2022 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

To solve this equation, multiply both sides by $(8k+2)$: $1 = 8k+2$. Then subtract 2 from both sides: $-1 = 8k$. Finally divide both sides by 8: $k = -\frac{1}{8}$. Therefore, $k = -4$.

To solve this, first convert $101_2$ to decimal: $101_2 = 5_{10}$. Then cube it: $5^3 = 125_{10}$. Finally convert 125 to binary: $125_{10} = 1111101_2$

-The shaded/stripped portion in the diagram is set F only.
-This implies that sets E and G though present in the Universal set, are not included in the expression.

-Mathematically; we have E’ n F n G’

For at least one head, we can find 1 minus probability of no heads. P(no heads) = $(\frac{1}{2})^2 = \frac{1}{4}$. Therefore, P(at least one head) = $1 – \frac{1}{4} = \frac{3}{4} = 0.75$

Since mean = 6, we can write: $\frac{2+5+(x+1)+(x+2)+7+9}{6} = 6$. Solving gives x = 4. Arranging numbers: 2,5,5,6,7,9. Median = $\frac{5+6}{2} = 5.5$

In the third quadrant, both sine and cosine are negative. Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we get $\cos^2 \theta = 1 – \frac{9}{25} = \frac{16}{25}$. Since it’s in third quadrant, $\cos \theta = -\frac{4}{5}$

Rationalize the denominator by multiplying numerator and denominator by $(3+\sqrt{2})$: $\frac{1}{3-\sqrt{2}} \cdot \frac{3+\sqrt{2}}{3+\sqrt{2}} = \frac{3+\sqrt{2}}{9-2} = \frac{3+\sqrt{2}}{7}$

Using Pythagorean theorem, if h=3cm and r=5cm, then half-chord length = $\sqrt{r^2-h^2} = \sqrt{25-9} = \sqrt{16} = 4$. Full chord length = 8cm

To find the angle, multiply the percentage by 360°: $28% \times 360° = 100.8°$

Volume of pyramid = $\frac{1}{3} \times$ base area $\times$ height = $\frac{1}{3} \times 24 \times 7.5 = 60$ cm³

Let d be the common difference. Using $a_{10} = a_1 + 9d$ and $a_4 = a_1 + 3d$: $32 = \frac{3}{2} + 9d$, solving gives $d = \frac{61}{18}$. Therefore $a_4 = \frac{3}{2} + 3(\frac{61}{18}) = \frac{35}{2}$

Using line LZ with angles sum = 180º

: ∠MNY = 180º – (g + h)º or 180º – gº – hº

This is a permutation with repetition. ZOOLOGY has 7 letters with O repeating twice, G appearing once, L appearing once, Y appearing once, Z appearing once. Formula: $\frac{7!}{2!}$ = 2520/3 = 840

Using power rule: $\int(2x+1)^3 dx = \frac{(2x+1)^4}{8} + C$

Solution:

To find \( AB \), we perform matrix multiplication. Let \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} \) and \( B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \). The resulting matrix \( AB \) is given by:

\[

AB = \begin{bmatrix} 

a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ 

a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \\ 

a_{31}b_{11} + a_{32}b_{21} & a_{31}b_{12} + a_{32}b_{22} 

\end{bmatrix}.

\]

Substituting \( A = \begin{bmatrix} 2 & 1 \\ 2 & 3 \\ 1 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 2 \\ 4 & 2 \end{bmatrix} \):

 Step 1: Solve the first row of \( AB \)

– 1st row, 1st column:

\[

(2)(3) + (1)(4) = 6 + 4 = 10.

\]

– 1st row, 2nd column:

\[

(2)(2) + (1)(2) = 4 + 2 = 6.

\]

 Step 2: Solve the second row of \( AB \)

– 2nd row, 1st column:

\[

(2)(3) + (3)(4) = 6 + 12 = 18.

\]

– 2nd row, 2nd column:

\[

(2)(2) + (3)(2) = 4 + 6 = 10.

\]

 Step 3: Solve the third row of \( AB \)

– 3rd row, 1st column:

\[

(1)(3) + (2)(4) = 3 + 8 = 11.

\]

– 3rd row, 2nd column:

\[

(1)(2) + (2)(2) = 2 + 4 = 6.

\]

 Final Result:

\[

AB = \begin{bmatrix} 10 & 6 \\ 18 & 10 \\ 11 & 6 \end{bmatrix}.

\]

Thus, the correct answer is:

\[

\text{D. } \begin{bmatrix} 10 & 6 \\ 18 & 10 \\ 11 & 6 \end{bmatrix}.

