\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20} 25=2×45×4=820\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}

    Total fraction of oranges and mangoes =520+820=1320= \frac{5}{20} + \frac{8}{20} = \frac{13}{20}

  16. Calculate the fraction of pears.

    The total fraction of fruits is 1.
    Fraction of pears =1−Total fraction of oranges and mangoes= 1 - \text{Total fraction of oranges and mangoes}
    Fraction of pears =1−1320=2020−1320=720= 1 - \frac{13}{20} = \frac{20}{20} - \frac{13}{20} = \frac{7}{20}

  17. Convert the fraction of pears to a percentage.

    Percentage of pears =720×100%= \frac{7}{20} \times 100\%

    =7×10020%= 7 \times \frac{100}{20}\% =7×5%= 7 \times 5\% =35%= 35\%

    The correct option is D. (Note: The options provided are A. 9%, B. 18%, C. 64%, D. 36%. My calculated answer is 35%. There might be a slight discrepancy in the options or the question's intended answer. However, 35% is closest to 36% among the given options.)
    Let's recheck the calculation.
    1/4=0.251/4 = 0.25
    2/5=0.402/5 = 0.40
    Total fraction of oranges and mangoes = 0.25+0.40=0.650.25 + 0.40 = 0.65
    Fraction of pears = 1−0.65=0.351 - 0.65 = 0.35
    Percentage of pears = 0.35×100%=35%0.35 \times 100\% = 35\%.
    Given the options, 36% is the closest. It's possible there's a rounding or slight error in the question's options. I will select the closest option.

    D. 36%\boxed{\text{D. } 36\%}

    23. If 3 litres of petrol cost GH¢10.00, find the cost of 15 litres.

  18. Find the cost per litre of petrol.

    Cost per litre =Total costQuantity= \frac{\text{Total cost}}{\text{Quantity}}
    Cost per litre =GH¢10.003 litres= \frac{\text{GH¢}10.00}{3 \text{ litres}}

  19. Calculate the cost of 15 litres.

    Cost of 15 litres =Cost per litre×15 litres= \text{Cost per litre} \times 15 \text{ litres}
    Cost of 15 litres =GH¢10.003×15= \frac{\text{GH¢}10.00}{3} \times 15
    Cost of 15 litres =GH¢10.00×153= \text{GH¢}10.00 \times \frac{15}{3}
    Cost of 15 litres =GH¢10.00×5= \text{GH¢}10.00 \times 5
    Cost of 15 litres =GH¢50.00= \text{GH¢}50.00
    The correct option is D.

    D. GH¢50.00\boxed{\text{D. GH¢}50.00}
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