Mathematics 2005 Jamb Examination

With replacement, \(P(\text{blue})=\dfrac{5}{10}=\dfrac{1}{2}\), \(P(\text{red})=\dfrac{3}{10}\).

Required probability = \(P(\text{blue})^2 + P(\text{red})^2 \)
\( = \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{3}{10}\right)^2 \)
\(= \dfrac{1}{4} + \dfrac{9}{100} \)
\(= \dfrac{25}{100} + \dfrac{9}{100} \)
\(= \dfrac{34}{100} = \dfrac{17}{50}.\)

Therefore option A is correct.

An integer is divisible by both 2 and 3 iff divisible by lcm(2,3)=6. The numbers in \(1..20\) divisible by 6 are \(6,12,18\) — three numbers.

Total possible integers = 20, so probability = \(\dfrac{3}{20}\).

Therefore option C is correct.

FROM the diagram above

100 + a = 180 (sum of angles on a straight line)

a = 180 - 100 = 80º

Z = 80 + 40 = 120º (exterior angles of a triangle = sum of the two opp. interior angles)

Z = x = 120º (corresponding angles are equal).

Since the bearing of S from R is 44^o, we use the alternate interior angles property between the north-south parallel lines to find that the bearing of R from S is 180^o+44^o=224^o.

Let the monthly income be \( I \). Total allowance = \( 0.20I \). The second child (middle child) received the remaining percentage of the allowance: \[ 100\% - 45\% - 25\% = 30\% \] We can now solve for \( I \): \[ 0.30 \times (0.20I) = 3000 \Rightarrow 0.06I = 3000 \] \[ I = \frac{3000}{0.06} = \boxed{50,000} \] Therefore, the father's monthly income is ₦50,000 (option C).

$$ \text{By long division:} $$ \[ \begin{array}{r|l}
      & 3x - 5 \\
2x - 1 & 6x^2 - 13x + 5 \\
      & -(6x^2 - 3x) \\
      & \underline{-10x + 5} \\
      & -(-10x + 5) \\
      & \underline{0}
\end{array}\]

Compute determinant by expansion along the first row:

\[ \det(P)=1\begin{vmatrix}5 & -1\\[4pt]2 & 0\end{vmatrix} -3\begin{vmatrix}4 & -1\\[4pt]-3 & 0\end{vmatrix} +2\begin{vmatrix}4 & 5\\[4pt]-3 & 2\end{vmatrix}. \]

Evaluate minors:

\(\begin{vmatrix}5 & -1\\2 & 0\end{vmatrix}=5\cdot0 -(-1)\cdot2 = 2.\)

\(\begin{vmatrix}4 & -1\\-3 & 0\end{vmatrix}=4\cdot0 -(-1)(-3)= -3.\)

\(\begin{vmatrix}4 & 5\\-3 & 2\end{vmatrix}=4\cdot2 -5(-3)=8+15=23.\)

So \(\det(P)=1\cdot2 -3(-3) +2\cdot23 \)\(= 2 +9 +46 = 57.\)

Elements of \(U\) are \(\{1,2,3,4,5,6,7,8,9\}\). Prime factors of 30 are \(2,3,5\), so \(P=\{2,3,5\}\).

The complement \(U\setminus P = \{1,4,6,7,8,9\}\), which is option A.

Midpoint formula: for points \((x_1,y_1)\) and \((x_2,y_2)\), midpoint is \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).

Here \(\dfrac{3+(-1)}{2}=\dfrac{2}{2}=1\) and \(\dfrac{-4+10}{2}=\dfrac{6}{2}=3\). So midpoint is \((1,3)\).

Compute \(2M\) and \(3N\) elementwise:

\(2M=\begin{pmatrix}2 & 6 & 4\\ 8 & 10 & -2\\ -6 & 4 & 0\end{pmatrix},\quad 3N=\begin{pmatrix}3 & -6 & 9\\ 12 & -3 & 15\\ 6 & -9 & -3\end{pmatrix}.\)

Subtract: \(2M-3N=\begin{pmatrix}2-3 & 6-(-6) & 4-9\\ 8-12 & 10-(-3) & -2-15\\ -6-6 & 4-(-9) & 0-(-3)\end{pmatrix} \)
\(= \begin{pmatrix}-1 & 12 & -5\\ -4 & 13 & -17\\ -12 & 13 & 3\end{pmatrix}.\)