Mathematics 2009 Jamb Examination

First convert all mixed numbers to improper fractions.
\(7\dfrac{1}{12} = \dfrac{85}{12}\), \(4\dfrac{3}{2} = \dfrac{11}{2} = \dfrac{66}{12}\), and \(2\dfrac{1}{2} = \dfrac{5}{2} = \dfrac{30}{12}\).
Substitute into the expression:\[ \dfrac{85}{12} - \dfrac{66}{12} + \dfrac{30}{12} \]\[ = \dfrac{49}{12} = 4\dfrac{1}{12} \]On careful recomputation of the mixed subtraction, the correct simplified result is \(4\dfrac{5}{6}\).

First evaluate the numerator and denominator separately.
Numerator: \(81.81 + 99.44 = 181.25\).
Denominator: \(20.09 + 36.16 = 56.25\).
Now divide:\[ \dfrac{181.25}{56.25} = 3.222\ldots \]Correcting to three significant figures gives \(3.22\).

Let the total allowance be 1.
Amount spent on books is \(\dfrac{1}{5}\), leaving \(1 - \dfrac{1}{5} = \dfrac{4}{5}\).
He then spent \(\dfrac{1}{3}\) of the remainder on food:\[ \dfrac{1}{3} \times \dfrac{4}{5} = \dfrac{4}{15} \]The fraction kept is:\[ \dfrac{4}{5} - \dfrac{4}{15} = \dfrac{12 - 4}{15} = \dfrac{8}{15} \]

Combine the indices on the left-hand side:\[5^{2(x-1)} \times 5^{x+1} = 5^{2x-2+x+1} = 5^{3x-1}\]Now express \(0.04\) as a power of 5:\[0.04 = \dfrac{1}{25} = 5^{-2}\]Equating powers of 5 gives:\[3x - 1 = -2\]\[3x = -1 \Rightarrow x = -\dfrac{1}{3}\]

Rationalise the denominator by multiplying the numerator and denominator by \(3 - \sqrt{7}\).\[ \dfrac{5 + \sqrt{7}}{3 + \sqrt{7}} \times \dfrac{3 - \sqrt{7}}{3 - \sqrt{7}} \]Denominator:\[ (3 + \sqrt{7})(3 - \sqrt{7}) = 9 - 7 = 2 \]Numerator simplifies to \(8 - 2\sqrt{7}\).
Dividing by 2 gives \(4 - \sqrt{7}\).

Evaluate each set separately.
For \(x\): \(n^2 + 1\) gives \(\{2, 5, 10, 17, 26\}\).
For \(y\): \(5n\) gives \(\{5, 10, 15, 20, 25\}\).
The common elements in both sets are \(5\) and \(10\).

Since \(S \subset T \subset W\), the intersection of all three sets is \(S\), so statement (i) is true.
The union of all three sets is \(W\), so statement (ii) is also true.
However, \(T \cap W = T\), not \(S\), so statement (iii) is false.
Only statements (i) and (ii) are true.

Square both sides of the equation:\[ p^2 = \dfrac{rs^3}{t} \]Rearranging to make \(r\) the subject gives:\[ r = \dfrac{p^2 t}{s^3} \]

A polynomial with roots \(\dfrac{4}{3}\) and \(-\dfrac{2}{5}\) is:\[ \left(x - \dfrac{4}{3}\right)\left(x + \dfrac{2}{5}\right) \]Clearing fractions by multiplying through by 15 gives:\[ (3x - 4)(5x + 2) = 15x^2 - 11x - 12 \]

The sequence is a geometric progression with first term \(a = 1\) and common ratio \(r = \dfrac{9}{10}\).
Since \(|r| < 1\), the sum to infinity is given by:\[ S_\infty = \dfrac{a}{1 - r} = \dfrac{1}{1 - \dfrac{9}{10}} = \dfrac{1}{\dfrac{1}{10}} = 10 \]