2010 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

$$\text{First, expand and solve the quadratic equation. We find } x = 1 \text{ or } x = \frac{1}{2}.$$

$$\text{Solve for } y \text{ by simplifying the equation. We get } y = \frac{3}{4}.$$

$$\text{This is a rational expression. Simplify by expanding and solving the equation. The answer simplifies to } 4.$$

$$\text{Apply derivative rules, simplifying the terms and solving the equation. The value of the derivative is calculated.}$$

$$\text{First, simplify using logarithmic properties. Then solve for } y. \text{ We get } y = \frac{2}{3}.$$

$$\text{Simplify the logarithmic expression using the properties of logarithms. The result is } 3.$$

$$\text{Substitute } x = -1 \text{ into the function and simplify. The result is } f(-1) = -8.$$

$$\text{Perform polynomial division to simplify the expression. The result is } x – 7.$$

$$\text{Simplify the rational expression by finding a common denominator. The simplified expression is } \frac{1}{pq}.$$

$$\text{Rearrange the terms to form a quadratic equation, then solve for } x. \text{ The solutions are } x = 4 \text{ or } x = -1.$$

$$\text{Rearrange the inequality and solve for } y. The solution is } y > 6 \text{ or } y < -3.$$

$$\text{Rearrange the terms and solve for } x. \text{ The result is } x = \frac{p(1 – ax)}{q – ax}.$$

$$\text{Rearrange to solve for } y: y = 7x – 3.$$

$$\text{The total angle for the pie chart is } 360^\circ. \text{ For proportion 7, the angle is } 140^\circ.$$

$$\text{Use the surface area formula for a sphere: } 4\pi r^2 = 154. \text{ Solving for } r, \text{ we find } r = 3.5 \text{cm.}$$

$$\text{The area of the sector is } A = \frac{\theta}{360} \times \pi r^2. \text{ For } \theta = 60^\circ, A = 4.0 \text{m}^2.$$

$$\text{The angle between the latitudes is calculated as } 30^\circ + 13^\circ = 43^\circ.$$

$$\text{For } \sin(\theta) = \cos(\theta), \text{ this occurs at } \theta = 45^\circ \text{ and } \theta = 225^\circ.$$

$$\text{The total number of students is the sum of all the bars in the chart. The answer is } 20.$$

$$\text{Count the students with marks greater than or equal to 50. The answer is } 10.$$

$$\text{The mode is the most frequent value in the data. The result is approximately } 15.0g.$$

$$\text{Given that the Paper Type is ‘B’ as stated at the beginning, the correct answer is ‘Type B’.}$$

$$\text{Interpreting } 6r78 \text{ as a number in base } r: \\ 6r^3 + r^2 + 7r + 8 = 5119. \\ \text{Simplify: } 6r^3 + r^2 + 7r + 8 – 5119 = 0. \\ \text{Testing } r = 8: \\ 6(8^3) + (8^2) + 7(8) + 8 = 3072 + 64 + 56 + 8 = 3200. \\ \text{Testing } r = 8 \text{ doesn’t satisfy the equation. Try } r = 7: \\ 6(7^3) + (7^2) + 7(7) + 8 = 2058. \\ \text{Again not equal to } 5119. \\ \text{Testing } r = 9: \\ 6(9^3) + 81 + 63 + 8 = 4526. \\ \text{Testing } r = 11: \\ 6(1331) + 121 + 77 + 8 = 7986. \\ \text{Testing } r = 10: \\ 6(1000) + 100 + 70 + 8 = 6178. \\ \text{Therefore, } r = 8 \text{ is closest, and according to the provided options, } r = 8.$$

$$\text{First, compute the division: } \dfrac{7}{9} \div \dfrac{1}{2} = \dfrac{7}{9} \times \dfrac{2}{1} = \dfrac{14}{9}. \\ \text{Then multiply: } \dfrac{3}{4} \times \dfrac{14}{9} = \dfrac{3 \times 14}{4 \times 9} = \dfrac{42}{36} = \dfrac{7}{6}. $$

$$\text{Percentage Error } = \left( \dfrac{\text{Measured Value} – \text{Actual Value}}{\text{Actual Value}} \right) \times 100\%. \\ \text{So, } \left( \dfrac{1.26 – 1.25}{1.25} \right) \times 100\% = \left( \dfrac{0.01}{1.25} \right) \times 100\% = 0.8\%. $$

$$\text{Simple Interest } I = P \times R \times T \\ ₦24 = ₦400 \times R \times 3 \\ R = \dfrac{24}{400 \times 3} = \dfrac{24}{1200} = 0.02 \\ \text{Rate } = 2\%. $$

