2021 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

To solve this equation: 1) Find a common denominator, 2) Convert all terms, 3) Multiply both sides to eliminate denominators, 4) Solve the resulting quadratic equation. After solving, we get two values of r. Only $r = \frac{5}{2}$ satisfies all constraints when checked in original equation.

This is a combination problem where we need to find $C(5,2)$ or $\binom{5}{2}$. The formula is $\frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$. This represents all possible unique ways to select 2 students from 5.

Volume of hemisphere is $V = \frac{2}{3}\pi r^3$. Differentiating with respect to r: $\frac{dV}{dr} = 2\pi r^2$. When $r = 2$: $\frac{dV}{dr} = 2\pi(2^2) = 8\pi$

To find maximum value: 1) Find derivative $\frac{dy}{dx} = 6x + 5$ 2) Set equal to zero: $6x + 5 = 0$ 3) Solve: $x = -\frac{5}{6}$ 4) Substitute back into original equation to find maximum value = 2

Using trapezium area formula: $A = \frac{h(a+b)}{2}$ where h is height and a,b are parallel sides. $91 = \frac{h(5+9)}{2}$ $91 = 7h$ Therefore $h = 13$

For parallel lines, slopes must be equal. Slope of $2y+8x-17=0$ is $-4$. Using point slope formula between given points: $\frac{-p-2}{-1-(-2)} = -4$ Solving for p gives $p = -\frac{2}{7}$

For similar solids, ratio of volumes is cube of length ratio. If length ratio is 2:3, volume ratio is $2^3:3^3 = 8:27$. If larger volume is 351, smaller volume is $\frac{8}{27} \times 351 = 104$

Expand: $y = 4x^3 – 2x^2$ Differentiate: $\frac{dy}{dx} = 12x^2 – 4x$ At $x = -1$: $12(-1)^2 – 4(-1) = 12 + 4 = 16$

Calculate: $241.34 \times 9 \times 10^{-6} = 2.17206 \times 10^{-3}$ To 4 significant figures = $0.002172$

First find the mean:

Mean = (1 + 2 + 3 + 4) ÷ 4 = 10 ÷ 4 = 2.5

Find deviations from mean for each number:

|1 – 2.5| = 1.5
|2 – 2.5| = 0.5
|3 – 2.5| = 0.5
|4 – 2.5| = 1.5

Calculate mean deviation:

Mean deviation = (1.5 + 0.5 + 0.5 + 1.5) ÷ 4
= 4 ÷ 4
= 1.0

Using Simple Interest formula: $I = PRT$, where $7.50 = 100 \times R \times 5$. Solving for R: $R = \frac{7.50}{500} = 0.015 = 1.5%$

First child: $\frac{1}{4}$. Remainder: $\frac{3}{4}$. Second child: $\frac{3}{4}$ of $\frac{3}{4} = \frac{9}{16}$. Third child gets: $\frac{3}{4} – \frac{9}{16} = \frac{3}{16}$

Convert to standard form and multiply/divide: $\frac{2.75 \times 10^{-3} \times 6.40 \times 10^{-3}}{2.5 \times 10^{-2} \times 8.0 \times 10^{-2}} = 8.8 \times 10^{-1}$

$\sqrt[3]{27} = 3$ and $\sqrt[3]{3} = \sqrt[3]{3}$. Therefore $3 + \sqrt[3]{3} = \frac{3\sqrt[3]{3} + (\sqrt[3]{3})^4}{\sqrt[3]{3}} = \frac{4\sqrt[3]{3}}{\sqrt[3]{3}} = 4$

First gets $\frac{1}{3}$. Of remaining $\frac{2}{3}$, second gets $\frac{2}{3}$ ($\frac{4}{9}$ of total). Third gets $\frac{2}{9}$ which is N12000. Total = N54000

Diameter is distance between points. Using distance formula: $\sqrt{(6-(-6))^2 + (6-1)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$. Radius is half of diameter = 6.5

