Mathematics 2021 Jamb Examination

We use the simple interest formula to find the rate:\[I = \frac{P \times R \times T}{100}\]Given that \(P = 100\), \(T = 5\), and \(I = 7.50\):\[7.50 = \frac{100 \times R \times 5}{100}\]\[7.50 = 5R\]\[R = \frac{7.50}{5} = 1.5\%\]The rate of interest per annum is 1.5%.

Let the total number of mangoes be 1.
The first child takes \(\frac{1}{4}\), so the remainder is:\[1 - \frac{1}{4} = \frac{3}{4}\]The second child takes \(\frac{3}{4}\) of this remainder:\[\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\]To find what the third child takes, subtract the shares of the first and second children from the total:\[1 - \left( \frac{1}{4} + \frac{9}{16} \right)\]\[1 - \left( \frac{4 + 9}{16} \right) = 1 - \frac{13}{16} = \frac{3}{16}\]

Convert each number to standard form first:\[\frac{(2.75 \times 10^{-3}) \times (6.4 \times 10^{-3})}{(2.5 \times 10^{-2}) \times (8 \times 10^{-2})}\]\[= \frac{17.6 \times 10^{-6}}{20 \times 10^{-4}}\]\[= \frac{17.6}{20} \times 10^{-6 - (-4)}\]\[= 0.88 \times 10^{-2}\]To express in correct standard form (one non-zero digit before the decimal):\[= 8.8 \times 10^{-3}\]

Simplify each term in the expression:
For the first term:\[\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\]For the second term, rationalise the denominator by multiplying by \(\frac{\sqrt{3}}{\sqrt{3}}\):\[\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}\]Now add the two results together:\[3\sqrt{3} + \sqrt{3} = 4\sqrt{3}\]

Let the total profit be \(x\).
The first brother gets \(\frac{1}{3}x\). The remainder is \(x - \frac{1}{3}x = \frac{2}{3}x\).
The second brother gets \(\frac{2}{3}\) of that remainder:\[\frac{2}{3} \times \frac{2}{3}x = \frac{4}{9}x\]The third brother receives the rest:\[x - \left( \frac{1}{3}x + \frac{4}{9}x \right) = 12000\]\[x - \left( \frac{3x + 4x}{9} \right) = 12000\]\[x - \frac{7x}{9} = 12000\]\[\frac{2x}{9} = 12000\]\[2x = 108000\]\[x = 54000\]The total profit shared was ₦54,000.00.

First, find the length of the diameter using the distance formula between P and Q:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]\[d = \sqrt{(6 - (-6))^2 + (6 - 1)^2}\]\[d = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ units}\]The radius is half the diameter:\[r = \frac{13}{2} = 6.5 \text{ units}\]

The perimeter of a sector is given by: \(2r + \text{Arc Length}\).
First, calculate the arc length \(L\):\[L = \frac{\theta}{360} \times 2\pi r\]\[L = \frac{48}{360} \times 2 \times \frac{22}{7} \times 10.5\]\[L = \frac{2}{15} \times 2 \times 22 \times 1.5 = 8.8 \text{ cm}\]Now add the two radii to the arc length:\[\text{Perimeter} = 2(10.5) + 8.8 = 21 + 8.8 = 29.8 \text{ cm}\]

In a rhombus, the diagonals bisect each other at right angles (90°). This forms four right-angled triangles inside the rhombus.
The legs of these triangles are half the lengths of the diagonals: \(\frac{6}{2} = 3\) cm and \(\frac{8}{2} = 4\) cm.
Using Pythagoras' theorem to find the side (hypotenuse) \(s\):\[s^2 = 3^2 + 4^2\]\[s^2 = 9 + 16 = 25\]\[s = \sqrt{25} = 5 \text{ cm}\]

It is easier to find the number of sides using the exterior angle. The sum of an interior and its corresponding exterior angle is 180°.
Exterior angle = \(180^o - 140^o = 40^o\).
The sum of exterior angles of any regular polygon is 360°. To find the number of sides (n):\[n = \frac{360^o}{\text{Exterior angle}}\]\[n = \frac{360}{40} = 9\]The polygon has 9 sides.

The volume of metal is the difference between the external volume and the internal volume of the cylinder.
Internal radius \(r = 10\) cm.
Thickness = 3 cm, so External radius \(R = 10 + 3 = 13\) cm.
Height (length) of the pipe \(h = 3\) m = 300 cm.
Volume of metal = \(\pi R^2h - \pi r^2h = \pi h(R^2 - r^2)\):\[\text{Volume} = \pi \times 300 \times (13^2 - 10^2)\]\[\text{Volume} = 300\pi \times (169 - 100)\]\[\text{Volume} = 300\pi \times 69 = 20700\pi \text{ cm}^3\]