2023 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

In SYLLABUS, there are 8 letters total with S appearing twice and L appearing twice. This is a permutation with repetition problem. The formula is $\frac{n!}{n_1!n_2!…n_k!}$ where n is total letters and n_i is repetition of each letter. Here $\frac{8!}{2!2!}$

Multiplying decimal numbers: $160.16 \times 160.04 = 25,641.0464$ Then multiply by 20.2: $25,641.0464 \times 20.2 = 517,949.13728$ Rounded to nearest whole number = 517,949

First evaluate inside parentheses: $-4 * 2 = (-4)^2(2) = 32$ $7 * -1 = 7^2(-1) = -49$ Then $32 \wedge -49 = 2(32) + (-49) = 15$

Using antiderivative: $\int (4x-6x^2)\sqrt{3}dx = \sqrt{3}(2x^2-2x^3) + C$ Evaluate from 0 to 1: $\sqrt{3}[(2(1)^2-2(1)^3) – (2(0)^2-2(0)^3)] = 85$

To find percentage decrease: $\frac{\text{Decrease}}{\text{Original}} \times 100 = \frac{1230-1040}{1230} \times 100 = \frac{190}{1230} \times 100 = 15.45%$

For a regular polygon: $\text{Interior angle} = \frac{(n-2)\times 180}{n}$ $\text{Exterior angle} = \frac{360}{n}$ If interior is 5 times exterior: $\frac{(n-2)\times 180}{n} = 5(\frac{360}{n})$ Solving gives n = 12 (dodecagon)

Using $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$, where $\frac{dA}{dt} = 1.5$ and $A = \pi r^2 = 2$ Solving for $\frac{dr}{dt}$ gives 0.299 cms⁻¹

Cross multiply: $y(4x+5d) = 3x-9c$ Expand: $4xy+5dy = 3x-9c$ Collect x terms: $4xy-3x = -9c-5dy$ Factor x: $x(4y-3) = -9c-5dy$ Therefore $x = \frac{9c-5dy}{4y-3}$

Expand: $3x – 3 ≤ 2x – 6$ Subtract 2x from both sides: $x – 3 ≤ -6$ Add 3 to both sides: $x ≤ -3$

Since tangent is perpendicular to radius, ∠PCS = 90°. As ∠PSR = 36°, and angle in semicircle is 90°, ∠PQR = 2 × 36° = 72°

Using speed formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$ Speed = $\frac{225}{4.5} = 50$ km/hr. For 150 km: $\text{Time} = \frac{150}{50} = 3$ hours

Converting to decimal: $1101001_2 = 105_{10}$, $101_2 = 5_{10}$. Division: $105 ÷ 5 = 21$. Converting 21 to binary: $21_{10} = 10101_2$

Simplify $3^{-3}\sqrt{2} = \frac{\sqrt{2}}{27}$. Group terms: $\frac{\sqrt{2}}{27}+3\sqrt{3} = a+b\sqrt{3}$. Compare coefficients: $a = \frac{\sqrt{2}}{27}$, $b = 3$

Using distance formula: $\sqrt{(t-(-3))^2+(-5-(-14))^2}=9$. Simplify: $\sqrt{(t+3)^2+81}=9$. Solve: $t = 3$

Let principal be P. Using SI formula: After 10 years: $P(1+10r)=3P$, so $10r=2$. For 5P: $P(1+tr)=5P$. Solve for t: $t=20$ years

As x→2, both numerator and denominator approach 0. Using L’Hôpital’s rule: $\lim_{x \to 2}\frac{2x+4}{2x-2}=4$

Using midpoints: $\frac{5.5(29)+7.5(40)+9.5(38)}{107}=7.7$ years

Majority means at least 3 women. Calculate: $\binom{4}{3}\binom{6}{2} + \binom{4}{4}\binom{6}{1} = 60$

To find the determinant of the matrix \(D\):

\[
D =
\begin{bmatrix}
2 & -1 & 3 \\
4 & 1 & 2 \\
1 & -3 & 1
\end{bmatrix},
\]

we use the formula for the determinant of a 3×3 matrix:

\[
|D| = a(ei − fh) − b(di − fg) + c(dh − eg),
\]

where the matrix is

\[
\begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}.
\]

