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Naledi buys 5 packets, each containing 12 items. Every item is sold for R2, and the total profit is R30. Let x be the cost of one packet. Which equation models the situation?

Mathematics

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Step-by-step solution

7 steps

Part 6· 3 steps

  1. Calculate the total number of items and total revenue.

    Naledi buys 5 packets, each with 12 items, so total items = 5×12=605 \times 12 = 60 items.
    Each item is sold for R2, so total revenue = 60×R2=R12060 \times R2 = R120.

  2. Define total cost and profit.

    Let xx be the cost of one packet. Naledi buys 5 packets, so total cost = 5x5x.
    Profit is given as R30.

  3. Use the formula Profit = Revenue - Total Cost.

    30=120−5x30 = 120 - 5x

    Rearranging this to match the options, we get 120−5x=30120 - 5x = 30.

    The correct equation is C. 120−5x=30, using profit = revenue - total cost\boxed{\text{C. } 120 - 5x = 30, \text{ using profit = revenue - total cost}}.

Part 7

To verify a proposed solution to an equation, substitute the proposed value into the original equation and check if both sides are equal.
The equation is 3x+5=293x + 5 = 29, and the proposed solution is x=8x = 8.
Substitute x=8x=8 into the left side of the equation:

3(8)+5=24+5=293(8) + 5 = 24 + 5 = 29

Since the left side (29) equals the right side (29), the proposed solution x=8x=8 is correct.

The correct calculation is D. 3(8)+5=29; both sides have the same value, so x=8 satisfies the equation\boxed{\text{D. } 3(8)+5= 29; \text{ both sides have the same value, so } x = 8 \text{ satisfies the equation}}.

Part 8· 4 steps

  1. Identify the initial amount and the amount set aside.

    A teacher has 53 counters. 5 counters are set aside.

  2. Determine the remaining amount.

    The remaining amount is 53−553 - 5.

  3. Represent the sharing of the remaining amount.

    The remaining amount is shared equally among 6 groups, and xx is the number in each group. This means 6×x6 \times x counters are shared.

  4. Formulate the equation.

    The total number of counters is 53. This total is made up of the 5 counters set aside and the 6x6x counters distributed among the groups.
    So, 6x+5=536x + 5 = 53.

    The correct number sentence is B. 6x+5=53\boxed{\text{B. } 6x+5 = 53}.

Part 9

Andile's model is 7+x+4=607 + x + 4 = 60.
The problem states "7 equal boxes of xx pencils". This means 7×x7 \times x, or 7x7x, not 7+x7 + x. The learner is incorrectly treating multiplication as addition. The correct model should be 7x+4=607x + 4 = 60.

The response that best diagnoses and repairs the model is A. The learner treated 7 equal groups as addition; write 7x+4=60 and connect 7x to repeated groups\boxed{\text{A. The learner treated 7 equal groups as addition; write } 7x + 4 = 60 \text{ and connect } 7x \text{ to repeated groups}}.

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