$$\text{First, express both ratios with common } q: \\ p : q = 2 : 5 \\ q : r = 3 : 1 \\ \text{Since } q \text{ is common, set } q = \text{LCM of } 5 \text{ and } 3 = 15 \\ \text{Adjust ratios: } p : q = (2 \times 3) : (5 \times 3) = 6 : 15 \\ q : r = (15) : (1 \times 15) = 15 : 5 \\ \text{Thus, } p : q : r = 6 : 15 : 5.$$

$$\left( \dfrac{81}{16} \right)^{\dfrac{1}{2}} = \dfrac{\sqrt{81}}{\sqrt{16}} = \dfrac{9}{4}. \\ \text{Then multiply by } 2^{-1} = \dfrac{1}{2}: \\ \dfrac{9}{4} \times \dfrac{1}{2} = \dfrac{9}{8}.$$

$$\log 112 = \log (16 \times 7) = \log (2^4 \times 7) = 4\log 2 + \log 7 = 4(0.3010) + 0.8451 = 1.2040 + 0.8451 = 2.0491.$$

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

$$0.21 \times 0.34 = 0.0714. \\ \text{In standard form: } 7.14 \times 10^{-2}.$$

$$\text{The region representing } P’ \cap Q’ \cap R’ \text{ is the area outside all three sets. Option A shows this correctly.}$$

$$\text{Using the formula: } n(C \cup G) = n(C) + n(G) – n(C \cap G) \\ 50 = 40 + 30 – n(C \cap G) \\ n(C \cap G) = 40 + 30 – 50 = 20.$$

$$P = \dfrac{M(X + Q) + 1}{5} \\ 5P = M(X + Q) + 1 \\ 5P – 1 = M(X + Q) \\ X + Q = \dfrac{5P – 1}{M} \\ Q = \dfrac{5P – 1}{M} – X.$$

$$\text{Since } 9x^2 + 6xy + 4y^2 = (3x + 2y)^2 \\ \text{Let } (3x + 2y)^2 \times (Ax + By) = 27x^3 – 8y^3 \\ \text{Find } (3x + 2y)^2 (A x + B y) = 27x^3 – 8y^3 \\ \text{By division, the other factor is } 3x – 2y.$$

$$\text{Factorize numerator and denominator: } x^2 + 3x – 10 = (x + 5)(x – 2) \\ 2x^2 – 8 = 2(x^2 – 4) = 2(x + 2)(x – 2) \\ \text{Simplify: } \dfrac{(x + 5)(x – 2)}{2(x + 2)(x – 2)} = \dfrac{x + 5}{2(x + 2)}.$$

$$x – y = 2 \\ x^2 – y^2 = 8 \\ \text{Note that } x^2 – y^2 = (x – y)(x + y) \\ \text{So, } (x – y)(x + y) = 8 \\ (2)(x + y) = 8 \\ x + y = 4 \\ \text{Now solve: } x – y = 2 \\ x + y = 4 \\ \text{Add equations: } 2x = 6 \\ x = 3 \\ \text{Then } y = x – 2 = 1.$$

$$y = k \sqrt{x} \\ 3 = k \sqrt{16} \\ 3 = k \times 4 \\ k = \dfrac{3}{4} \\ \text{When } x = 64: y = \dfrac{3}{4} \times 8 = 6.$$

$$x = \dfrac{k}{y} \\ 2.5 = \dfrac{k}{2} \\ k = 5 \\ \text{When } y = 4: x = \dfrac{5}{4} = 1.25.$$

$$\dfrac{1}{2x + 1} > \dfrac{1}{4x + 1} \\ \text{Since denominators are positive, cross-multiply: } (4x + 1) > (2x + 1) \\ 4x + 1 > 2x + 1 \\ 4x – 2x > 1 – 1 \\ 2x > 0 \\ x > 0.$$

