WAEC Further Mathematics Past Questions & Answers - Page 133

The word has 8 letters with one letter repeated 3 times, therefore we have:

\(\frac{8!}{3!} = 6720\) ways.

Given the equation of the circle \(x^{2} + y^{2} - 8x - 2y + 1 = 0\).

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding, we have \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2} \equiv x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)

Comparing the RHS of the equation above with the equation rewritten as \(x^{2} + y^{2} - 8x - 2y = -1\), we have

\(-2a = -8; -2b = -2 \implies a = 4, b = 1\)

\(\therefore r^{2} - 4^{2} - 1^{2} = -1 \implies r^{2} = -1 + 16 + 1 = 16\)

\(r = \sqrt{16} = 4\)

Given \(\sin \theta = \frac{3}{5}  \implies opp = 3, hyp = 5\)

Using Pythagoras' Theorem, we have \( adj = \sqrt{5^{2} - 3^{2}} = \sqrt{16}  = 4\)

\(\therefore \cos \theta = \frac{4}{5}, 0° < \theta < 90°\)

In the quadrant where \(180° - \theta\) lies is the 2nd quadrant and here, only \(\sin \theta = +ve\).

\(\therefore \cos (180 - \theta) = -ve = \frac{-4}{5}\)

Recall, the terms of a GP is given by \(T_{n} = ar^{n - 1}\)

Given, \(T_{1} = a = \frac{3}{4}\) and \(T_{2} \times T_{3} = ar \times ar^{2} = 972\)

\(\implies \frac{3}{4}r \times \frac{3}{4}r^{2} = \frac{9}{16}r^{3} = 972\)

\(r^{3} = \frac{972 \times 16}{9} = 1728  \implies r = \sqrt[3]{1728} = 12\)

Note that every 4- digit and 5- digit numbers formed is greater than 150. Therefore, \(^{5}P_{5} + ^{5}P_{4} = \frac{5!}{(5 - 5)!} + \frac{5!}{(5 - 4)!} = 120 + 120 = 240\) numbers are greater than 150 already.

Among the 3- digit numbers, the numbers 123, 124, 125, 132, 134, 135, 142, 143 and 145 are removed from the ones that meet the criterion so we have:

\(^{5}P_{3} - 9 = \frac{5!}{(5 - 3)!} = 60 - 9 = 51 \implies 240 + 51 = 291\) numbers are greater than 150.