Mathematics 2015 Jamb Examination

To solve the quadratic equation \(3x^2 - 4x - 5 = 0\), we use the quadratic formula because the expression is not easily factorable:\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]From the equation, we have \(a = 3\), \(b = -4\), and \(c = -5\). Substituting these into the formula:\[ x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-5)}}{2(3)} \]\[ x = \dfrac{4 \pm \sqrt{16 + 60}}{6} \]\[ x = \dfrac{4 \pm \sqrt{76}}{6} \]\[ x = \dfrac{4 \pm 8.7178}{6} \]For the first root:\[ x_1 = \dfrac{4 + 8.7178}{6} = \dfrac{12.7178}{6} \approx 2.1196 \]For the second root:\[ x_2 = \dfrac{4 - 8.7178}{6} = \dfrac{-4.7178}{6} \approx -0.7863 \]Rounding to two decimal places, we get \(x = 2.12\) or \(x = -0.79\).

In an Arithmetic Progression (AP), the sum of the first \(n\) terms is given by the formula:\[ S_n = \dfrac{n}{2}(a + l) \]Where:
\(a\) = first term = 6
\(l\) = last term = 10
\(S_n\) = sum = 40
Substituting these values into the formula to find \(n\):\[ 40 = \dfrac{n}{2}(6 + 10) \]\[ 40 = \dfrac{n}{2}(16) \]\[ 40 = 8n \]\[ n = \dfrac{40}{8} = 5 \]The sequence contains 5 terms.

A line passing through the origin has a y-intercept \(c = 0\). Its equation is in the form \(y = mx\).
The gradient \(m\) is calculated using the origin (0, 0) and the given point (-3, -4):\[ m = \dfrac{y_2 - y_1}{x_2 - x_1} \]\[ m = \dfrac{-4 - 0}{-3 - 0} = \dfrac{-4}{-3} = \dfrac{4}{3} \]Substituting the gradient back into the equation:\[ y = \dfrac{4}{3}x \text{ or } y = \dfrac{4x}{3} \]

Given the compound interest formula:\[ A = P\left(1 + \dfrac{r}{100}\right)^n \]Substitute the values:
\(P = 126\)
\(r = 4\)
\(n = 2\)\[ A = 126\left(1 + \dfrac{4}{100}\right)^2 \]\[ A = 126(1 + 0.04)^2 \]\[ A = 126(1.04)^2 \]\[ A = 126(1.0816) \]\[ A = 136.2816 \]Rounding to two decimal places for currency, we get ₦136.30.

To make \(x\) the subject, we isolate it step-by-step:
Start with: \(s = 2 + \dfrac{t}{5}(x + \dfrac{3y}{5})\)
Step 1: Subtract 2 from both sides:\[ s - 2 = \dfrac{t}{5}(x + \dfrac{3y}{5}) \]Step 2: Multiply by 5 and divide by \(t\):\[ \dfrac{5(s - 2)}{t} = x + \dfrac{3y}{5} \]Step 3: Subtract \(3y/5\) from both sides:\[ x = \dfrac{5(s - 2)}{t} - \dfrac{3y}{5} \]Looking at the options, we can rewrite \(\dfrac{5(s - 2)}{t}\) as \(5[(s - 2) \div t]\). Thus, option A matches.

To find the value of \(x = \log_5 20\), we use the change of base formula:\[ \log_a b = \dfrac{\log_{10} b}{\log_{10} a} \]\[ x = \dfrac{\log_{10} 20}{\log_{10} 5} \]Using common logarithm values:
\(\log_{10} 20 \approx 1.30103\)
\(\log_{10} 5 \approx 0.69897\)\[ x = \dfrac{1.30103}{0.69897} \approx 1.86135 \]Alternatively:\[ \log_5 20 = \log_5 (4 \times 5) = \log_5 4 + \log_5 5 = \log_5 4 + 1 \]\[ \log_5 4 = \dfrac{\log_{10} 4}{\log_{10} 5} \approx \dfrac{0.60206}{0.69897} \approx 0.861 \]\[ x = 0.861 + 1 = 1.861 \]

First, we find the interest (I) gained:\[ I = \text{Amount} - \text{Principal} \]\[ I = 588 - 525 = 63 \]The formula for Simple Interest is:\[ I = \dfrac{P \times R \times T}{100} \]Where:
\(P = 525\)
\(T = 3\) years
\(I = 63\)
We need to find \(R\):\[ 63 = \dfrac{525 \times R \times 3}{100} \]\[ 6300 = 1575R \]\[ R = \dfrac{6300}{1575} = 4 \]The rate per annum is 4%.

The union of sets \((A \cup B) \cup C\) is a set containing all unique elements that appear in set A, set B, or set C.
Set A = {1, 5, 7}
Set B = {3, 9, 12, 15}
Set C = {2, 4, 6, 8}
First, find \(A \cup B\):\[ A \cup B = \{1, 3, 5, 7, 9, 12, 15\} \]Now, find the union of this result with set C:\[ (A \cup B) \cup C = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15\} \]The final set includes all the elements from the three separate sets arranged in ascending order.

This problem describes Hooke's Law, where extension (e) is directly proportional to tension (T).\[ e \propto T \Rightarrow e = kT \]First, find the constant \(k\) using the initial data (\(T = 8\) N, \(e = 2\) cm):\[ 2 = k(8) \]\[ k = \dfrac{2}{8} = 0.25 \]Now, find the new extension when the tension is 12 N:\[ e = 0.25 \times 12 \]\[ e = 3 \, \text{cm} \]Alternatively, since it is a direct proportion, we can use the ratio method:\[ \dfrac{e_1}{T_1} = \dfrac{e_2}{T_2} \Rightarrow \dfrac{2}{8} = \dfrac{e_2}{12} \]\[ 8e_2 = 24 \]\[ e_2 = 3 \, \text{cm} \]

To factorise the quadratic trinomial \(x^2 - 2x - 15\), we need to find two numbers that multiply to give −15 and add to give −2.
Pairs of factors of −15:
1 and −15 (Sum = −14)
−1 and 15 (Sum = 14)
3 and −5 (Sum = −2)
−3 and 5 (Sum = 2)
The pair that satisfies both conditions is +3 and −5.
We rewrite the middle term and factorise by grouping:\[ x^2 - 5x + 3x - 15 \]\[ x(x - 5) + 3(x - 5) \]\[ (x - 5)(x + 3) \]