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Calculate Mary's total earnings including commission and find the sum of June and Peter's ages given June's age is 12.

Business/Management

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4 parts
44

KSh 55,000\text{KSh } 55,000

a

3j

b

36 years36 \text{ years}

46

A=110∘,B=70∘,C=110∘A = 110^\circ, B = 70^\circ, C = 110^\circ

Step-by-step solution

3 steps

Here are the solutions for questions 44, 45, 46, and 47.

Part 44· 3 steps

  1. Calculate the sales amount on which commission is paid.

    Commission is paid on sales above KSh 50,000.

    Sales for commission=Total sales−Threshold\text{Sales for commission} = \text{Total sales} - \text{Threshold} Sales for commission=KSh 350,000−KSh 50,000=KSh 300,000\text{Sales for commission} = \text{KSh } 350,000 - \text{KSh } 50,000 = \text{KSh } 300,000
  2. Calculate the commission earned.

    Commission rate is 15%.

    Commission=15%×KSh 300,000\text{Commission} = 15\% \times \text{KSh } 300,000 Commission=15100×300,000=15×3,000=KSh 45,000\text{Commission} = \frac{15}{100} \times 300,000 = 15 \times 3,000 = \text{KSh } 45,000
  3. Calculate Mary's total earnings for the month.

    Total earnings = Fixed monthly salary + Commission

    Total earnings=KSh 10,000+KSh 45,000=KSh 55,000\text{Total earnings} = \text{KSh } 10,000 + \text{KSh } 45,000 = \text{KSh } 55,000

    Mary earned KSh 55,000\boxed{\text{KSh } 55,000} that month.

Part 45

Let June's age be jj.
Peter is twice as old as June, so Peter's age is 2j2j.

Part a3j

Write an expression for the sum of their ages.

Sum of ages=June’s age+Peter’s age\text{Sum of ages} = \text{June's age} + \text{Peter's age} Sum of ages=j+2j=3j\text{Sum of ages} = j + 2j = 3j

The expression for the sum of their ages is 3j\boxed{3j}.

Part b

36 years36 \text{ years}

Work out the sum of their ages if June's age is 12 years.

Substitute j=12j = 12 into the expression from part (a).

Sum of ages=3×12=36 years\text{Sum of ages} = 3 \times 12 = 36 \text{ years}

The sum of their ages is 36 years\boxed{36 \text{ years}}.

Part 46

A=110∘,B=70∘,C=110∘A = 110^\circ, B = 70^\circ, C = 110^\circ

Since the diagram is not fully visible, I will assume a standard configuration for parallel lines intersected by a transversal, where the 70∘70^\circ angle is an acute angle in the upper-left position of the top intersection.

Let the 70∘70^\circ angle be ∠X\angle X.

  • Angle A: Assume A is the angle adjacent to ∠X\angle X on a straight line.

    A=180∘−70∘=110∘A = 180^\circ - 70^\circ = 110^\circ

    Reason: Angles on a straight line sum to 180∘180^\circ.

  • Angle B: Assume B is the corresponding angle to ∠X\angle X.

    B=70∘B = 70^\circ

    Reason: Corresponding angles are equal when lines are parallel.

  • Angle C: Assume C is the co-interior angle with ∠X\angle X.

    C=180∘−70∘=110∘C = 180^\circ - 70^\circ = 110^\circ

    Reason: Co-interior angles (consecutive interior angles) sum to 180∘180^\circ when lines are parallel.

The values are A=110∘,B=70∘,C=110∘\boxed{A = 110^\circ, B = 70^\circ, C = 110^\circ}.

Part 47

To plot the points A(-2,4), B(0,0), C(1,-1), and D(3,1) on the graph grid:

  • Point A (-2, 4): Start at the origin (0,0). Move 2 units to the left along the x-axis, then 4 units up parallel to the y-axis. Mark this point.
  • Point B (0, 0): This is the origin itself. Mark this point.
  • Point C (1, -1): Start at the origin. Move 1 unit to the right along the x-axis, then 1 unit down parallel to the y-axis. Mark this point.
  • Point D (3, 1): Start at the origin. Move 3 units to the right along the x-axis, then 1 unit up parallel to the y-axis. Mark this point.

After plotting all four points, use a ruler to draw straight lines connecting them in order:

  1. Join point A to point B.
  2. Join point B to point C.
  3. Join point C to point D.

This will form a path on the graph.

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