NECO 2024 · Paper 2 · Q6

  1. (a)

    Use mathematical tables to evaluate 40004×75.95.61×4.39\dfrac{\sqrt[4]{4000} \times 75.9}{5.61 \times 4.39}.

  2. (b)

    Express 4553six4553_{\text{six}} in base three.

Worked solution (try it first)

(a)

  1. Set out the working with logarithms from four-figure tables:
  2. No. Log
    40004\sqrt[4]{4000} 3.6021÷4=0.90053.6021 \div 4 = 0.9005
    75.975.9 1.88021.8802
    numerator 2.78072.7807
    5.615.61 0.74900.7490
    4.394.39 0.64250.6425
    denominator 1.39151.3915
    result 2.7807−1.3915=1.38922.7807 - 1.3915 = 1.3892
  3. The antilog of .3892.3892 is 2450, and the characteristic 1 means ×10\times 10: the answer is about 24.5024.50.

(b)

  1. First change to base ten: 4553six=4×216+5×36+5×6+34553_{\text{six}} = 4 \times 216 + 5 \times 36 + 5 \times 6 + 3
    =864+180+30+3= 864 + 180 + 30 + 3
    =1077= 1077.
  2. Then divide by 3 repeatedly, keeping the remainders: 1077÷3=3591077 \div 3 = 359 r 0.
  3. 359÷3=119359 \div 3 = 119 r 2.
  4. 119÷3=39119 \div 3 = 39 r 2.
  5. 39÷3=1339 \div 3 = 13 r 0.
  6. 13÷3=413 \div 3 = 4 r 1.
  7. 4÷3=14 \div 3 = 1 r 1.
  8. 1÷3=01 \div 3 = 0 r 1.
  9. Read the remainders from the bottom up: 1110220three1110220_{\text{three}}.

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