Mathematics 2021 Jamb Examination

The locus of a point that maintains an equal distance from two intersecting straight lines consists of the angle bisectors of the angles formed by the intersection of those lines. Every point on an angle bisector is at the same perpendicular distance from both lines that form the angle. Therefore, the set of all such points defines the angle bisector.

First, calculate the total sum of all the numbers in the data set:\[\text{Total} = 4 + 16 + 30 + 20 + 10 + 14 + 26 = 120\]Next, identify the numbers in the set that are equal to or greater than 16:
These are 16, 30, 20, and 26.
Sum of these selected numbers:\[16 + 30 + 20 + 26 = 92\]To find the sum of the angles of the sectors for these numbers in a pie chart (where the total circle is 360^o), we use the formula:\[\text{Sum of angles} = \frac{\text{Sum of selected numbers}}{\text{Total sum}} \times 360^o\]\[\text{Sum of angles} = \frac{92}{120} \times 360^o\]\[= 92 \times 3 = 276^o\]

The mean is the sum of the numbers divided by the count. For the first ten numbers:\[\frac{\sum_{i=1}^{10} x_i}{10} = 16 \implies \sum_{i=1}^{10} x_i = 160\]Let the eleventh number be \(y\). When it is added, the new total count of numbers is 11 and the new mean is 18:\[\frac{160 + y}{11} = 18\]Multiply both sides by 11 to find the new total sum:\[160 + y = 18 \times 11\]\[160 + y = 198\]Subtract 160 from both sides to find the eleventh number:\[y = 198 - 160 = 38\]

First, list all possible pairs that can be selected from the set {1, 2, 3, 4}. The total number of ways to choose 2 numbers from 4 is given by \(^4C_2\):\[^4C_2 = \frac{4 \times 3}{2 \times 1} = 6 \text{ pairs}\]The possible pairs are:
{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}.
Now, find the sum of each pair:
1 + 2 = 3 (Odd)
1 + 3 = 4 (Even)
1 + 4 = 5 (Odd)
2 + 3 = 5 (Odd)
2 + 4 = 6 (Even)
3 + 4 = 7 (Odd)
The successful outcomes (where the sum is even) are {1, 3} and {2, 4}. There are 2 such outcomes.
Probability is given by:\[P(\text{Even Sum}) = \frac{\text{Successful outcomes}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3}\]

First, determine the total number of integers in the range from 41 to 56 inclusive:\[\text{Total numbers} = (56 - 41) + 1 = 15 + 1 = 16\]Next, identify the multiples of 9 within this range (41, 42, ..., 56):
9 × 5 = 45
9 × 6 = 54
There are exactly 2 multiples of 9 in this set (45 and 54).
The probability is calculated as:\[P(\text{multiple of 9}) = \frac{\text{Count of multiples of 9}}{\text{Total count of numbers}} = \frac{2}{16} = \frac{1}{8}\]

First, calculate the total interest accumulated over 4 months using the simple interest formula:\[I = \frac{P \times R \times T}{100}\]Here, the Principal (P) is 10.00, the Rate (R) is 2% per month, and the Time (T) is 4 months:\[I = \frac{10 \times 2 \times 4}{100} = \frac{80}{100} = 0.80\]The total amount owed before repayment is the Principal plus the Interest:\[\text{Total amount} = 10.00 + 0.80 = 10.80\]After repaying 8.00, the remaining balance is: 10.80 - 8.00 = ₦2.80

Apply logarithm laws: \(n \log a = \log a^n\), \(\log a - \log b = \log \frac{a}{b}\), and \(\log a + \log b = \log(ab)\).
Step 1: Apply the power rule to the first term:\[2 \log \frac{2}{5} = \log \left(\frac{2}{5}\right)^2 = \log \frac{4}{25}\]Step 2: Combine the terms using subtraction and addition rules:\[\log \frac{4}{25} - \log \frac{72}{125} + \log 9 = \log \left( \frac{4}{25} \div \frac{72}{125} \times 9 \right)\]Step 3: Perform the arithmetic inside the logarithm:\[\log \left( \frac{4}{25} \times \frac{125}{72} \times 9 \right)\]\[\log \left( \frac{1}{1} \times \frac{5}{18} \times 9 \right) = \log \left( \frac{45}{18} \right) = \log \left( \frac{5}{2} \right)\]Step 4: Expand the result to match options:\[\log 5 - \log 2\]Since \(\log 5 = \log \frac{10}{2} = \log 10 - \log 2 = 1 - \log 2\):\[(1 - \log 2) - \log 2 = 1 - 2\log 2\]

Average speed is defined as the total distance divided by the total time taken.
1. Total Distance = \(P + Q\)
2. Time for the first leg (Calabar to Enugu): \(t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{P}{U}\)
3. Time for the second leg (Enugu to Benin): \(t_2 = \frac{Q}{W}\)
4. Total Time = \(t_1 + t_2 = \frac{P}{U} + \frac{Q}{W}\)
To add the times, find a common denominator (UW):\[\text{Total Time} = \frac{PW + QU}{UW}\]Now, calculate Average Speed:\[\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{P + Q}{\frac{PW + QU}{UW}}\]\[= \frac{UW(P + Q)}{PW + QU}\]

Inverse variation means \(w = \frac{k}{\text{variable}}\). In this case:\[w = \frac{k}{\left(\frac{uv}{u + v}\right)} = \frac{k(u + v)}{uv}\]Substitute the given values (w = 8, u = 2, v = 6) to find the constant \(k\):\[8 = \frac{k(2 + 6)}{2 \times 6}\]\[8 = \frac{8k}{12}\]\[8 = \frac{2k}{3} \implies 24 = 2k \implies k = 12\]Substitute \(k = 12\) back into the equation:\[w = \frac{12(u + v)}{uv}\]Multiply both sides by \(uv\) to clear the fraction:\[uvw = 12(u + v)\]

First, evaluate \(g(x + 1)\) by substituting \((x + 1)\) into the function wherever \(x\) appears:\[g(x + 1) = (x + 1)^2 + 3(x + 1)\]Expand the terms:\[g(x + 1) = (x^2 + 2x + 1) + (3x + 3)\]\[g(x + 1) = x^2 + 5x + 4\]Now, find the difference \(g(x + 1) - g(x)\):\[(x^2 + 5x + 4) - (x^2 + 3x)\]\[= x^2 + 5x + 4 - x^2 - 3x\]\[= 2x + 4\]Factor out the common term 2:\[= 2(x + 2)\]