Question

1. (a) Simplify, without using tables or calculator: (3⁄₄)(3³⁄₈+1⁵⁄₈) ÷ (2⅛−1½)
(b) Given that log₁₀2 = 0.3010 and log₁₀3 = 0.4771, evaluate, correct to 2 significant figures and without using tables or calculator, log₁₀1.125.

Accepted Answer

(a) 3⁄₄(3³⁄₈+1⁵⁄₈) = 3⁄₄(27⁄₈+13⁄₈) — wait, using the value as printed: 3⁄₄(3³⁄₈+1⁵⁄₈) = 3⁄₄ × (17⁄₈+3⁄₂) ... simplifying the bracket to 40⁄₈=5, gives 3⁄₄×5 = 15⁄₄. Dividing by (2⅛−1½) = 5⁄₈: (15⁄₄)÷(5⁄₈) = (15⁄₄)×(8⁄₅) = 30⁄₅ = 6.

(b) 1.125 = 1125⁄1000 = 9⁄₈ = 3²⁄₂³. log₁₀1.125 = log₁₀9 − log₁₀8 = 2log₁₀3 − 3log₁₀2 = 2(0.4771) − 3(0.3010) = 0.9542 − 0.9030 = 0.0512 ≈ 0.051 (2 s.f.).

WAEC Mathematics 2012 Theory — Question 1

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