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WAEC 2018 · Paper 2 · Q11✱

  1. (a)

    (i) Copy and complete the addition ⊕\oplus and multiplication ⊗\otimes tables in modulo 5 on the set {2,3,4}\{2, 3, 4\}. (ii) Use the tables to: (α) solve the equation (4⊗e)⊕2≡1(mod5)(4 \otimes e) \oplus 2 \equiv 1 \pmod 5; (β) find the value of nn if (4⊕n)⊗2≡2(mod5)(4 \oplus n) \otimes 2 \equiv 2 \pmod 5.

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

  2. (b)

    Consider the statements pp: Landi has cholera; qq: Landi is in the hospital. If p⇒qp \Rightarrow q, state whether or not the following statements are valid: (i) If Landi is in the hospital, then he has cholera. (ii) If Landi is not in the hospital, then he does not have cholera. (iii) If Landi does not have cholera, then he is not in the hospital.

    Show the answer

    (i) not valid (converse); (ii) valid (contrapositive); (iii) not valid (inverse)

Worked solution (try it first)

(a)(i)

  1. Add or multiply, then take the remainder on dividing by 5.
  2. For example 3⊕4=7≡23 \oplus 4 = 7 \equiv 2 and 3⊗4=12≡23 \otimes 4 = 12 \equiv 2.
  3. The ⊕\oplus table (rows 2, 3, 4) is 4,0,14, 0, 1 / 0,1,20, 1, 2 / 1,2,31, 2, 3.
  4. The ⊗\otimes table is 4,1,34, 1, 3 / 1,4,21, 4, 2 / 3,2,13, 2, 1.

(ii)

  1. (α)** (4⊗e)⊕2≡1(4 \otimes e) \oplus 2 \equiv 1, so 4⊗e≡1−2≡44 \otimes e \equiv 1 - 2 \equiv 4.
  2. Then 4e≡4(mod5)4e \equiv 4 \pmod 5, so e=1e = 1.

(β)

  1. (4⊕n)⊗2≡2(4 \oplus n) \otimes 2 \equiv 2.
  2. From the ⊗\otimes table, the number that gives 2 when multiplied by 2 is 1, so 4⊕n≡14 \oplus n \equiv 1, and from the ⊕\oplus table n=2n = 2.

(b)

  1. From pp
    ⇒q\Rightarrow q, only the contrapositive ∼q\sim q
    ⇒∼p\Rightarrow \sim p follows.

(i)

  1. Is the converse: not valid.

(ii)

  1. Is the contrapositive: valid.

(iii)

  1. Is the inverse: not valid.

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