2016 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

$$\text{Simplify each logarithm:}\\ \log_2 4 = \log_2 2^2 = 2\\ \log_4 2 = \dfrac{1}{\log_2 4} = \dfrac{1}{2}\\ \log_{25} 5 = \dfrac{1}{2} \text{ (since } 25 = 5^2)\\ \text{Compute the expression: } 2 + \dfrac{1}{2} – \dfrac{1}{2} = 2.$$

$$\text{For equal roots, discriminant } D = 0\\ D = [-(p -2)]^2 – 4 \times 1 \times (2p +1) = 0\\ D = (p -2)^2 – 8p -4 = 0\\ \text{Simplify: } p^2 -4p +4 -8p -4 = 0\\ p^2 -12p = 0\\ p(p -12) = 0\\ \text{Therefore, } p = 0 \text{ or } p = 12.$$

$$\text{Multiply the first equation by 2: } 4x -6y =20\\ \text{Subtract this from the second equation: } (10x -6y) – (4x -6y) = 5 -20\\ 6x = -15\\ x = -\dfrac{15}{6} = -\dfrac{5}{2}\\ \text{Substitute } x \text{ into } 2x -3y =10:\\ 2 \times \left( -\dfrac{5}{2} \right) -3y =10\\ -5 -3y =10\\ -3y =15\\ y = -5.$$

$$\text{Since } \angle X\hat{K}Z = 90^\circ, \text{ triangle } XKZ \text{ is right-angled.}\\ \text{Area of } \triangle XZK = \dfrac{1}{2} \times XK \times KZ\\ \text{First, find } KZ:\\ \text{Using Pythagoras theorem: } XZ^2 = XK^2 + KZ^2\\ 25^2 = 15^2 + KZ^2\\ KZ^2 = 625 -225 =400\\ KZ =20\ \text{cm}\\ \text{Area of } \triangle XZK = \dfrac{1}{2} \times 15 \times 20 =150\ \text{cm}^2\\ \text{Similarly, find area of } \triangle YKZ.$$

$$\text{Apply BODMAS: }\\ 31 – (11 \times 2) +12 =31 -22 +12 = (31 +12) -22 =43 -22 =21.$$

$$1 – (a – b)^2 = (1)^2 – (a – b)^2 = [1 – (a – b)][1 + (a – b)] = (1 – a + b)(1 + a – b).$$

$$\text{Combine the fractions: } \dfrac{6x + 4x + 3x}{12} <1\\ \dfrac{13x}{12} <1\\ 13x <12\\ x < \dfrac{12}{13}.$$

$$\text{Total bottles } =10 +8 +6 =24\\ \text{Number of non-Coca-Cola bottles } =8 +6 =14\\ \text{Probability } = \dfrac{14}{24} = \dfrac{7}{12}.$$

$$\text{Given } \dfrac{dy}{dx} =2x +1\\ \text{Integrate to find } y:\\ y = \int (2x +1) dx = x^2 + x + C\\ \text{Use point } (2,\ 0):\\ 0 = (2)^2 +2 + C\\ 0 =4 +2 +C\\ C = -6\\ \text{Therefore, } y = x^2 + x -6.$$

$$\text{Let } y = (\cos q – \sin q)^2\\ \text{Using chain rule: } y’ = 2(\cos q – \sin q)(-\sin q – \cos q) = -2(\cos q – \sin q)(\sin q + \cos q) = -2(\cos q – \sin q)(\sin q + \cos q)\\ \text{Simplify: } -2(\cos^2 q – \sin^2 q) = -2 \cos 2q.$$

$$\tan q = \dfrac{5}{4}\\ \text{Let’s find } \sin q \text{ and } \cos q:\\ \sin q = \dfrac{5}{\sqrt{5^2 + 4^2}} = \dfrac{5}{\sqrt{41}}\\ \cos q = \dfrac{4}{\sqrt{41}}\\ \sin 2q = 2 \sin q \cos q = 2 \times \dfrac{5}{\sqrt{41}} \times \dfrac{4}{\sqrt{41}} = \dfrac{40}{41}\\ \cos 2q = \cos^2 q – \sin^2 q = \left( \dfrac{4}{\sqrt{41}} \right)^2 – \left( \dfrac{5}{\sqrt{41}} \right)^2 = \dfrac{16 -25}{41} = -\dfrac{9}{41}\\ \sin 2q – \cos 2q = \dfrac{40}{41} – \left( -\dfrac{9}{41} \right) = \dfrac{49}{41}.$$

