Mathematics 2015 Jamb Examination

The coordinates of the mid-point \((x, y)\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) are found using the formula:\[ x = \dfrac{x_1 + x_2}{2}, \quad y = \dfrac{y_1 + y_2}{2} \]Given \(A = (-3, 5)\) and \(B = (4, -1)\):\[ x = \dfrac{-3 + 4}{2} = \dfrac{1}{2} \]\[ y = \dfrac{5 + (-1)}{2} = \dfrac{4}{2} = 2 \]Thus, the mid-point is \(\left(\dfrac{1}{2}, 2\right)\).

Simple Interest (I) is calculated using the formula:\[ I = \dfrac{P \times R \times T}{100} \]Where \(P\) is the Principal, \(R\) is the Rate, and \(T\) is the Time in years.
Given:
\(P = 325\)
\(T = 5\)
\(R = 3\)\[ I = \dfrac{325 \times 3 \times 5}{100} \]\[ I = \dfrac{4875}{100} = 48.75 \]The simple interest is ₦48.75.

To square 0.09, we multiply the number by itself:\[ 0.09 \times 0.09 = 0.0081 \]In terms of significant figures, the leading zeros are not significant. The digits '8' and '1' are the first and second significant figures. To express this to 4 significant figures, we include two trailing zeros:\[ 0.008100 \]However, among the provided options, 0.0081 represents the simplified value correctly.

First, factorise the denominators:\[ x^2 + 3x + 2 = (x + 1)(x + 2) \]\[ x^2 + 5x + 6 = (x + 2)(x + 3) \]The expression becomes:\[ \dfrac{1}{(x + 1)(x + 2)} + \dfrac{1}{(x + 2)(x + 3)} \]Find the Lowest Common Multiple (LCM) of the denominators, which is \((x + 1)(x + 2)(x + 3)\):\[ \dfrac{(x + 3) + (x + 1)}{(x + 1)(x + 2)(x + 3)} \]\[ = \dfrac{2x + 4}{(x + 1)(x + 2)(x + 3)} \]\[ = \dfrac{2(x + 2)}{(x + 1)(x + 2)(x + 3)} \]Cancel out the common factor \((x + 2)\):\[ = \dfrac{2}{(x + 1)(x + 3)} \]

Probability of an event is given by the number of successful outcomes divided by the total outcomes. Since the draws from the two bags are independent, the probability of both being blue is the product of their individual probabilities.
Bag 1: Total balls = \(3 + 5 = 8\). Probability of blue \(P(B_1) = \dfrac{3}{8}\).
Bag 2: Total balls = \(2 + 4 = 6\). Probability of blue \(P(B_2) = \dfrac{2}{6} = \dfrac{1}{3}\).\[ P(\text{Both Blue}) = P(B_1) \times P(B_2) \]\[ P(\text{Both Blue}) = \dfrac{3}{8} \times \dfrac{1}{3} = \dfrac{3}{24} = \dfrac{1}{8} \]Based on the options, 3/24 is the expected form.

The intersection of two sets \(A \cap B\) consists only of elements that are present in both set A and set B.
Set A contains: 3, 4, 1, 10, ⅓
Set B contains: 4, 3, ⅓, 7
Comparing the two sets, the common elements are 3, 4, and ⅓. Therefore, \(A \cap B = \{3, 4, ⅓\}\).

Convert both sides to base 10:
For 132_x: \(1(x^2) + 3(x^1) + 2(x^0) = x^2 + 3x + 2\)
For 70_eight: \(7(8^1) + 0(8^0) = 56 + 0 = 56\)
Equate them:\[ x^2 + 3x + 2 = 56 \]\[ x^2 + 3x - 54 = 0 \]Factorise the quadratic equation:\[ (x + 9)(x - 6) = 0 \]So, \(x = -9\) or \(x = 6\). Since a base cannot be negative, \(x = 6\).

The Selling Price (S.P.) is given as ₦420,000, and the gain is 15%. This means the selling price is 115% of the Cost Price (C.P.).\[ 115\% \text{ of C.P.} = 420,000 \]\[ 1.15 \times \text{C.P.} = 420,000 \]\[ \text{C.P.} = \dfrac{420,000}{1.15} \]\[ \text{C.P.} = 365,217.39 \]Rounding to the nearest whole number gives ₦365,217.

Express all terms as powers of 4 (or 2). Using base 4:\[ 64 = 4^3 \]\[ 16 = 4^2 \]The equation becomes:\[ \dfrac{1}{(4^3)^{(x + 2)}} = \dfrac{4^{(x - 3)}}{(4^2)^x} \]\[ 4^{-3(x + 2)} = 4^{(x - 3) - 2x} \]Equate the exponents:\[ -3x - 6 = x - 3 - 2x \]\[ -3x - 6 = -x - 3 \]\[ -6 + 3 = 3x - x \]\[ -3 = 2x \]\[ x = -\dfrac{3}{2} \]

To solve the quadratic equation \(x^2 - 2x - 3 = 0\), we look for two numbers that multiply to −3 and add to −2. These numbers are −3 and +1.\[ x^2 - 3x + x - 3 = 0 \]\[ x(x - 3) + 1(x - 3) = 0 \]\[ (x + 1)(x - 3) = 0 \]Therefore, \(x + 1 = 0 \Rightarrow x = -1\) or \(x - 3 = 0 \Rightarrow x = 3\).