WAEC 2009 · Paper 2 · Q6✱

  1. (a)

    By how much is 11000211000_2 greater than or less than 1112×112111_2 \times 11_2? Give your answer in base two.

    Model answer

    1112×112=101012111_2 \times 11_2 = 10101_2 and 110002−101012=11211000_2 - 10101_2 = 11_2, so 11000211000_2 is greater by 11211_2 (3 in base ten).

  2. (b)

    A shopkeeper has 20 television sets in stock. He sells 18 of them at a profit of 15% and the remaining two at a loss of 5%. Find his percentage profit on the 20 sets.

Worked solution (try it first)

(a)

  1. Multiply in base two: 1112×112=1112×102+1112111_2 \times 11_2 = 111_2 \times 10_2 + 111_2
    =11102+1112= 1110_2 + 111_2
    =101012= 10101_2.
  2. Subtract: 110002−101012=11211000_2 - 10101_2 = 11_2 (check in base ten: 24−21=324 - 21 = 3).
  3. So 11000211000_2 is greater by 11211_2.

(b)

  1. Let each set cost pp.
  2. The 18 sets sell for 18×1.15p=20.7p18 \times 1.15p = 20.7p.
  3. The other 2 sell for 2×0.95p=1.9p2 \times 0.95p = 1.9p.
  4. Total sales =22.6p= 22.6p.
  5. Total cost =20p= 20p.
  6. Profit =2.6p= 2.6p.
  7. Percentage profit =2.6p20p×100=13%= \frac{2.6p}{20p} \times 100 = 13\%.

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