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1990 WAEC Mathematics Theory (a) Simplify \(\frac{0.016 \times 0.084}{0.48}\) [Leave your answer in standard form]. (b) Eight wooden poles...

Mathematics
WAEC 1990

(a) Simplify \(\frac{0.016 \times 0.084}{0.48}\) [Leave your answer in standard form].

(b) Eight wooden poles are to be used for pillars and the lengths of the poles form an Arithmetic Progression (A.P). If the second pole is 2m and the sixth is 5m, give the lengths of the poles, in order.

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Explanation

(a) \(\frac{0.016 \times 0.084}{0.48}\)

= \(\frac{16 \times 10^{-3} \times 84 \times 10^{-3}}{48 \times 10^{-2}}\)

= \(28 \times 10^{-3 - 3 - (-2)}\)

= \(28 \times 10^{-4} = 2.8 \times 10^{-3}\)

(b) \(T_{n} = a + (n - 1)d\) (Terms of an A.P)

\(T_{2} = a + d = 2 ... (1)\)

\(T_{6} = a + 5d = 5 ... (2)\)

\((2) - (1) : 4d = 3 \implies d = \frac{3}{4} = 0.75\)

Put the value of d in (1), so we have

\(a + 0.75 = 2 \implies a = 1.25\)

Therefore, the length of the poles in metres = \(1.25, 2, 2.75, 3.5, 4.25, 5, 5.75, 6.5\)


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