\]

Square both sides: $r^2 = \frac{3v}{\pi h}$. Multiply both sides by $\pi h$: $\pi r^2h = 3v$. Therefore $v = \frac{\pi r^2h}{3}$

Determinant = $ad-bc$ where $a=2$, $b=1$, $c=3$, $d=3$. Therefore: $(2)(3)-(1)(3) = 6-3 = 3$

$\log_2 8\sqrt{2} = \log_2 8 + \log_2 \sqrt{2} = 3 + \frac{1}{2} = 3.5$

A circle is defined as the set of all points in a plane that are equidistant from a fixed point called the center

New numerator = $96(1-0.375) = 60$. New denominator = $50(1-0.2) = 40$. Therefore $\frac{60}{40} = \frac{3}{2} = 1.5$

One crate = 30 eggs. Sold eggs = $5.5 \times 30 = 165$ eggs. Remaining = $200-165 = 35$ eggs. Percentage = $\frac{35}{200} \times 100 = 17.5%$

Using identity $\sec^2 \theta = 1 + \tan^2 \theta$, we get $1 + \tan^2 \theta + \tan^2 \theta = 3$, so $2\tan^2 \theta = 2$, therefore $\tan \theta = 1$, which occurs at 45°

Using L’Hôpital’s rule or factoring: As $x \to 1$, both numerator and denominator → 0. Factoring gives $(x-1)(x^2+x+6)/(x-1)$. When $x = 1$, value = 8

Parallel lines have the same slope. Original slope = 2. Using point-slope form with y-intercept = 5: $y = 2x + 5$ or $2x – y = -5$

Arrange in order: 10, 10, 13, 14, 17, 18. With 6 numbers, median = average of 3rd and 4th numbers = $\frac{13+14}{2} = 13.5$

Converting: $1\times2^4 + 0\times2^3 + 1\times2^2 + 1\times2^1 + 0\times2^0 + 0\times2^{-1} + 1\times2^{-2} + 1\times2^{-3} = 16+0+4+2+0+0+0.25+0.125 = 22.375$

Using direct proportion: $\frac{C_1}{N_1} = \frac{C_2}{N_2}$. Therefore $\frac{10000}{20} = \frac{1000000}{x}$. Solving gives $x = 2000$

Mid-point formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{-3+2}{2}, \frac{5+10}{2}) = (-\frac{1}{2}, \frac{15}{2})$

When n = 3, we have $^{10}C_8$. Using formula: $\frac{10!}{8!(10-8)!} = \frac{10 \times 9}{2 \times 1} = 45$

Initial velocity is the first derivative of displacement when t = 0. $v = \frac{dS}{dt} = 6t + 5$. When t = 0, v = 5 m/s

Given:

• hyp = 5cm
• adj = 4cm
• opp = x cm

Using Pythagoras theorem: $$\text{hyp}^2 = \text{opp}^2 + \text{adj}^2$$ $$5^2 = x^2 + 4^2$$ $$25 = x^2 + 16$$ $$x^2 = 25 – 16$$ $$x^2 = 9$$ $$x = \sqrt{9}$$ $$x = 3\text{ cm}$$

Substituting values: $3 * 5 = 3 – 5 + (3 \times 5)^2 = 3 – 5 + 225 = 223$

Using inverse variation: $xy = k$. When $x = 3, y = 4$, so $k = 12$. When $y = 12$, $x = \frac{k}{y} = \frac{12}{12} = 1$

Common difference $d = \frac{26-8}{3} = 6$. Therefore $p = 8 + d = 14$ and $q = p + d = 20$

Using mean formula: $\frac{2x + 4x + (2x-13) + 6x}{4} = 4$. Simplifying: $\frac{14x-13}{4} = 4$. Therefore $14x-13 = 16$, so $x = \frac{29}{14} \approx 2.0$

This is a difference of squares: $4a^2 – 9b^2 = (2a+3b)(2a-3b)$

Using the formula: n(P∪Q) = n(P) + n(Q) – n(P∩Q). Therefore 40 = 20 + 30 – n(P∩Q). Solving gives n(P∩Q) = 10

Converting to algebra: $3x – 2 < x - 6$. Solving: $2x < -4$, therefore $x < -2$

The arrow pointing to the right from 2 with a closed circle at 2 indicates numbers greater than or equal to 2.

x ≥ 2 is represented by an arrow with an filled-in dot above the 2.

The arrow points in the direction of all the numbers that are greater than 2.

If #3800 is 80% of original price, then original price = $\frac{3800}{0.8} = #4750$