2020 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

To solve this, we need to convert each number from base 3 to base 10. $(212)_3 = 2(3^2) + 1(3^1) + 2(3^0) = 18 + 3 + 2 = 23$, $(121)_3 = 1(3^2) + 2(3^1) + 1(3^0) = 9 + 6 + 1 = 16$, $(222)_3 = 2(3^2) + 2(3^1) + 2(3^0) = 18 + 6 + 2 = 26$. Therefore, $23 – 16 + 26 = 33$, which is $(313)_3$

Using the difference of squares formula $(a^2 – b^2) = (a+b)(a-b)$, let $a = 4a + 3$ and $b = 3a – 2$. Expanding gives $(a + 5)(7a + 1)$

To find the median, arrange numbers in ascending order: 89, 100, 108, 119, 120, 131, 131, 130, 141, 161. With 10 numbers, median is average of 5th and 6th numbers: $(120 + 131)/2 = 125.5$

Multiply both sides by 15 to eliminate fractions: $5(x + 1) – 15 > 3(x + 4)$. Simplify: $5x + 5 – 15 > 3x + 12$. Solve: $5x – 10 > 3x + 12$, $2x > 22$, $x > 11$

Rounding to 2 s.f.: 0.02174 ≈ 0.022, 1.2047 ≈ 1.2, 0.023789 ≈ 0.024. Then calculate $\frac{0.022 \times 1.2}{0.024} = 1.1$

Let rate of each girl be x. Then rate of each boy is $\frac{5x}{4}$. Total work = 1 unit. Equation: $4(\frac{5x}{4}) + 10x = \frac{1}{5}$. Solve for new number of boys needed in 3 hours.

At the equator (0°), the circumference is $2\pi R$. This is because the radius at any latitude $\theta$ is $R\cos\theta$, and at 0°, $\cos(0°) = 1$

Using 3D Pythagoras theorem, diagonal = $\sqrt{12^2 + 9^2 + 8^2} = \sqrt{289} = 17$m. Floor diagonal = $\sqrt{12^2 + 9^2} = 15$m. $\cos\theta = \frac{15}{17}$

Multiply each score by its frequency and sum: $0(4) + 1(3) + 2(15) + 3(16) + 4(1) + 5(0) + 6(1) = 0 + 3 + 30 + 48 + 4 + 0 + 6 = 91$

In a pie chart, $360°$ represents all students (150). Therefore, $12°$ represents $\frac{12}{360} \times 150 = 5$ students

$\log_{10}4.5 = \log_{10}(4.5) = \log_{10}(\frac{9}{2}) = \log_{10}9 – \log_{10}2 = 2\log_{10}3 – \log_{10}2 = 2(0.4771) – 0.3010 = 0.6532$

Factor numerator: $324-4x^2 = -2(2x^2-162) = -2(2x+18)(x-9)$. Cancel common factor $(2x+18)$ to get $-2(x-9)$

Total weight = 240g. Angle for meat = $\frac{105}{240} \times 360° = 157.5°$

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

Simplify: $\frac{27}{5} \times \frac{1}{27} \times 5 = \frac{27}{5} \times \frac{5}{27} = 1$

Total bottles = 24. Not Coca-Cola = Fanta + Sprite = 14. Probability = $\frac{14}{24} = \frac{7}{12}$

Since PQRS is a quadrilateral:

$$2y + 2x + QPS = 360°$$ $$\text{i.e. } (y + x) + QPS = 360°$$ $$QPS = 360° – 2(y + x)$$

But $$x + y + 68° = 180°$$

Therefore: $$x + y = 180° – 68° = 112°$$

$$QPS = 360° – 2(112°)$$ $$= 360° – 224° = 136°$$

Gradient = $\frac{y_2-y_1}{x_2-x_1} = \frac{-4-0}{0-(-2)} = \frac{-4}{2} = -2$

Parallel lines have same gradient. Original gradient = $-\frac{7}{5}$. Use point-slope form: $y – 7 = -\frac{7}{5}(x – 5)$. Simplify to get 7x + 5y = 70

Using SI formula: $I = \frac{PRT}{100}$. 27 = $\frac{225 \times 4 \times x}{100}$. Solve for x = 3 years

Mean = 10. Sum of squared deviations = 28. SD = $\sqrt{\frac{28}{7}} = 2$

Prime numbers between 20 and 30: 23, 29. Total numbers = 11. Probability = $\frac{2}{11}$

The gradient of line = $\frac{\text{Change in y}}{\text{Change in x}}$ = $\frac{y_2-y_1}{x_2-x_1}$
Given:

$$y_2 = 0$$
$$y_1 = 4$$
$$x_2 = 3$$
$$x_1 = 0$$

Therefore:
$$\frac{y_2-y_1}{x_2-x_1}=\frac{0-4}{3-0}=\frac{-4}{3}$$
Equation of straight line: $y = mx + c$
Where:

$$m = \text{gradient} = -\frac{4}{3}$$
$$c = \text{y-intercept} = 4$$

Therefore:
$$y = -\frac{4}{3}x + 4$$
Multiplying through by 3:
$$3y = -4x + 12$$

0.0014 × 0.011 = 0.0000154 = 1.54 × $10^{-5}$

Using $\tan 60° = \frac{h}{30}$. $h = 30\tan 60° = 30\sqrt{3}$

Using SI formula: $\frac{I \times 100}{P \times T} = R$. Interest = 240, Principal = 1000, Time = 3. R = 8%

For the first part (angle calculation): $$360° – (60° + 60° + 67° + 50° = 237°)$$ $$360° – 237° = 123°$$

For the salary calculation: $$\text{Salary} = \frac{123,360 \times N60,001}{123,360}$$ $$= N2,050$$

Multiply all terms by 12x: $4x + 6x > 3x$. Simplify: $7x > 0$. Therefore $x > 0$. Then check denominator restrictions to get $0 < x < 6$

Let $y = \sqrt{x+\sqrt{2}}$. Then $y = \frac{1}{x-\sqrt{2}}$. Cross multiply and solve to get $x = 4 + 2\sqrt{2}$

$a^2 = 2 – a$. Rearrange to $a^2 + a – 2 = 0$. Factorize: $(a + 2)(a – 1) = 0$. Therefore $a = 1$ or $a = -2$

When chord equals radius, it subtends 60° at center. Arc length = $\frac{\theta}{360°} \times 2\pi r = \frac{60}{360} \times 2\pi r = \frac{\pi r}{3}$

Arc length becomes circumference of base. $\frac{300}{360} \times 2\pi \times 7.2 = 2\pi r$. Solve for r = 6cm

Volume = $\pi r^2h$. $3080 = \pi \times 7^2 \times h$. Solve for h = 20m

Total parts = 30. Angle for cassava = $\frac{3}{30} \times 360° = 36°$

Let r be common ratio. Then $\frac{n}{n-2} = \frac{n+3}{n} = r$. Solve quadratic equation to get $r = \frac{3}{2}$

Let x be number taking both. Using sets: $32 + 24 – x + 4 = 40$. Solve for x = 20

Sum of interior angles of pentagon = $540°$. Therefore $6x + 6y = 540$. Solve for y = $90 – x$

Let n be number of students. $\frac{15n + 45}{n + 1} = 18$. Solve to get n = 9

Vertical height = opposite = $500 \sin 30° = 500 \times \frac{1}{2} = 250$m

Expand determinant: $x^3 – x – 2 = 0$. Factor to get $(x – \sqrt{2})(x + \sqrt{2})(x + 1) = 0$. Therefore $x = \sqrt{2}$