For matrix \(D\), we identify:

\[
a = 2, \, b = -1, \, c = 3, \, d = 4, \, e = 1, \, f = 2, \, g = 1, \, h = -3, \, i = 1.
\]

Substitute these values into the formula:

1. Compute \( ei − fh \):
\[
ei − fh = (1)(1) − (2)(-3) = 1 + 6 = 7.
\]

2. Compute \( di − fg \):
\[
di − fg = (4)(1) − (2)(1) = 4 − 2 = 2.
\]

3. Compute \( dh − eg \):
\[
dh − eg = (4)(-3) − (1)(1) = -12 − 1 = -13.
\]

Now, substitute these results into the determinant formula:

\[
|D| = 2(7) − (-1)(2) + 3(-13).
\]

Simplify:

\[
|D| = 14 + 2 − 39 = -23.
\]

Thus, \(|D| = -23\).

**Answer**: C. \(-23\)

Using trigonometry: $\tan \theta = \frac{6}{8}$, $\theta = 37°$. Bearing = $323°$

Total outcomes = $2^3 = 8$. Probability = 1 – P(no heads) = $1 – \frac{1}{8} = \frac{7}{8}$

Perpendicular slope = $\frac{2}{3}$. Using point-slope form: $y-3 = \frac{2}{3}(x-2)$. Simplify to $3y = 2x + 5$

Using product rule and chain rule: $\frac{dy}{dx} = \frac{10x^{5/3}}{3}-\frac{8x^{2/3}}{3}$

The first five prime numbers are $2, 3, 5, 7, 11$. The mean is $\frac{28}{5} = 5.6$. The absolute deviations are $|2-5.6|, |3-5.6|, |5-5.6|, |7-5.6|, |11-5.6|$. Mean deviation = $\frac{3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = 2.72$

Find intersection points, then integrate: $\int_{x_1}^{x_2} [(4x+16)-(2x^2+10)]dx = 18$

Convert $40_{2y}$ to base 10: $4(2y)^1 + 0(2y)^0 = 102$. Solve: $4(2y) = 102$, so y = 3

Using log properties: $\log[(y+8)(y-8)] = \log(225)$. Therefore $y^2-64 = 225$, $y = \pm17$

Using set theory: $n(E) + n(F) – n(E\cap F) = n(E\cup F) \leq 500$. Solve: $350 + 400 – x \leq 500$, x = 250

Using difference of squares twice: $(4x^2+y^2)(4x^2-y^2)$ then $(4x^2+y^2)(2x+y)(2x-y)$

A-B = {1,3,5}, B-A = {8}. Therefore $(A-B)\cup(B-A)$ = {1,3,5,8}

Let x be white balls. $\frac{x}{8+x}=\frac{1}{2}\cdot\frac{8}{8+x}$. Solve: x = 4. Probability = $\frac{8}{12}\cdot\frac{4}{11}=\frac{8}{33}$

Total = 107 students. Median position = 54th. In 7-8 class (40 students), 29+20=49 so median = 7.5

Solution:

To find \( A \), we solve the given matrix equation:

\[

A \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}.

\]

To isolate \( A \), we multiply both sides by the inverse of the matrix \(\begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix}\).

 Step 1: Find the inverse of \(\begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix}\)

The inverse of a \( 2 \times 2 \) matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is given by:

\[

\text{Inverse} = \frac{1}{ad – bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

\]

Here, \( a = 0 \), \( b = 1 \), \( c = 2 \), and \( d = -1 \). The determinant is:

\[

\text{Determinant} = (0)(-1) – (2)(1) = -2.

\]

The inverse is:

\[

\begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix}^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & -1 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{bmatrix}.

\]

 Step 2: Solve for \( A \)

Multiply both sides of the equation by the inverse:

\[

A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{bmatrix}.

