Mathematics 2020 Jamb Examination

To solve this, we can either perform the arithmetic directly in base 3 or convert to base 10 and then convert the result back to base 3.
Converting to base 10:
\((212)_3 = (2 \times 3^2) + (1 \times 3^1) + (2 \times 3^0) = 18 + 3 + 2 = 23_{10}\)
\((121)_3 = (1 \times 3^2) + (2 \times 3^1) + (1 \times 3^0) = 9 + 6 + 1 = 16_{10}\)
\((222)_3 = (2 \times 3^2) + (2 \times 3^1) + (2 \times 3^0) = 18 + 6 + 2 = 26_{10}\)
Now perform the operations in base 10:\[23 - 16 + 26 = 7 + 26 = 33_{10}\]Now convert \(33_{10}\) back to base 3:\[33 / 3 = 11 \text{ remainder } 0\]\[11 / 3 = 3 \text{ remainder } 2\]\[3 / 3 = 1 \text{ remainder } 0\]\[1 / 3 = 0 \text{ remainder } 1\]Reading the remainders from bottom to top, we get \(1020_3\).

This expression is in the form of the difference of two squares, which follows the identity:\[x^2 - y^2 = (x + y)(x - y)\]In this case, let \(x = (4a + 3)\) and \(y = (3a - 2)\).
Adding the terms:\[(4a + 3) + (3a - 2) = 4a + 3 + 3a - 2 = 7a + 1\]Subtracting the terms:\[(4a + 3) - (3a - 2) = 4a + 3 - 3a + 2 = a + 5\]Multiplying the results together gives the factored form:\[(7a + 1)(a + 5)\]

The median is the middle value of an ordered data set. First, we arrange the numbers in ascending order:
89, 100, 108, 119, 120, 130, 131, 131, 141, 161.
Since there are 10 numbers (an even number), the median is the average of the two middle terms (the 5th and 6th terms).
The 5th term is 120 and the 6th term is 130.
The median is calculated as:\[\text{Median} = \frac{120 + 130}{2} = \frac{250}{2} = 125\]

To solve the inequality, we first clear the fractions by multiplying through by the lowest common multiple of the denominators 3 and 5, which is 15:\[15 \left[ \frac{1}{3}(x + 1) - 1 \right] > 15 \left[ \frac{1}{5}(x + 4) \right]\]\[5(x + 1) - 15 > 3(x + 4)\]Expanding the brackets:\[5x + 5 - 15 > 3x + 12\]\[5x - 10 > 3x + 12\]Collect like terms:\[5x - 3x > 12 + 10\]\[2x > 22\]Divide both sides by 2:\[x > 11\]

First, round each component of the expression to two significant figures:
0.02174 becomes 0.022 (the first significant figure is 2, the second is 1, followed by 7, so we round up).
1.2047 becomes 1.2 (the first is 1, second is 2, followed by 0, so it stays 2).
0.023789 becomes 0.024 (first is 2, second is 3, followed by 7, so we round up).
Now evaluate the expression with these rounded values:\[\frac{0.022 \times 1.2}{0.024}\]Multiplying the numerator:\[0.022 \times 1.2 = 0.0264\]Dividing by the denominator:\[\frac{0.0264}{0.024} = 1.1\]

Let the rate of work for a girl be \(g\). According to the question, the rate of work for a boy is \(\frac{5}{4}g\).
The combined rate of 4 boys and 10 girls is:\[4\left(\frac{5}{4}g\right) + 10g = 5g + 10g = 15g\]The total work to be done (cutting the field) is the rate multiplied by the time taken:\[\text{Work} = 15g \times 5 = 75g\]Now, we need to find how many boys (let this be \(B\)) are needed to complete the same work (75g) in 3 hours. The rate of \(B\) boys is \(B \times \frac{5}{4}g\).\[\text{Rate} \times \text{Time} = \text{Work}\]\[\left(B \times \frac{5}{4}g\right) \times 3 = 75g\]Divide both sides by \(g\) and simplify:\[\frac{15}{4}B = 75\]Multiply by 4 and divide by 15:\[15B = 300\]\[B = 20\]

The radius \(r\) of a circle of latitude \(\theta\) on the earth's surface is related to the radius of the earth \(R\) by the formula:\[r = R \cos \theta\]The circumference \(C\) of any circle is given by \(2\pi r\). Therefore, the circumference of the latitude \(\theta\) is:\[C = 2\pi(R \cos \theta) = 2\pi R \cos \theta\]

Let the length \(l = 12\), width \(w = 9\), and height \(h = 8\).
The diagonal of the floor \(d_f\) is found using Pythagoras:\[d_f = \sqrt{l^2 + w^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{m}\]The room diagonal \(D\) is found using the floor diagonal and the height:\[D = \sqrt{d_f^2 + h^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{m}\]The angle \(\alpha\) made by the room diagonal with the floor forms a right-angled triangle where the floor diagonal is the adjacent side and the room diagonal is the hypotenuse.\[\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{d_f}{D} = \frac{15}{17}\]

To find the total number of goals scored by all teams, we multiply each goal value by its corresponding frequency and sum the results (\(\sum fx\)):\[(0 \times 4) + (1 \times 3) + (2 \times 15) + (3 \times 16) + (4 \times 1) + (5 \times 0) + (6 \times 1)\]\[= 0 + 3 + 30 + 48 + 4 + 0 + 6\]\[= 91\]

A pie chart represents data proportional to the angle of each sector relative to the total angle of a circle, which is 360^o. To find the number of students, we use the ratio:\[\text{Number of students} = \frac{\text{Sector angle}}{\text{Total angle}} \times \text{Total number of students}\]\[\text{Physics students} = \frac{12}{360} \times 150\]Simplifying the fraction:\[\frac{12}{360} = \frac{1}{30}\]Now multiply:\[\frac{1}{30} \times 150 = 5\]