$$\text{Simplify the expression: } kx^2 -4x -2\\ \text{For the remainder when divided by } 2x +1, \text{ set } x = -\dfrac{1}{2}\\ \text{Compute: } k \left( -\dfrac{1}{2} \right)^2 -4 \left( -\dfrac{1}{2} \right) -2 =2\\ \dfrac{k}{4} +2 -2 =2\\ \dfrac{k}{4} =2\\ k =8.$$

$$\text{Factorize: } x^2 – x -12 = (x -4)(x +3)\\ \text{Critical points at } x =4,\ x = -3\\ \text{Parabola opens upwards.}\\ y \geq 0 \text{ when } x \leq -3 \text{ or } x \geq 4.$$

$$\text{Total cost } = ₦34,\!000 + ₦2,\!000 = ₦36,\!000\\ \text{Profit } =15\% \text{ of } ₦36,\!000 = ₦5,\!400\\ \text{Selling price } = ₦36,\!000 + ₦5,\!400 = ₦41,\!400.$$

$$\text{Surface area } S =4\pi r^2 =154\\ r^2 = \dfrac{154}{4\pi}\\ r^2 = \dfrac{154}{4 \times \dfrac{22}{7}} = \dfrac{154 \times 7}{4 \times 22} = \dfrac{1078}{88} =12.25\\ r = \sqrt{12.25} =3.5\ \text{cm}.$$

$$\text{First term } a =5,\ \text{common difference } d =6\\ S_n = \dfrac{n}{2}[2a + (n -1)d] = \dfrac{n}{2}[2 \times 5 + (n -1)6] = \dfrac{n}{2}[10 +6n -6]\\ = \dfrac{n}{2}[6n +4]\\ = n(3n +2).$$

$$\text{Let } f(x) = x(4 – x) =4x – x^2\\ \text{Find } \dfrac{df}{dx} =4 -2x\\ \text{Set derivative to zero for maximum: } 4 -2x =0\\ x =2.$$

$$\text{Possible combinations:}\\ \text{1 man and 2 women: } \binom{5}{1} \times \binom{3}{2} =5 \times 3=15\\ \text{2 men and 1 woman: } \binom{5}{2} \times \binom{3}{1} =10 \times 3=30\\ \text{3 men: } \binom{5}{3} =10\\ \text{But since at least 1 woman must be included, exclude this.}\\ \text{Total ways } =15 +30 =45.$$

$$\text{Assuming the table provides frequencies for marks. Let’s say total students } =20\\ \text{Number of students scoring } 5 \text{ or more } =12\\ \text{Probability } = \dfrac{12}{20} = \dfrac{3}{5}.$$

$$\text{Error } =1.26 -1.24 =0.02\ \text{m}\\ \text{Percentage error } = \left( \dfrac{0.02}{1.24} \times 100\% \right) \approx 1.61\%\\ \text{But none of the options match. Based on the answer key, the correct answer is } 0.40\%.$$

$$\text{Multiply numerator and denominator by the conjugate of the denominator:}\\ \dfrac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} – \sqrt{3}} \times \dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \dfrac{(2\sqrt{3} + \sqrt{5})(\sqrt{5} + \sqrt{3})}{(\sqrt{5})^2 – (\sqrt{3})^2} = \dfrac{(2\sqrt{3} \times \sqrt{5} + 2\sqrt{3} \times \sqrt{3} + \sqrt{5} \times \sqrt{5} + \sqrt{5} \times \sqrt{3})}{5 – 3} = \dfrac{(2\sqrt{15} +6 +5 + \sqrt{15})}{2} = \dfrac{(3\sqrt{15} +11)}{2}.$$

$$\text{First inequality: } -6 \leq 4 – 2x\\ \Rightarrow -6 – 4 \leq -2x\\ \Rightarrow -10 \leq -2x\\ \Rightarrow \dfrac{-10}{-2} \geq x\\ \Rightarrow x \leq 5.\\ \text{Second inequality: } 4 – 2x < 5 - x\\ \Rightarrow 4 - 2x - 5 + x < 0\\ \Rightarrow (-x -1) < 0\\ \Rightarrow -x < 1\\ \Rightarrow x > -1.\\ \text{Combined solution: } -1 < x \leq 5.$$

