2019 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

$$\text{Possible outcomes for at least 10: }(4,6),(5,5),(6,4). \text{ Total }=3.$$ \text{Total possible outcomes }=36.$$ \text{Probability }= \frac{3}{36}=\frac{1}{12}.$$

$$Possible pairs: (1,2)=3, (1,3)=4, (1,4)=5, (2,3)=5, (2,4)=6, (3,4)=7.$$ \text{Even sums: }4 \text{ and }6.$$ \text{Total favorable pairs }=2.$$ \text{Total possible pairs }=6.$$ \text{Probability }= \frac{2}{6}=\frac{1}{3}.$$

$$\text{Scores: }0,1,2,3,4,5,6$$ \text{Frequency: }7,11,6,7,5,3,0.$$ \text{Mode is the highest frequency: }1 \text{ with frequency }11.$$ \text{Median: Arrange the data in order and find the middle value. Total frequencies }=7+11+6+7+5+3=39.$$ \text{The 20th value lies in score }1.$$

$$\text{Total angle }=360^\circ. \text{ Number of students offering Physics }= \frac{12}{360} \times150=5.$$

$$Volume of hemisphere: \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (3)^3 =18\pi.$$ \text{Volume of cone: } \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 \times6=18\pi.$$ \text{Total volume }=18\pi +18\pi=36\pi.$$

$$\text{External radius } R =6 \text{ cm}, \text{ internal radius } r = R -1=5 \text{ cm}. \text{ Volume } = \pi (R^2 – r^2) \times \text{length} =\pi (36 -25) \times10 = \pi \times11 \times10=110\pi \text{ cm}^3.$$

$$Using the Law of Cosines to find PR: PR^2 = PQ^2 + QR^2 – 2 \times PQ \times QR \times \cos120^\circ =1 +4 -2 \times1 \times2 \times (-0.5) =5 +2 =7 \Rightarrow PR= \sqrt{7}.$$

$$\cot q = \frac{x}{y} = \frac{\cos q}{\sin q} \Rightarrow \sin q = \frac{y}{\sqrt{x^2 + y^2}}.$$ \text{Thus, } \csc q = \frac{1}{\sin q} = \frac{\sqrt{x^2 + y^2}}{y}.$$

$$Using the Law of Sines or Cosines in navigation bearings to calculate the bearing of Q from P as 52°8′.$$

$$Let height of tree be h. \text{ Distance from point } =14 \text{ meters. \tan 30^\circ = \frac{h}{14} \Rightarrow h=14 \times \frac{\sqrt{3}}{3} \approx8.08 \text{ meters. Adding 2 meters gives } h=10.08 \text{ meters}.$$

$$Triangles involving circle properties and isosceles triangles. \angle ZYQ =135^\circ.$$

$$\text{Sum of interior angles } = (2k +1 -2) \times 180 = (2k -1) \times 180.$$ \text{Each interior angle } =140 \Rightarrow \frac{(2k -1) \times180}{2k +1} =140.$$ \text{Solving: } (2k -1) \times180 =140(2k +1) \Rightarrow 360k -180 =280k +140 \Rightarrow80k=320 \Rightarrowk=4.$$

$$Since } PS = QS, \text{ triangle } QPS \text{ is isosceles. \angle QSR =50^\circ \Rightarrow \angle QPS = \frac{180 -50}{2} =65^\circ.$$

$$\text{Perimeter } = \text{arc length} + 2 \times \text{radius}.$$ \text{Arc length } = \frac{48}{360} \times 2\pi \times 10.5 = \frac{48}{360} \times 21\pi = \frac{7}{5} \pi.$$ \text{Thus, perimeter } = \frac{7}{5} \pi + 2 \times10.5 = \frac{7}{5} \pi +21.$$

$$\text{Area of rectangle } PQRS = PS \times QR = 73.5 \text{ cm}^2.$$ \text{Given } QR = \frac{1}{3} PS, \text{ so } PS \times \frac{1}{3} PS = 73.5 \Rightarrow PS^2 = 73.5 \times 3 = 220.5 \Rightarrow PS = \sqrt{220.5} \approx 14.86 \text{ cm}.$$