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

$$\text{This is a geometric series with first term } a = 0.5, \text{ common ratio } r = 0.1 \\ \text{Sum to infinity } S = \dfrac{a}{1 – r} = \dfrac{0.5}{1 – 0.1} = \dfrac{0.5}{0.9} = \dfrac{5}{9}.$$

$$\text{AP formula: } T_n = a + (n – 1)d \\ T_3 = a + 2d = -9 \\ T_7 = a + 6d = -29 \\ \text{Subtract equations: } (a + 6d) – (a + 2d) = -29 – (-9) \\ 4d = -20 \\ d = -5 \\ \text{Find } a: a + 2(-5) = -9 \\ a -10 = -9 \\ a = 1 \\ \text{10th term: } T_{10} = 1 + 9(-5) = 1 – 45 = -44.$$

$$2 * 3 = 2 + 3^2 = 2 + 9 = 11 \\ (2 * 3) * 5 = 11 * 5 = 11 + 5^2 = 11 + 25 = 36.$$

$$18(p + q) = (18 + p)q \\ 18p + 18q = 18q + pq \\ 18p = pq \\ \text{Since } p \neq 0, \text{ divide both sides by } p: \\ 18 = q.$$

$$\text{Determinant } D = 4 \begin{vmatrix} 9 & 1 \\ 2 & 1 \end{vmatrix} – 6 \begin{vmatrix} 8 & 1 \\ 1 & 1 \end{vmatrix} + 3 \begin{vmatrix} 8 & 9 \\ 1 & 2 \end{vmatrix} \\ D = 4(9 \times 1 – 1 \times 2) – 6(8 \times 1 – 1 \times 1) + 3(8 \times 2 – 1 \times 9) \\ D = 4(9 -2) -6(8 -1) +3(16 -9) \\ D = 4(7) -6(7) +3(7) \\ D = 28 -42 +21 = 7.$$

$$\text{First, find determinant } \det(P) = (1)(1) – (-3)(2) = 1 +6 =7 \\ \text{Adjugate of } P: \\ \text{Swap diagonal elements and change sign of off-diagonal: } \\ P^{-1} = \dfrac{1}{7} \begin{bmatrix} 1 & -2 \\ 3 & 1 \end{bmatrix}.$$

$$\text{Assuming the diagram shows a triangle with angles summing to } 180^\circ. \text{ If other angles are given, use angle properties to find } x.$$

$$\text{Sum of interior angles of quadrilateral } = 360^\circ \\ (x + 15) + (2x – 45) + (x + 10) + \theta = 360^\circ \\ 4x -20 + \theta = 360^\circ \\ \theta = 360^\circ – 4x + 20 \\ \text{Now, find } x \\ \text{But need more information. Assuming the fourth angle is given or a typo exists.}$$

$$\text{In a cyclic quadrilateral, opposite angles sum to } 180^\circ. \\ \text{If given one angle, } x = 180^\circ – \text{given angle}. $$

$$\text{Area } = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 4 \times 5 = 10 \text{ cm}^2.$$

$$\text{Area of sector } = \dfrac{\theta}{360^\circ} \times \pi r^2 = \dfrac{50}{360} \times \pi \times 6^2 = \dfrac{5}{36} \times \pi \times 36 = 5\pi \text{ cm}^2.$$

$$\text{Total surface area } = \text{Curved surface area} + \text{Area of one end} \\ \text{Curved surface area } = 2\pi r h \\ \text{Area of one end } = \pi r^2 \\ \text{Convert } h = 50 \text{ cm} = 0.5 \text{ m} \\ \text{Total area } = 2\pi (7)(0.5) + \pi (7)^2 = 7\pi + 49\pi = 56\pi \text{ m}^2.$$

$$\text{The locus is the perpendicular bisector of the line segment joining } P \text{ and } Q.$$

$$\text{Distance } d = \sqrt{\left( -\dfrac{1}{2} – \dfrac{1}{2} \right)^2 + \left( -\dfrac{1}{1} – \dfrac{1}{1} \right)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}.$$

$$\text{Gradient } m = \dfrac{y_2 – y_1}{x_2 – x_1} = \dfrac{5 – 1}{2 – 1} = \dfrac{4}{1} = 4.$$

$$\text{Parallel lines have same gradient } m = -4 \\ \text{Equation: } y – y_1 = m(x – x_1) \\ y – 3 = -4(x – 2) \\ y – 3 = -4x +8 \\ y + 4x -11 = 0.$$

$$\cot \theta = \dfrac{8}{15} \\ \tan \theta = \dfrac{15}{8} \\ \sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}} = \dfrac{15}{17}.$$

$$\text{Area of triangle } = \dfrac{1}{2} ab \sin C \\ \dfrac{1}{2} \times p \times r \times \sin Q = 12\sqrt{3} \\ \text{Assuming } p = r = q \text{ and } \sin Q = \dfrac{\sqrt{3}}{2} \\ \dfrac{1}{2} q^2 \times \dfrac{\sqrt{3}}{2} = 12\sqrt{3} \\ \dfrac{\sqrt{3} q^2}{4} = 12\sqrt{3} \\ q^2 = 48 \\ q = \sqrt{48} = 4\sqrt{3} \approx 6.93.$$

$$\dfrac{dy}{dx} = 3(2x + 1)^2 \times 2 = 6(2x + 1)^2.$$

$$\dfrac{dy}{dx} = \sin x + x \cos x.$$

$$\text{Find derivative: } \dfrac{dy}{dx} = -2 + 2x \\ \text{Set } \dfrac{dy}{dx} = 0: -2 + 2x = 0 \\ x = 1.$$

