2023 - JAMB Mathematics Past Questions and Answers - page 2

Let's denote the length of the longer side of the rectangle as \(L\) and the length of the shorter side as \(L - 6\).

The area of the rectangle is given by \(A = L \times (L - 6)\).

According to the problem, if we add 2 cm to each side, the new dimensions will be \(L + 2\) and \((L - 6) + 2\) (or \(L - 4\)).

The new area will be \((L + 2) \times (L - 4)\).

The problem states that the area increases by 68 cm\(^2\), so we can set up the equation:

\[(L + 2) \times (L - 4) = L \times (L - 6) + 68\]

Now, let's solve for \(L\):

\[L^2 - 4L + 2L - 8 = L^2 - 6L + 68\]

Combine like terms:

\[L^2 - 2L - 8 = L^2 - 6L + 68\]

Subtract \(L^2\) from both sides:

\[-2L - 8 = -6L + 68\]

Add \(6L\) to both sides:

\[4L - 8 = 68\]

Add 8 to both sides:

\[4L = 76\]

Divide by 4:

\[L = 19\]

Now that we have the length of the longer side (\(L = 19\)), we can find the length of the shorter side (\(L - 6\)):

\[L - 6 = 19 - 6 = 13\]

Therefore, the correct answer is 13 cm

To estimate the median for grouped data, we can use the cumulative frequency. The median is the middle value, and for grouped data, it falls within the class interval where the cumulative frequency exceeds half of the total frequency.

Let's calculate the median:

1. Find the median position, which is \( \frac{n}{2} \), where \( n \) is the total frequency.

\[ \text{Median Position} = \frac{29 + 40 + 38}{2} = \frac{107}{2} = 53.5 \]

2. Identify the class interval where the cumulative frequency exceeds the median position. In this case, it is the 7-8 age group.

3. Now, use the formula for estimating the median in a grouped frequency distribution:

\[ \text{Median} = L + \frac{\left(\frac{n}{2}\right) - F}{f} \times w \]

where:
- \( L \) is the lower class boundary of the median class,
- \( F \) is the cumulative frequency of the class before the median class,
- \( f \) is the frequency of the median class,
- \( w \) is the width of the median class interval.

In the 7-8 age group:
- \( L = 7 - 0.5 = 6.5 \) (assuming the classes are inclusive),
- \( F = 29 \) (cumulative frequency of the class before the median class),
- \( f = 40 \) (frequency of the median class),
- \( w = 8 - 7 = 1 \) (width of the median class interval).

\[ \text{Median} = 6.5 + \frac{53.5 - 29}{40} \times 1 \]

\[ \text{Median} = 6.5 + \frac{24.5}{40} \]

\[ \text{Median} = 6.5 + 0.6125 \]

\[ \text{Median} \approx 7.1125 \]

Therefore, the closest option is 7.725

Let's denote the initial deposit as \(P\) and the interest rate as \(r\).

After 10 years at simple interest, the amount in the bank account is three times the initial deposit:

\[ A_{10} = P + 10Pr = 3P \]

Now, we can solve for \(r\):

\[ P + 10Pr = 3P \]

\[ 10Pr = 2P \]

\[ r = \frac{2}{10} = 0.2 \]

Now, we want to find out after how many years the amount will be five times the initial deposit:

\[ A_t = P + tPr = 5P \]

Substitute the known values:

\[ P + t(0.2)P = 5P \]

\[ 1 + 0.2t = 5 \]

\[ 0.2t = 4 \]

\[ t = \frac{4}{0.2} \]

\[ t = 20 \]

Therefore, after 20 years, the amount in the bank account will be five times the initial deposit.

So, the correct answer is 20 years

Given \(T_2 = -\frac{2}{3}\) and \(S_\infty = \frac{3}{2}\) for a geometric series, let's find the common ratio \(r\).

The formula for the second term (\(T_2\)) in a geometric series is given by \(T_2 = ar\), where \(a\) is the first term and \(r\) is the common ratio.

We are given \(T_2 = -\frac{2}{3}\), so \(ar = -\frac{2}{3}\) ---equation (i).

The formula for the sum to infinity (\(S_\infty\)) in a geometric series is given by \(S_\infty = \frac{a}{1 - r}\).

We are given \(S_\infty = \frac{3}{2}\), so \(\frac{a}{1 - r} = \frac{3}{2}\) ---equation (ii).

Now, let's solve for \(a\) in terms of \(r\) using equation (i):

\[ar = -\frac{2}{3}\]

\[a = -\frac{2}{3r}\] ---equation (iii).