\]

Perform the matrix multiplication:

\[

A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{bmatrix}.

\]

– 1st row, 1st column:

\[

(2)(\frac{1}{2}) + (-1)(1) = 1 – 1 = 0.

\]

– 1st row, 2nd column:

\[

(2)(\frac{1}{2}) + (-1)(0) = 1 + 0 = 1.

\]

– 2nd row, 1st column:

\[

(1)(\frac{1}{2}) + (0)(1) = \frac{1}{2} + 0 = \frac{1}{2}.

\]

– 2nd row, 2nd column:

\[

(1)(\frac{1}{2}) + (0)(0) = \frac{1}{2} + 0 = \frac{1}{2}.

\]

Thus:

\[

A = \begin{bmatrix} 0 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}.

\]

 Final Answer:

The correct option is:

\[

\text{B. } \begin{bmatrix} 0 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}.

\]

From the graph shown, count students who scored 25% or higher: 16 students

Using polynomial remainder theorem, substitute x = -1: $-2(-1)^3+6(-1)^2+17(-1)-21 = 2+6-17-21 = -30$

Commutativity means order doesn’t matter: $ab = ba$ for all a,b in set A

Rearrange: $x^2-x-6 \leq 0$. Solve: $(x-3)(x+2) \leq 0$. Therefore $-2 \leq x \leq 3$

Common difference = 5. First term = 3. General term = $a_n = a_1 + (n-1)d = 3 + 5(n-1) = 5n-2$

The bisectors of the angles between two lines are the loci of points equidistant from both lines

Let third number be x. First = 1.35x, Second = 1.8x. Ratio = $\frac{1.35x}{1.8x} = \frac{27}{36} = \frac{3}{4}$

Using trigonometry: Height = $24 + 24\tan 30° – 24\tan 60° = 24\sqrt{3}$ meters

Using Pythagorean theorem and trigonometry in the triangles shown: $AB = \sqrt{x^2 + y^2} = 36.2$ cm

Cost price = $\frac{230}{1.15} = 200$. For ₦180: Loss% = $\frac{200-180}{200} \times 100 = 10%$

Arc length = $r\theta$ where $\theta$ is in radians. Converting angle and calculating: 30.5 cm

Range: $(37.5 \times 51.5)$ to $(38.5 \times 52.5)$. Therefore: $1931.25 \leq A < 2021.25$

Total items = $81 \times \frac{9}{4} = 182$. Brand B = $\frac{5}{8} \times \frac{8}{9} \times 182 = 135$. Difference = 54

Side lengths: 40, 60, 80 cm. Using Heron’s formula: $A = \sqrt{s(s-a)(s-b)(s-c)} = 1161$ cm²

CI = $P(1+r)^n – P = 15700(1.08)^2 – 15700 = 2612.48$

Explanation:

The volume of the cylinder is given by:

\[
\text{Volume of the cylinder} = \frac{\theta}{360} \times \pi r^2 h
\]

Given:

\[
\theta = 360^\circ – 90^\circ = 270^\circ
\]

Substitute the values into the formula:

\[
\text{Volume of the cylinder} = \frac{270}{360} \times \frac{22}{7} \times 14^2 \times 32
\]

Simplify step-by-step:

\[
\text{Volume of the cylinder} = \frac{270}{360} \times \frac{22}{7} \times \frac{196}{1} \times \frac{32}{1}
\]

\[
= \frac{270 \times 22 \times 196 \times 32}{360 \times 7}
\]

Calculate the numerator and denominator:

\[
\text{Numerator: } 270 \times 22 \times 196 \times 32 = 37,255,680
\]

\[
\text{Denominator: } 360 \times 7 = 2,520
\]

Divide:

\[
\text{Volume of the cylinder} = \frac{37,255,680}{2,520} = 14,784 \, \text{cm}^3
\]

—

Final Answer:

\[
\boxed{14,784 \, \text{cm}^3}
\]

Common difference = 3. First term = 0. Using arithmetic sequence sum formula: $S_{12} = \frac{n}{2}(2a + (n-1)d) = 198$