(a) Without using tables or calculator, simplify : \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\), leaving the answer in standard form (scientific notation).
(b)
In the diagram, \(\overline{EF}\) is parallel to \(\overline{GH}\). If \(< AEF = 3x°, < ABC = 120°\) and \(< CHG = 7x°\), find the value of \(< GHB\).
(b)
In the diagram, \(\overline{EF}\) is parallel to \(\overline{GH}\). If \(< AEF = 3x°, < ABC = 120°\) and \(< CHG = 7x°\), find the value of \(< GHB\).
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Correct Answer: Option n
Explanation:
(a) \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\)
= \(\frac{6 \times 10^{-1} \times 32 \times 4 \times 10^{-3}}{12 \times 10^{-1} \times 8 \times 10^{-3} \times 16 \times 10^{-2}}\)
= \(\frac{4}{2 \times 4 \times 10^{-2}}\)
= \(\frac{1}{2} \times \frac{1}{10^{-2}\)
= \(0.5 \times 10^{2} \)
= \(5 \times 10^{1}\)
(b)
In the diagram above, \(< GHB = 180 - 7x°\) (angles on a straight line)
\(< HBN = 180 - 7x°\) (alternate angles)
\(< EBN = 3x°\) (corresponding angles)
Then \(3x° + (180 - 7x)° = 120°\)
\(\implies 3x° - 7x° + 180° = 120°\)
\(\implies 180° - 120° = 7x° - 3x°\)
\(60° = 4x°\)
\(x = 15°\)
Hence, \(< GHB = 180° - (7 \times 15°) = 180° - 105°\)
= \(75°\).
(a) \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\)
= \(\frac{6 \times 10^{-1} \times 32 \times 4 \times 10^{-3}}{12 \times 10^{-1} \times 8 \times 10^{-3} \times 16 \times 10^{-2}}\)
= \(\frac{4}{2 \times 4 \times 10^{-2}}\)
= \(\frac{1}{2} \times \frac{1}{10^{-2}\)
= \(0.5 \times 10^{2} \)
= \(5 \times 10^{1}\)
(b)
In the diagram above, \(< GHB = 180 - 7x°\) (angles on a straight line)
\(< HBN = 180 - 7x°\) (alternate angles)
\(< EBN = 3x°\) (corresponding angles)
Then \(3x° + (180 - 7x)° = 120°\)
\(\implies 3x° - 7x° + 180° = 120°\)
\(\implies 180° - 120° = 7x° - 3x°\)
\(60° = 4x°\)
\(x = 15°\)
Hence, \(< GHB = 180° - (7 \times 15°) = 180° - 105°\)
= \(75°\).