Substitute equation (iii) into equation (ii):

\[\frac{-\frac{2}{3r}}{1 - r} = \frac{3}{2}\]

Solve for \(r\):

\[-\frac{2}{3r(1 - r)} = \frac{3}{2}\]

Cross-multiply:

\[-4 = 9r(1 - r)\]

Expand and rearrange:

\[9r^2 - 9r - 4 = 0\]

Now, solve this quadratic equation. The solutions for \(r\) are:

\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

where \(a = 9\), \(b = -9\), and \(c = -4\).

\[r = \frac{9 \pm \sqrt{81 + 144}}{18}\]

\[r = \frac{9 \pm \sqrt{225}}{18}\]

\[r = \frac{9 \pm 15}{18}\]

Two possible values for \(r\) are:

\[r_1 = \frac{24}{18} = \frac{4}{3}\]

\[r_2 = \frac{-6}{18} = -\frac{1}{3}\]

For a geometric series, the common ratio (\(|r|\)) must be less than 1 for the series to converge to a finite value. Therefore, \(r = -\frac{1}{3}\) is the valid solution.

Hence, the correct answer is \(-\frac{1}{3}\)

The area \(A\) of a rectangle is given by the formula \(A = \text{length} \times \text{width}\).

Given that the length is 38 m and the width is 52 m, we can calculate the area:

\[A = 38 \times 52\]

\[A = 1976 \, \text{m}^2\]

Now, let's consider the possible range of values for the area due to the measurements being correct to the nearest meter.

For the lower bound, we can calculate the area using the lower limits of the lengths:

\[A_{\text{lower}} = (38 - 0.5) \times (52 - 0.5)\]

\[A_{\text{lower}} = 37.5 \times 51.5\]

\[A_{\text{lower}} = 1931.25 \, \text{m}^2\]

For the upper bound, we can calculate the area using the upper limits of the lengths:

\[A_{\text{upper}} = (38 + 0.5) \times (52 + 0.5)\]

\[A_{\text{upper}} = 38.5 \times 52.5\]

\[A_{\text{upper}} = 2021.25 \, \text{m}^2\]

Therefore, the range of possible values for the area is \(1931.25 \, \text{m}^2 \leq A < 2021.25 \, \text{m}^2\).

So, the correct answer is \(1931.25 \, \text{m}^2 \leq A < 2021.25 \, \text{m}^2\)

To express \(16.54 \times 10^{-5} - 6.76 \times 10^{-8} + 0.23 \times 10^{-6}\) in standard form, we need to perform the arithmetic and then rewrite the result in standard form.

\[16.54 \times 10^{-5} - 6.76 \times 10^{-8} + 0.23 \times 10^{-6}\]

First, combine the like terms:

\[= 16.54 \times 10^{-5} + 0.23 \times 10^{-6} - 6.76 \times 10^{-8}\]

Now, let's add the coefficients:

\[= (16.54 + 0.23) \times 10^{-5} - 6.76 \times 10^{-8}\]

\[= 16.77 \times 10^{-5} - 6.76 \times 10^{-8}\]

Now, rewrite in standard form:

\[= 1.677 \times 10^{-4} - 6.76 \times 10^{-8}\]

Since the second term is much smaller than the first term, we can disregard it for the purpose of expressing the result in standard form.

Therefore, the expression \(16.54 \times 10^{-5} - 6.76 \times 10^{-8} + 0.23 \times 10^{-6}\) in standard form is approximately \(1.677 \times 10^{-4}\).

So, the correct answer is \(1.66 \times 10^{-4}\)

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a plane is given by the distance formula:

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

In this case, the coordinates of points P and Q are \((-3, -14)\) and \((t, -5)\) respectively.

The distance between P and Q is given as 9 units. Substituting the coordinates into the formula:

\[9 = \sqrt{(t - (-3))^2 + ((-5) - (-14))^2}\]

\[9 = \sqrt{(t + 3)^2 + 9^2}\]

\[81 = (t + 3)^2 + 81\]

Now, let's solve for \(t\):

\[0 = (t + 3)^2\]

The only solution to this quadratic equation is \(t = -3\).

Therefore, the correct answer is -3

The given binary operations are:

\[a * b = a^2b\]
\[a ^ b = 2a + b\]

Now, let's evaluate \((-4 * 2) ^ (7 * -1)\):

\[(-4 * 2) = (-4)^2 \times 2 = 16 \times 2 = 32\]

\[ (7 * -1) = 7^2 \times (-1) = 49 \times (-1) = -49\]

Now, substitute these values into \(a ^ b = 2a + b\):

\[ (32) ^ (-49) = 2 \times 32 + (-49) = 64 - 49 = 15\]

So, the correct answer is 15

Let's evaluate the definite integral \(\int_0^1 (4x - 6 \cdot 3\sqrt{x^2}) \, dx\).

First, simplify the integrand:

\[4x - 6 \cdot 3\sqrt{x^2} = 4x - 18\sqrt{x^2}.\]

Now, integrate each term separately:

\[ \int_0^1 (4x - 18\sqrt{x^2}) \, dx = \left[2x^2 - 6x\sqrt{x^2}\right]_0^1.\]

Evaluate the expression at the upper and lower limits:

\[ \left[2(1)^2 - 6(1)\sqrt{1^2}\right] - \left[2(0)^2 - 6(0)\sqrt{0^2}\right] \]

Simplify further:

\[ (2 - 6) - 0 = -4.\]

Therefore, the correct answer is \(-\frac{8}{5}\)