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The distance between Ogoja and Nbube is 24km. If a driver covered the distance in 1½hours, what was his average speed for the journey?

Mathematics

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C

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Here are the solutions to the questions:

Part 24· 3 steps

The distance between Ogoja and Nbube is 24km. If a driver covered the distance in 1121\frac{1}{2} hours, what was his average speed for the journey?

  1. Identify the given distance and time.

    Distance (DD) = 24 km
    Time (TT) = 1121\frac{1}{2} hours = 1.5 hours

  2. Use the formula for average speed.

    Average Speed=DistanceTime\text{Average Speed} = \frac{\text{Distance}}{\text{Time}}

  3. Substitute the values and calculate the speed.

    Average Speed=24 km1.5 hours=16 km/h\text{Average Speed} = \frac{24 \text{ km}}{1.5 \text{ hours}} = 16 \text{ km/h}

    The average speed for the journey is 16 km/h16 \text{ km/h}.

    The correct option is (c).

    c) 16 km/h\boxed{\text{c) } 16 \text{ km/h}}

Part 25· 3 steps

Find the breadth of a rectangular field with area 648cm2648 \text{cm}^2 and length 40cm40 \text{cm}.

  1. Identify the given area and length.

    Area (AA) = 648 cm2648 \text{ cm}^2
    Length (LL) = 40 cm40 \text{ cm}

  2. Use the formula for the area of a rectangle, A=L×BA = L \times B, where BB is the breadth.

  3. Rearrange the formula to solve for breadth and substitute the values.

    B=AL=648 cm240 cmB = \frac{A}{L} = \frac{648 \text{ cm}^2}{40 \text{ cm}} B=16.2 cmB = 16.2 \text{ cm}

    The breadth of the rectangular field is 16.2 cm16.2 \text{ cm}.

    The correct option is (b).

    b) 16.2 cm\boxed{\text{b) } 16.2 \text{ cm}}

Part 26· 4 steps

Find the difference between the LCM and HCF of 10 and 12.

  1. Find the prime factorization of 10 and 12.

    10=2×510 = 2 \times 5
    12=22×312 = 2^2 \times 3

  2. Calculate the HCF (Highest Common Factor).

    HCF(10, 12) = 2

  3. Calculate the LCM (Least Common Multiple).

    LCM(10, 12) = 22×3×5=4×3×5=602^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60

  4. Find the difference between the LCM and HCF.

    Difference = 60−2=5860 - 2 = 58
    The difference is 58.

    The correct option is (d).

    d) 58\boxed{\text{d) } 58}

Part 27· 2 steps

Ayobami used 587kg of sugar to bake a cake and mixed 140kg sugar for coating the cake. How much of sugar content is used?

  1. Identify the amounts of sugar used.

    Sugar for baking = 587 kg
    Sugar for coating = 140 kg

  2. Add the amounts to find the total sugar content.

    Total sugar = 587 kg+140 kg=727 kg587 \text{ kg} + 140 \text{ kg} = 727 \text{ kg}
    The total sugar content used is 727 kg727 \text{ kg}.

    The correct option is (b).

    b) 727 kg\boxed{\text{b) } 727 \text{ kg}}

Part 28· 4 steps

Simplify 4+9−25\sqrt{4} + \sqrt{9 - \sqrt{25}}.

  1. Simplify the innermost square root.

    25=5\sqrt{25} = 5

  2. Substitute and simplify the next part of the expression.

    9−5=4\sqrt{9 - 5} = \sqrt{4}

  3. Simplify the remaining square roots.

    4+4=2+2\sqrt{4} + \sqrt{4} = 2 + 2

  4. Perform the final addition.

    2+2=42 + 2 = 4

    The simplified value is 4.

    The correct option is (a).

    a) 4\boxed{\text{a) } 4}

Part 29· 5 steps

Simplify 513−214+125\frac{1}{3} - 2\frac{1}{4} + \frac{1}{2}.

  1. Convert mixed numbers to improper fractions.

    513=1635\frac{1}{3} = \frac{16}{3}
    214=942\frac{1}{4} = \frac{9}{4}

  2. Rewrite the expression.

    163−94+12\frac{16}{3} - \frac{9}{4} + \frac{1}{2}

  3. Find the LCM of the denominators (3, 4, 2), which is 12.

    Convert fractions to have this common denominator.

    16×43×4−9×34×3+1×62×6=6412−2712+612\frac{16 \times 4}{3 \times 4} - \frac{9 \times 3}{4 \times 3} + \frac{1 \times 6}{2 \times 6} = \frac{64}{12} - \frac{27}{12} + \frac{6}{12}

  4. Perform the operations.

    64−27+612=37+612=4312\frac{64 - 27 + 6}{12} = \frac{37 + 6}{12} = \frac{43}{12}

  5. Convert the improper fraction back to a mixed number.

    4312=3712\frac{43}{12} = 3\frac{7}{12}

    The simplified value is 37123\frac{7}{12}.

