WAEC 2016 · Paper 2 · Q13

  1. (a)

    Find the equation of a straight line which passes through the point (2,−3)(2, -3) and is parallel to the line 2x+y=62x + y = 6.

    Show the answer

    2x+y−1=02x + y - 1 = 0

  2. (b)

    The operation Δ\Delta is defined on the set T={2,3,5,7}T = \{2, 3, 5, 7\} by x Δ y=(x+y+xy) mod 8x\,\Delta\,y = (x + y + xy) \bmod 8. (i) Construct the modulo 8 table for the operation Δ\Delta on the set TT. (ii) Use the table to find: I. 2 Δ (5 Δ 7)2\,\Delta\,(5\,\Delta\,7); II. nn if 2 Δ n=5 Δ 72\,\Delta\,n = 5\,\Delta\,7.

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

Worked solution (try it first)

(a)

  1. 2x+y=62x + y = 6 is y=−2x+6y = -2x + 6, with gradient −2-2.
  2. A parallel line has the same gradient: y=−2x+cy = -2x + c.
  3. Through (2,−3)(2, -3): −3=−4+c-3 = -4 + c, so c=1c = 1.
  4. The line is y=−2x+1y = -2x + 1, or 2x+y−1=02x + y - 1 = 0.

(b)(i)

  1. Each entry is x+y+xyx + y + xy reduced modulo 8.
  2. For example 2 Δ 3=2+3+6=11≡32\,\Delta\,3 = 2 + 3 + 6 = 11 \equiv 3 and 5 Δ 7=5+7+35=47≡75\,\Delta\,7 = 5 + 7 + 35 = 47 \equiv 7.
  3. Δ\Delta 2 3 5 7
    2 0 3 1 7
    3 3 7 7 7
    5 1 7 3 7
    7 7 7 7 7

(ii)

  1. I.
  2. 5 Δ 7=75\,\Delta\,7 = 7, so 2 Δ (5 Δ 7)=2 Δ 72\,\Delta\,(5\,\Delta\,7) = 2\,\Delta\,7
    =7= 7.
  3. II. 2 Δ n=5 Δ 7=72\,\Delta\,n = 5\,\Delta\,7 = 7: in the row of 2, the 7 is in the column of 7, so n=7n = 7.

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