$$\text{Vertex form: } x = -\frac{b}{2a} = -\frac{1}{2}.$$

$$Assuming the graph passes through points corresponding to the quadratic equation } y = x^2 + x -2.$$

$$This is a geometric progression with first term } a=3, r=2.$$ \text{Sum } S_{18}=3(2^{18} -1).$$

$$Total cost: #40d + #10(d-10) = 2000 \Rightarrow40d +10d -100 =2000 \Rightarrow50d=2100 \Rightarrowd=42.$$ \text{Assistant worked }42-10=32 \text{ days. Earnings: }10 \times32=#320.$$

$$This is a geometric progression with first term } \log a \text{ and common ratio } \log a.$$ \text{Sum } S_{20} = \log a \times \frac{1 – (\log a)^{20}}{1 – \log a}.$$

$$7^2 +3^2 +z^2=194 \Rightarrow 49 +9 +z^2=194 \Rightarrow z^2=136 \Rightarrow z= \sqrt{136} \approx 11.66.$$

$$\frac{x}{x+y} + \frac{y}{x-y} – \frac{x^2}{(x+y)(x-y)} = \frac{x(x-y) + y(x+y) -x^2}{(x+y)(x-y)} = \frac{x^2 -xy + xy + y^2 -x^2}{(x+y)(x-y)} = \frac{y^2}{(x+y)(x-y)} = \frac{y^2}{x^2 – y^2}.$$

$$Let length=l and width=w. \ l + w =12, \ l \times w=35.$$ \text{Solving } l=12-w, \ (12-w)w=35 \Rightarrow w^2 -12w +35=0 \Rightarrow (w-5)(w-7)=0 \Rightarrow w=5 \text{ or }7.$$

$$Assuming the hypotenuse is } (3x +1), \text{ then } x^2 + (3x -1)^2 = (3x +1)^2.$$ \text{Expand: } x^2 +9x^2 -6x +1 =9x^2 +6x +1.$$ \text{Simplify: }10x^2 -6x +1 =9x^2 +6x +1 \Rightarrow x^2 -12x=0 \Rightarrow x(x -12)=0 \Rightarrow x=12.$$

$$x -120 =0 \Rightarrow x=120.$$

$$4x^2 -4 +1 = (2x)^2 -2^2 +1 = (2x)^2 -1.$$ \text{To make it a perfect square, add 1: } 4x^2 -4 +1 =4x^2 -3 \text{ (Not a perfect square).} \text{Alternatively, } 4x^2 -4 =4(x^2 -1)=4(x-1)(x+1). \text{ To make } 4x^2 -4 \text{ a perfect square, add }1 \text{ to get } (2x)^2 -4x +1=(2x-1)^2.$$`

$$2 + \frac{1}{4} = 2.25.$$

$$\text{Using synthetic division: } (x^3 – 2x^2 -5x +6) \div (x -1) = x^2 -x -6.$$

$$y^3 + y^2 -4xy -4y = y^2(y + 1) -4y(x +1) = y(y+1)(y -4x -4).$$

$$f(x-4) = x^2 + 2x + 3. \text{ Let } x-4 = t \Rightarrow x = t + 4.$$ \text{Thus, } f(t) = (t+4)^2 + 2(t+4) + 3 = t^2 + 8t + 16 + 2t + 8 + 3 = t^2 + 10t + 27.$$ \text{Therefore, } f(2) = 2^2 + 10 \times 2 + 27 = 4 + 20 + 27 = 51.$$

$$P = \frac{r}{r + q} \Rightarrow P(r + q) = r \Rightarrow Pr + Pq = r \Rightarrow Pr – r = -Pq \Rightarrow r(P – 1) = -Pq \Rightarrow r = \frac{Pq}{1 – P}.$$

$$Y \propto \frac{1}{x^2} \text{ and } X \propto Z^2. \text{ If constants are involved, the relationship simplifies to } Y = \frac{C}{Z}.$$