    The correct option is (c).

    c) 3712\boxed{\text{c) } 3\frac{7}{12}}

Part 30· 3 steps

Add 3min. 25secs. and 2mins. 44secs.

  1. Add the minutes and seconds separately.

    Minutes: 3+2=5 min3 + 2 = 5 \text{ min}
    Seconds: 25+44=69 secs25 + 44 = 69 \text{ secs}

  2. Convert the total seconds into minutes and remaining seconds.

    69 secs=1 min 9 secs69 \text{ secs} = 1 \text{ min } 9 \text{ secs} (since 1 min = 60 secs)

  3. Add the converted minutes to the total minutes.

    Total time = 5 min+1 min 9 secs=6 min 9 secs5 \text{ min} + 1 \text{ min } 9 \text{ secs} = 6 \text{ min } 9 \text{ secs}
    The sum is 6 mins. 9 secs.

    The correct option is (b).

    b) 6mins. 9secs.\boxed{\text{b) } 6 \text{mins. } 9 \text{secs.}}

Part 31· 3 steps

What is the volume of a cylinder with base diameter 14cm and height 10cm?

  1. Identify the given values and calculate the radius.

    Diameter (dd) = 14 cm   ⟹  \implies Radius (rr) = 142=7 cm\frac{14}{2} = 7 \text{ cm}
    Height (hh) = 10 cm

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

    Use π=227\pi = \frac{22}{7}.

  3. Substitute the values and calculate the volume.

    V=227×(7 cm)2×10 cmV = \frac{22}{7} \times (7 \text{ cm})^2 \times 10 \text{ cm} V=227×49 cm2×10 cmV = \frac{22}{7} \times 49 \text{ cm}^2 \times 10 \text{ cm} V=22×7 cm2×10 cmV = 22 \times 7 \text{ cm}^2 \times 10 \text{ cm} V=1540 cm3V = 1540 \text{ cm}^3

    The volume of the cylinder is 1540 cm31540 \text{ cm}^3.

    The correct option is (b).

    b) 1540cm3\boxed{\text{b) } 1540 \text{cm}^3}

Part 32· 4 steps

Find the value of P in the figure below.

  1. Recognize that angles on a straight line sum to 180∘180^\circ.

  2. Set up the equation based on the given angles.

    110∘+2P=180∘110^\circ + 2P = 180^\circ

  3. Solve for $2P$.

    2P=180∘−110∘2P = 180^\circ - 110^\circ 2P=70∘2P = 70^\circ

  4. Solve for P.

    P=70∘2P = \frac{70^\circ}{2} P=35∘P = 35^\circ

    The value of P is 35∘35^\circ.

    The correct option is (c).

    c) 35∘\boxed{\text{c) } 35^\circ}

Part 33· 2 steps

Change 3% to fraction in its lowest term.

  1. Convert the percentage to a fraction by dividing by 100.

    3%=31003\% = \frac{3}{100}

  2. Check if the fraction can be simplified.

    The numerator (3) is a prime number. The denominator (100) is not divisible by 3.
    Therefore, the fraction is already in its lowest term.

    The correct option is (a).

    a) 3100\boxed{\text{a) } \frac{3}{100}}

Part 34· 1 step

A line from the centre of a circle to any point on a circumference is called ---

  1. Recall the definition of a radius.

    A radius is a line segment that connects the center of a circle to any point on its circumference.

    The correct option is (b).

    b) radius\boxed{\text{b) radius}}

Part 35· 3 steps

Write 2019 in Roman numeral.

  1. Break down the number 2019.

    2019=2000+10+92019 = 2000 + 10 + 9

  2. Convert each part to its Roman numeral.

    2000=MM2000 = \text{MM}
    10=X10 = \text{X}
    9=IX9 = \text{IX}

  3. Combine the Roman numerals.

    2019=MMXIX2019 = \text{MMXIX}
    The Roman numeral for 2019 is MMXIX.

    The correct option is (b).

    b) MMXIX\boxed{\text{b) MMXIX}}

Part 36· 2 steps

Convert 121012_{10} to binary number.

  1. Perform repeated division by 2, noting the remainders.

    12÷2=612 \div 2 = 6 remainder 0
    6÷2=36 \div 2 = 3 remainder 0
    3÷2=13 \div 2 = 1 remainder 1
    1÷2=01 \div 2 = 0 remainder 1

  2. Read the remainders from bottom to top.

    The binary representation of 121012_{10} is 110021100_2.

    The correct option is (a).

    a) 11002\boxed{\text{a) } 1100_2}
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