WAEC 2023 · Paper 2 · Q6✱

  1. (a)(i)

    M={n:2n−3≤37}M = \{n : 2n - 3 \le 37\}, where nn is a counting number. Write down all the elements in MM.

    Show the answer

    {1,2,3,…,20}\{1, 2, 3, \dots, 20\}

  2. (a)(ii)

    If a number is selected at random from MM, what is the probability that it is a: (α\alpha) multiple of 3; (β\beta) factor of 10? Enter both.

    Separate values with commas, e.g. 3, −2

  3. (b)

    A shop owner gave an end-of-year bonus to two of his attendants, Kontor and Gapson, in the ratio of their ages. Kontor’s age is one and a half times that of Gapson, who is 20 years old. If Kontor received Le 200,000.00, find: (i) the total amount shared; (ii) Gapson’s share.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. Solve the inequality: 2n−3≤372n - 3 \le 37, so 2n≤402n \le 40 and n≤20n \le 20.
  2. Counting numbers start at 1, so M={1,2,3,…,20}M = \{1, 2, 3, \ldots, 20\}.

(ii)

  1. MM has 20 members.
  2. (α\alpha) Multiples of 3: 3,6,9,12,15,183, 6, 9, 12, 15, 18, which is 6 numbers, so P=620=310P = \frac{6}{20} = \frac{3}{10}.
  3. (β\beta) Factors of 10: 1,2,5,101, 2, 5, 10, which is 4 numbers, so P=420=15P = \frac{4}{20} = \frac15.

(b)(i)

  1. Kontor is 112×20=301\frac12 \times 20 = 30 years old, so the ratio of their ages is 30:20=3:230 : 20 = 3 : 2.
  2. Kontor's 3 parts are Le 200,000, so 1 part is 200 0003\frac{200\,000}{3} and the 5 parts total 200 0003×5≈Le 333,333.33\frac{200\,000}{3} \times 5 \approx \text{Le } 333,333.33.

(ii)

  1. Gapson's 2 parts: 200 0003×2≈Le 133,333.33\frac{200\,000}{3} \times 2 \approx \text{Le } 133,333.33.

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