(a) Without using tables or calculator, simplify :... | WAEC SSCE - General Mathematics - 2014 (Essay)

Question 1 Report

(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 \(< GHB\).

Answer

(a) Simplify in standard form.

Write every number in standard form first, then combine powers of 10.

Numerator:

\[0.6\times32\times0.004=(6\times10^{-1})(3.2\times10^{1})(4\times10^{-3})\]\[=(6\times3.2\times4)\times10^{-1+1-3}=76.8\times10^{-3}=7.68\times10^{-2}\]

Denominator:

\[1.2\times0.008\times0.16=(1.2\times10^{0})(8\times10^{-3})(1.6\times10^{-1})\]\[=(1.2\times8\times1.6)\times10^{0-3-1}=15.36\times10^{-4}=1.536\times10^{-3}\]

Divide:

\[\frac{7.68\times10^{-2}}{1.536\times10^{-3}}=\frac{7.68}{1.536}\times10^{-2-(-3)}=5\times10^{1}\]

So the value is \(\mathbf{5.0\times10^{1}}\) (that is, \(50\)).

(b) Find \(\angle GHB\).

From the diagram, \(\overline{EF}\parallel\overline{GH}\). The line \(A\,E\,B\) is straight, it bends at \(B\) where \(\angle ABC=120^{\circ}\), and the line then continues straight through to \(H\). The marked angles are \(\angle AEF=3x^{\circ}\) and \(\angle CHG=7x^{\circ}\).

Draw a line \(BK\) through \(B\) parallel to both \(\overline{EF}\) and \(\overline{GH}\).

Since \(A,E,B\) are collinear, \(\angle FEB=180^{\circ}-3x\). With \(EF\parallel BK\), the co-interior angles give

\[\angle EBK=180^{\circ}-\angle FEB=3x^{\circ}\]

With \(BK\parallel GH\) and \(BH\) a transversal, alternate angles give

\[\angle KBH=\angle GHB=7x^{\circ}\]

The ray \(BK\) lies inside \(\angle EBH\), so

\[\angle EBH=\angle EBK+\angle KBH\]\[120=3x+7x=10x\]\[x=12\]

Therefore

\[\angle GHB=7x=7(12)=\mathbf{84^{\circ}}\]

Read lesson note on Modular Arithmetic (WAEC)

Read lesson note on Angles And Intercepts On Parallel Lines (WAEC)

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