Perimeter = arc length + 2 radii = $\frac{48}{360} \times 2\pi r + 2r = \frac{48}{360} \times 2\pi \times 10.5 + 21 = 8.8 + 21 = 29.8$

In rhombus, if diagonals are d₁ and d₂, side length = $\sqrt{\frac{d_1^2 + d_2^2}{4}}$ = $\sqrt{\frac{36 + 64}{4}} = \sqrt{25} = 5$

For regular polygon with n sides, interior angle = $\frac{(n-2) \times 180}{n}$. Solving: $140 = \frac{180(n-2)}{n}$ gives n = 9

Volume = length × π(R² – r²) where R = 13cm, r = 10cm, length = 300cm. Volume = $300\pi(169-100) = 20700\pi$ cm³

The angle bisector of two intersecting lines is the locus of points equidistant from both lines. This is because any point on the angle bisector has equal perpendicular distances to both lines.

Total = 120. Numbers ≥ 16: 16, 30, 20, 26. Sum = 92. Angle = $\frac{92}{120} \times 360° = 276°$

Let x be 11th number. $(10 \times 16 + x) \div 11 = 18$ Solving: $160 + x = 198$ Therefore x = 38

Possible sums: (1,2)=3, (1,3)=4, (1,4)=5, (2,3)=5, (2,4)=6, (3,4)=7. Even sums: 4, 6. Total combinations: 6. Probability = $\frac{2}{6} = \frac{1}{3}$

Numbers from 41 to 56: 16 numbers. Multiples of 9 in range: 45, 54. Probability = $\frac{2}{16} = \frac{1}{8}$

Interest = $10 \times 0.02 \times 4 = 0.80$. Total owed = 10.80. Amount paid = 8.00. Still owes = 2.80

Convert all to same base, simplify using log laws. Final answer = $1 – 2\log 2$

Average speed = Total distance/Total time = $\frac{P+Q}{\frac{P}{U}+\frac{Q}{W}} = \frac{UW(P+Q)}{PW+QU}$

If $w \propto \frac{u+v}{uv}$, then $w = k\frac{u+v}{uv}$. Substitute values: $8 = k\frac{8}{12}$, so $k = 12$. Therefore $uvw = 12(u + v)$

g(x+1) = $(x+1)^2 + 3(x+1) = x^2 + 2x + 1 + 3x + 3$. g(x) = $x^2 + 3x$. Difference = $2x + 4 = 2(x + 2)$

Using factor theorem and polynomial division, factors are (m – 2), (m – 1), and (m + 1)

Sum of angles = 360°. Solve: $5x-30 + 4x+60 + 60-x + 3x+61 = 360$. Solving gives x = 42. Substitute to find smallest angle

Find differences: 4, 6, 8 shows second differences are constant = 2. Therefore quadratic sequence with n-th term = $n^2 + n$

For identity e: m $*$ e = m for all m. So me + m + e = m. Solve for e: e(m + 1) = 0 for all m. Therefore e = 0

Using difference of squares twice: $(9a^2 + 4b^2)(9a^2 – 4b^2)$ = $(9a^2 + 4b^2)(3a – 2b)(3a + 2b)$

$\log_9 x = 1.5$ means $x = 9^{1.5} = (9^{\frac{3}{2}}) = (3^3)^{\frac{3}{2}} = 3^{4.5} = 27$

Solve: -2 < 2x-6 < 4. Add 6: 4 < 2x < 10. Divide by 2: 2 < x < 5. Integers in this range are 3,4

Using right triangle: $\tan \theta = \frac{6}{YX}$. Using Pythagoras: $YX^2 + 6^2 = 12^2$. Solve for angle = 60°

Use inclusion-exclusion principle: Total = (12+10+14) – (5+4+2) + 3 = 36 – 11 – 3 = 18

Substitute x = -2 and x = 1 into expression and set equal to zero. Solve simultaneous equations for L and k. L = -2, k = -1