$$ab^2 = 2 \times (-2)^2 = 8, \ bc^2 = -2 \times \left(-\frac{1}{2}\right)^2 = -0.5. \ ab^2 – bc^2 = 8 – (-0.5) = 8.5.$$ \text{Similarly, } a^2c = 4 \times -\frac{1}{2} = -2, \ abc = 2 \times (-2) \times \left(-\frac{1}{2}\right) = 2.$$ \text{Thus, } a^2c – abc = -2 – 2 = -4.$$ \text{Product } = 8.5 \times (-4) = -34.$$

$$\left(1 + 1\right)^{-1} = 2^{-1} = \frac{1}{2}.$$

$$\text{Assuming all logs are base 10: } 3 \log 9 = \log 9^3 = \log 729, \log 64 = \log 2^6 = 6 \log 2, \log 72 = \log (8 \times 9) = \log 8 + \log 9.$$

$$\sqrt{27 + 1} = \sqrt{28} \approx 5.2915.$$

$$\text{Assuming } v \text{ cancels out, simplify to } \frac{13}{r}.$$

$$\text{Let total profit be } P. \text{ First brother gets } \frac{P}{3}, \text{ remaining } \frac{2P}{3}. \text{ Second gets } \frac{2}{3} \times \frac{2P}{3} = \frac{4P}{9}. \text{ Remaining for third: } P – \frac{P}{3} – \frac{4P}{9} = \frac{2P}{9} = #12,000. \text{ Thus, } P = #54,000.$$

$$\frac{0.00275 \times 0.00640}{0.025 \times 0.08} = \frac{0.0000176}{0.002} = 0.0088 = 8.8 \times 10^{-3}.$$

$$\frac{0.24633}{0.0306} \approx 8.05 \approx 8.0 \text{ (to one decimal place)}.$$

`$$\text{Percentage error} = \left

$$\text{Prime numbers between 1 and 6 are } 2, 3, 5. \text{ So, } x = 2 + 3 + 5 = 10. \text{ H.C.F. of } 6, 9, 15 \text{ is } 3. \text{ Thus, } x \times y = 10 \times 3 = 30.$$

$$\text{Simplify the expression step-by-step: } 2 – \frac{1}{32} = \frac{63}{32}, \text{ then } 1 + \frac{32}{63} = \frac{95}{63}. \text{ Multiply by } \frac{2}{4} \text{ to get } \frac{95}{126} \approx 0.754.$$

$$\text{Possible outcomes for at least 10: }(4,6),(5,5),(6,4). \text{ Total }=3.$$ \text{Total possible outcomes }=36.$$ \text{Probability }= \frac{3}{36}=\frac{1}{12}.$$

$$Possible pairs: (1,2)=3, (1,3)=4, (1,4)=5, (2,3)=5, (2,4)=6, (3,4)=7.$$ \text{Even sums: }4 \text{ and }6.$$ \text{Total favorable pairs }=2.$$ \text{Total possible pairs }=6.$$ \text{Probability }= \frac{2}{6}=\frac{1}{3}.$$

$$\text{Scores: }0,1,2,3,4,5,6$$ \text{Frequency: }7,11,6,7,5,3,0.$$ \text{Mode is the highest frequency: }1 \text{ with frequency }11.$$ \text{Median: Arrange the data in order and find the middle value. Total frequencies }=7+11+6+7+5+3=39.$$ \text{The 20th value lies in score }1.$$

$$\text{Total angle }=360^\circ. \text{ Number of students offering Physics }= \frac{12}{360} \times150=5.$$

$$Volume of hemisphere: \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (3)^3 =18\pi.$$ \text{Volume of cone: } \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 \times6=18\pi.$$ \text{Total volume }=18\pi +18\pi=36\pi.$$

$$\text{External radius } R =6 \text{ cm}, \text{ internal radius } r = R -1=5 \text{ cm}. \text{ Volume } = \pi (R^2 – r^2) \times \text{length} =\pi (36 -25) \times10 = \pi \times11 \times10=110\pi \text{ cm}^3.$$

$$Using the Law of Cosines to find PR: PR^2 = PQ^2 + QR^2 – 2 \times PQ \times QR \times \cos120^\circ =1 +4 -2 \times1 \times2 \times (-0.5) =5 +2 =7 \Rightarrow PR= \sqrt{7}.$$