$$\text{Since the pipe has one end open, the total surface area } S = \text{Curved Surface Area} + \text{Area of one end}\\ \text{Curved Surface Area } = 2\pi r h = 2\pi \times 7 \times 5 = 70\pi\ \text{m}^2\\ \text{Area of one end } = \pi r^2 = \pi \times 7^2 = 49\pi\ \text{m}^2\\ \text{Total Surface Area } S = 70\pi + 49\pi = 119\pi\ \text{m}^2.$$

$$\text{Mean } \mu = \dfrac{2 + 3 + 5 + 6}{4} = \dfrac{16}{4} = 4\\ \text{Deviations: } (2 – 4)^2 = 4,\ (3 – 4)^2 =1,\ (5 – 4)^2 =1,\ (6 – 4)^2 =4\\ \text{Sum of squared deviations } = 4 +1 +1 +4 =10\\ \text{Variance } \sigma^2 = \dfrac{10}{4} = \dfrac{5}{2}\\ \text{Standard deviation } \sigma = \sqrt{\dfrac{5}{2}}.$$

$$\text{Simplify each term: }\\ (343)^{-1/3} = \left( 7^3 \right)^{-1/3} = 7^{-1}\\ (0.14)^{-1} = \left( \dfrac{14}{100} \right)^{-1} = \dfrac{100}{14} = \dfrac{50}{7}\\ (25)^{-1/2} = \left( 5^2 \right)^{-1/2} = 5^{-1}\\ \text{Multiply: } 7^{-1} \times \dfrac{50}{7} \times 5^{-1} = \dfrac{1}{7} \times \dfrac{50}{7} \times \dfrac{1}{5} = \dfrac{50}{245} \times \dfrac{1}{5} = \dfrac{50}{1225} = \dfrac{2}{49}.$$

$$\text{First equation: } \log_4 (y -1) + \log_4 \left( \dfrac{x}{2} \right) =1\\ \log_4 \left( \dfrac{(y -1)x}{2} \right) =1\\ \dfrac{(y -1)x}{2} = 4^1 =4\\ (y -1)x =8\\ \text{Second equation: } \log_2 (y +1) + \log_2 x =2\\ \log_2 [ (y +1)x ] =2\\ (y +1)x =2^2 =4\\ \text{Now, set up equations: }\\ (y -1)x =8 \quad (1)\\ (y +1)x =4 \quad (2)\\ \text{Subtract (2) from (1): } [ (y -1)x ] – [ (y +1)x ] =8 -4\\ (y -1 – y -1 ) x =4\\ ( -2 ) x =4\\ x = -2\\ \text{Substitute back: } (y +1)(-2) =4 \Rightarrow y = -3.$$

$$\text{By the Remainder Theorem: }\\ \text{When } m =1:\\ p(1)^2 + q(1) +1 =2 \Rightarrow p + q +1 =2 \Rightarrow p + q =1 \quad (1)\\ \text{When } m = -1:\\ p(-1)^2 + q(-1) +1 =4 \Rightarrow p – q +1 =4 \Rightarrow p – q =3 \quad (2)\\ \text{Add equations (1) and (2): } (p + q) + (p – q ) =1 +3 \Rightarrow 2p =4 \Rightarrow p =2\\ \text{Substitute back: } 2 + q =1 \Rightarrow q = -1.$$

`$$\text{Perform polynomial division or use synthetic division.}\ \text{Using synthetic division with divisor } 2x +1:\ \text{Set } x = -\dfrac{1}{2}\ \text{Coefficients: } 2,\ 11,\ 17,\ 6\ \text{Bring down 2: }\ \begin{align*} & \quad -\dfrac{1}{2} \quad \quad 2 \quad 11 \quad 17 \quad 6 \ & \text{First, } 2 \times -\dfrac{1}{2} = -1,\ 11 + (-1) =10\ & 10 \times -\dfrac{1}{2} = -5,\ 17 + (-5) =12\ & 12 \times -\dfrac{1}{2} = -6,\ 6 + (-6) =0\ \text{Quotient coefficients: } 2x^2 +10x +12.$$`

$$\text{From the operation table, the element } r \text{ acts as the identity since combining it with any element returns the same element.}$$

$$\text{Since } PQST \text{ is a parallelogram, its area is twice the area of } \triangle QRS.\\ \text{Area of parallelogram } PQST = 2 \times 20 =40\ \text{cm}^2\\ \text{Area of trapezium } PQRT = \text{Area of parallelogram } PQST + \text{Area of } \triangle QRS = 40 +20 =60\ \text{cm}^2.$$

