2023 - JAMB Mathematics Past Questions and Answers - page 1

The given sequence is an arithmetic sequence with a common difference of 5.

The general term (nth term) of an arithmetic sequence is given by the formula \(a_n = a_1 + (n-1)d\), where \(a_n\) is the nth term, \(a_1\) is the first term, \(n\) is the term number, and \(d\) is the common difference.

For the given sequence, the first term \(a_1\) is 3, and the common difference \(d\) is 5.

So, the general term is \(a_n = 3 + (n-1) \times 5\), which simplifies to \(a_n = 5n - 2\).

Therefore, the correct option is \(5n - 2\).

Let's denote the total number of balls as \( T \), the number of red balls as \( R \), and the number of white balls as \( W \).

Given that the probability of drawing a white ball is half of the probability of drawing a red ball, we can set up the following equations:

\[ P(\text{White}) = \frac{1}{2} \times P(\text{Red}) \]

Since \( P(\text{White}) = \frac{W}{T} \) and \( P(\text{Red}) = \frac{R}{T} \), we have:

\[ \frac{W}{T} = \frac{1}{2} \times \frac{R}{T} \]

Now, we know that \( R = 8 \) (the number of red balls). We can substitute this value into the equation:

\[ \frac{W}{T} = \frac{1}{2} \times \frac{8}{T} \]

To find \( W \), we can solve for \( T \):

\[ W = \frac{1}{2} \times 8 = 4 \]

So, there are 4 white balls.

Now, for the probability of drawing a red ball and then a white ball without replacement:

\[ P(\text{Red and White}) = \frac{R}{T} \times \frac{W}{T-1} \]

Substituting the values:

\[ P(\text{Red and White}) = \frac{8}{T} \times \frac{4}{T-1} \]

Since \( T = 8 + 4 = 12 \), we can substitute this value:

\[ P(\text{Red and White}) = \frac{8}{12} \times \frac{4}{11} \]

Simplifying:

\[ P(\text{Red and White}) = \frac{2}{3} \times \frac{4}{11} \]

Therefore, the correct answer is \(\frac{8}{33}\)

Let's solve the logarithmic equation:

\[ \log_2(6 - x) = 3 - \log_2(x) \]

First, combine the logarithmic terms on the right side:

\[ \log_2(6 - x) + \log_2(x) = 3 \]

Now, use the product rule of logarithms (\(\log_a(b) + \log_a(c) = \log_a(b \cdot c)\)):

\[ \log_2[(6 - x) \cdot x] = 3 \]

Next, simplify the expression inside the logarithm:

\[ \log_2(6x - x^2) = 3 \]

Now, rewrite the equation in exponential form:

\[ 2^3 = 6x - x^2 \]

\[ 8 = 6x - x^2 \]

Bring all terms to one side to form a quadratic equation:

\[ x^2 - 6x + 8 = 0 \]

Now, factor the quadratic:

\[ (x - 4)(x - 2) = 0 \]

Set each factor equal to zero:

\[ x - 4 = 0 \] or \[ x - 2 = 0 \]

So, the solutions are:

\[ x = 4 \] or \[ x = 2 \]

Therefore, the correct answer is \(x = 4\) or \(x = 2\)

To find matrix \(A\), we need to solve for \(A\) in the equation \(A \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}\).

Let \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\). Then the equation becomes:

\[
\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\]

Now, perform the matrix multiplication:

\[
\begin{bmatrix} a(0) + b(2) & a(1) + b(-1) \\ c(0) + d(2) & c(1) + d(-1) \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\]

Simplify the matrix equation:

\[
\begin{bmatrix} 2b & a - b \\ 2d & c - d \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\]

Now, equate corresponding elements:

1. \(2b = 2\)  \(\Rightarrow b = 1\)
2. \(a - b = -1\) \(\Rightarrow a - 1 = -1\)  \(\Rightarrow a = 0\)
3. \(2d = 1\)  \(\Rightarrow d = \frac{1}{2}\)
4. \(c - d = 0\) \(\Rightarrow c - \frac{1}{2} = 0\)  \(\Rightarrow c = \frac{1}{2}\)

Therefore, the matrix \(A\) is:

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

The correct option is \(\begin{bmatrix} 0 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\)

To find the bearing of R from P, we need to determine the angle between the north direction and the line PR. 

The distance PQ is 8 km north, and QR is 6 km west. We can use trigonometry to find the angle.

\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \]

\[ \tan(\theta) = \frac{6}{8} \]

\[ \theta = \tan^{-1}\left(\frac{6}{8}\right) \]

\[ \theta \approx 36.87^\circ \]

Now, the bearing is measured clockwise from the north direction. So, the bearing of R from P is:

\[ \text{Bearing} = 360^\circ - 36.87^\circ \]

\[ \text{Bearing} \approx 323.13^\circ \]

The nearest degree is **323°**.