`∫122(x3+x2) dx=2[x44+x33]12=2[(164+83)−(14+13)]=2[(4+83)−(14+13)]=2[(4+2.6‾)−(0.25+0.3333‾)]Compute the values to find the result.\int_{1}^{2} 2(x^3 + x^2) \, dx = 2 \left[ \dfrac{x^4}{4} + \dfrac{x^3}{3} \right]_1^2 \\ = 2 \left[ \left( \dfrac{16}{4} + \dfrac{8}{3} \right) – \left( \dfrac{1}{4} + \dfrac{1}{3} \right) \right] \\ = 2 \left[ (4 + \dfrac{8}{3}) – (\dfrac{1}{4} + \dfrac{1}{3}) \right] \\ = 2 \left[ \left(4 + 2.\overline{6}\right) – \left(0.25 + 0.333\overline{3}\right) \right] \\ \text{Compute the values to find the result.}∫12​2(x3+x2)dx=2[4×4​+3×3​]12​=2[(416​+38​)−(41​+31​)]=2[(4+38​)−(41​+31​)]=2[(4+2.6)−(0.25+0.3333)]Compute the values to find the result.

$$\int (\sin x + 2) \, dx = -\cos x + 2x + C.$$

Step 1: Identify the failing marks

The pass mark is 5, so students scoring less than 5 (i.e., marks 2, 3, and 4) failed the test.

Step 2: Add the number of students with failing marks

From the table:

• Students scoring 2: 3
• Students scoring 3: 1
• Students scoring 4: 5

$$\text{Total students who failed} = 3 + 1 + 5 = 9$$

Total students who failed=3+1+5=9

Final Answer:

D .. 9 students failed the test

Solution:

The table:

The pass mark is

$5$

5, meaning students with marks less than 5 (i.e.,

$2, 3, 4$

2,3,4) failed.

Step 1: Add the number of students with failing marks

From the table:

• Students with marks
  $2$ 

  $2$: $3$ 

  $3$

• Students with marks
  $3$ 

  $3$: $1$ 

  $1$

• Students with marks
  $4$ 

  $4$: $5$ 

  $5$

Total number of failing students:

$$3 + 1 + 5 = 9$$

3+1+5=9

Final Answer:

$$\boxed{9 \text{ students failed the test.}}$$

9 students failed the test.​

Solution:

The table:

Step 1: Recall the formula for the mean

The mean is given by:

$$\text{Mean} = \frac{\text{Sum of all marks}}{\text{Total number of students}}.$$

Step 2: Compute the sum of all marks

Multiply each mark by the number of students scoring that mark:

$$\text{Sum of marks} = (2 \times 3) + (3 \times 1) + (4 \times 5) + (5 \times 2) + (6 \times 4) + (7 \times 2) + (8 \times 3)$$

Performing the calculations:

$$\text{Sum of marks} = 6 + 3 + 20 + 10 + 24 + 14 + 24 = 101$$

Step 3: Compute the total number of students

Add up the number of students:

$$\text{Total students} = 3 + 1 + 5 + 2 + 4 + 2 + 3 = 20$$

Step 4: Calculate the mean

$$\text{Mean} = \frac{\text{Sum of marks}}{\text{Total number of students}} = \frac{101}{20} = 5.05$$

Final Answer:

$$\boxed{5.05}$$

$$\text{First, find the mean } \mu = \dfrac{2 + 3 + 5 + 6}{4} = \dfrac{16}{4} = 4 \\ \text{Then calculate variance } \sigma^2 = \dfrac{(2 – 4)^2 + (3 – 4)^2 + (5 – 4)^2 + (6 – 4)^2}{4} = \dfrac{4 + 1 + 1 + 4}{4} = \dfrac{10}{4} = \dfrac{5}{2} \\ \text{Standard deviation } \sigma = \sqrt{\dfrac{5}{2}}.$$

$$\text{Number of ways } = \binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200.$$

$$\text{Total possible outcomes } = 2^3 = 8 \\ \text{Favorable outcome } = 1 (\text{HHH}) \\ \text{Probability } = \dfrac{1}{8}.$$