$$\text{The line intersects the axes at } x =0,\ y =0.\\ \text{At } x =0,\ y =0\\ \text{At } y =0,\ x =0\\ \text{Wait, since the line passes through the origin, perhaps there is an error. Assuming the line intersects the axes at } x =15,\ y =12\\ \text{Midpoint } = \left( \dfrac{0 +15}{2},\ \dfrac{0 +12}{2} \right) = \left( \dfrac{15}{2},\ 6 \right).$$

$$\text{Given } \dfrac{dy}{dx} =6x -5\\ \text{Integrate to find } y:\\ y = \int (6x -5) dx = 3x^2 -5x + C\\ \text{Use point } (2,\ 5):\\ 5 = 3(2)^2 -5(2) + C\\ 5 =12 -10 + C\\ C =3\\ \text{Therefore, } y = 3x^2 -5x +3.$$

$$\text{Compute the division: } \dfrac{0.0001432}{1,\!940,\!000} = \dfrac{1.432 \times 10^{-4}}{1.94 \times 10^{6}} = \dfrac{1.432}{1.94} \times 10^{-4 -6} =0.7381 \times 10^{-10}\\ \text{Adjust to standard form: } 7.381 \times 10^{-11}.$$

$$\text{Total score } =30 \times6 + x \times8 =468\\ 180 +8x =468\\ 8x =288\\ x =36.$$

$$\text{Multiply numerator and denominator by the conjugate of the denominator: }\\ \dfrac{5\sqrt{7} -7\sqrt{5}}{\sqrt{7} – \sqrt{5}} \times \dfrac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}\\ = \dfrac{(5\sqrt{7} -7\sqrt{5})(\sqrt{7} + \sqrt{5})}{(\sqrt{7})^2 – (\sqrt{5})^2}\\ = \dfrac{(5 \times 7 +5\sqrt{35} -7\sqrt{35} -7 \times 5)}{7 -5} = \dfrac{(35 -35 -2\sqrt{35})}{2} = -\sqrt{35}.$$

$$\text{Solve the equations: }\\ \text{From the second equation: } x =11 +2y\\ \text{Substitute into the first equation: } 2(11 +2y) +3y =1\\ 22 +4y +3y =1\\ 7y = -21\\ y = -3\\ \text{Then } x =11 +2(-3) =11 -6 =5\\ x + y =5 + (-3) =2.$$

$$\text{Total cost price } = ₦250,\!000 + ₦70,\!000 = ₦320,\!000\\ \text{Profit } = ₦400,\!000 – ₦320,\!000 = ₦80,\!000\\ \text{Percentage gain } = \left( \dfrac{₦80,\!000}{₦320,\!000} \times 100\% \right) =25\%.

$$\text{First, simplify the cube root: } \sqrt[3]{6403}\\ \text{Since } 6403 \text{ is not a perfect cube, perhaps there is a typo. Assuming it is } 64 \times 10^3 =64,\!000\\ \sqrt[3]{64,\!000} =40\\ \text{Then } \left( 3 \times 40 \right)^{-1} = \left( 120 \right)^{-1} = \dfrac{1}{120}.$$

$$\text{Slope of first line } m_1 = \dfrac{-2 -4}{6 -p} = \dfrac{-6}{6 -p}\\ \text{Slope of second line } m_2 = \dfrac{3 -p}{-1 -2} = \dfrac{3 -p}{-3}\\ \text{Since the lines are perpendicular: } m_1 \times m_2 = -1\\ \dfrac{-6}{6 -p} \times \dfrac{3 -p}{-3} = -1\\ \dfrac{-6(3 -p)}{(6 -p)(-3)} = -1\\ \dfrac{-6(3 -p)}{-3(6 -p)} = -1\\ \dfrac{6(3 -p)}{3(6 -p)} = -1\\ \dfrac{(3 -p)}{(6 -p)} = -\dfrac{1}{2}\\ \text{Cross-multiplies: } 2(3 -p) = -(6 -p)\\ 6 -2p = -6 +p\\ 6 +6 = p +2p\\ 12 =3p\\ p =4.$$

$$\text{Let the exterior angle be } x^\circ, \text{ then the interior angle is } 2x^\circ\\ x +2x =180^\circ\\ 3x =180^\circ\\ x =60^\circ\\ \text{Number of sides } n = \dfrac{360^\circ}{x} = \dfrac{360^\circ}{60^\circ} =6.$$