Therefore, the correct answer is 323°

Let's denote the cost price of the article as \(C\).

According to the given information, when the article is sold for ₦230.00, there is a 15% profit. Therefore,

\[ \text{Selling Price} = C + \text{Profit} \]

\[ 230.00 = C + 0.15C \]

\[ 230.00 = 1.15C \]

Now, we can find the cost price \(C\):

\[ C = \frac{230.00}{1.15} \]

\[ C \approx 200.00 \]

Now, when the article is sold for ₦180.00:

\[ \text{Profit or Loss} = \text{Selling Price} - \text{Cost Price} \]

\[ \text{Profit or Loss} = 180.00 - 200.00 \]

\[ \text{Profit or Loss} = -20.00 \]

Now, to find the percentage profit or loss:

\[ \text{Percentage Profit or Loss} = \left(\frac{\text{Profit or Loss}}{\text{Cost Price}}\right) \times 100 \]

\[ \text{Percentage Profit or Loss} = \left(\frac{-20.00}{200.00}\right) \times 100 \]

\[ \text{Percentage Profit or Loss} = -10\% \]

Therefore, the correct answer is a 10% loss

We know that the area \(A\) of a circle is given by the formula \(A = \pi r^2\), where \(r\) is the radius.

The problem states that the area \(A\) is increasing at a constant rate, which means \(dA/dt = 1.5 \, \text{cm}^2/\text{s}\).

We need to find the rate at which the radius \(r\) is increasing when the area of the circle is \(2 \, \text{cm}^2\), so we are looking for \(dr/dt\) when \(A = 2 \, \text{cm}^2\).

First, differentiate the area formula with respect to time:

\[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]

Now, substitute the given values:

\[ 1.5 = 2\pi (r) \frac{dr}{dt} \]

We are asked to find \(dr/dt\) when \(A = 2 \, \text{cm}^2\), so substitute \(A = 2\) into the equation:

\[ 1.5 = 2\pi (r) \frac{dr}{dt} \]

\[ 1.5 = 2\pi (2) \frac{dr}{dt} \]

Now, solve for \(\frac{dr}{dt}\):

\[ \frac{dr}{dt} = \frac{1.5}{4\pi} \]

Using a calculator, this is approximately 0.119 \(\text{cm/s}\).

Therefore, the correct answer is 0.299 cm/s (rounded to three significant figures).

The percent error is calculated using the formula:

\[ \text{Percent Error} = \left| \frac{\text{Measured Value} - \text{Accepted Value}}{\text{Accepted Value}} \right| \times 100 \]

Let's calculate the percent error using the given values:

\[ \text{Percent Error} = \left| \frac{18 - 18.4}{18.4} \right| \times 100 \]

\[ \text{Percent Error} = \left| \frac{-0.4}{18.4} \right| \times 100 \]

\[ \text{Percent Error} \approx 0.021739 \times 100 \]

\[ \text{Percent Error} \approx 2.174 \%\]

Rounded to two decimal places, the percent error is approximately \(2.17\%\).

Therefore, the correct answer is 2.17\%

To find the number of different 8-letter words using the letters of the word "SYLLABUS," we can use the formula for permutations of a multiset, where some elements are repeated.

The correct formula is:

\[ \frac{8!}{2! \cdot 2!} \]

This is because there are 8 positions, but the letter "S" appears twice, and the letter "L" appears twice. So, we divide by \(2!\) for each repeated letter to correct for the overcounting.

Therefore, the correct answer is \(\frac{8!}{2! \cdot 2!}\)

The formula for compound interest is given by:

\[ A = P \left(1 + \frac{r}{100}\right)^t \]

where:
- \(A\) is the amount after \(t\) years,
- \(P\) is the principal amount (initial amount),
- \(r\) is the annual interest rate, and
- \(t\) is the number of years.

The compound interest (\(CI\)) is then given by:

\[ CI = A - P \]

Let's calculate the compound interest for ₦15,700 at 8% per annum compounded annually for 2 years.

\[ A = 15,700 \left(1 + \frac{8}{100}\right)^2 \]

\[ A = 15,700 \times \left(1 + \frac{2}{25}\right)^2 \]

\[ A = 15,700 \times \left(\frac{27}{25}\right)^2 \]

\[ A = 15,700 \times \frac{729}{625} \]

\[ A = 18,312 \]

Now, calculate the compound interest:

\[ CI = A - P \]

\[ CI = 18,312 - 15,700 \]

\[ CI = 2,612 \]

Therefore, the correct answer is ₦2,612.48