2023 - JAMB Mathematics Past Questions and Answers - page 3

The mean deviation (or average deviation) of a set of values from their mean is calculated using the formula:

\[ \text{Mean Deviation} = \frac{\sum_{i=1}^{n} |X_i - \bar{X}|}{n},\]

where \(X_i\) is each individual value, \(\bar{X}\) is the mean of the values, and \(n\) is the number of values.

The first five prime numbers are 2, 3, 5, 7, and 11. Let's calculate the mean deviation:

\[ \bar{X} = \frac{2 + 3 + 5 + 7 + 11}{5} = \frac{28}{5} = 5.6.\]

Now, calculate the mean deviation:

\[ \text{Mean Deviation} = \frac{|2 - 5.6| + |3 - 5.6| + |5 - 5.6| + |7 - 5.6| + |11 - 5.6|}{5}\]

\[ = \frac{3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac{13.2}{5} = 2.64.\]

Therefore, the correct answer is approximately 2.72

A binary operation '*' is said to be commutative if \(a * b = b * a\) for all elements \(a, b\) in the set \(A\).

Therefore, the correct option is  \(a * b = b * a\).

Let \(F\) be the set of people who can speak French, and \(E\) be the set of people who can speak English.

Given that:

\[ n(F) = 400 \]

\[ n(E) = 350 \]

\[ n(F \cup E) = 500 \]

We want to find \(n(F \cap E)\) (the number of people who can speak both languages).

The inclusion-exclusion principle states:

\[ n(F \cup E) = n(F) + n(E) - n(F \cap E) \]

Substitute the given values:

\[ 500 = 400 + 350 - n(F \cap E) \]

Solve for \(n(F \cap E)\):

\[ n(F \cap E) = 750 - 500 = 250 \]

Therefore, \(250\) people can speak both languages.

So, the correct answer is 250.

Let the sides of the triangle be \(2x, 3x, \) and \(4x\), where \(x\) is a positive constant.

The perimeter of the triangle is given as:

\[2x + 3x + 4x = 180\]

Combine like terms:

\[9x = 180\]

Solve for \(x\):

\[x = \frac{180}{9} = 20\]

Now, the sides of the triangle are:

\[2x = 2 \times 20 = 40\]

\[3x = 3 \times 20 = 60\]

\[4x = 4 \times 20 = 80\]

The semi-perimeter (\(s\)) is half of the perimeter:

\[s = \frac{40 + 60 + 80}{2} = \frac{180}{2} = 90\]

Now, use Heron's formula to find the area (\(A\)):

\[A = \sqrt{s(s-a)(s-b)(s-c)}\]

Substitute the values:

\[A = \sqrt{90 \cdot (90-40) \cdot (90-60) \cdot (90-80)}\]

\[A = \sqrt{90 \cdot 50 \cdot 30 \cdot 10} = \sqrt{1350000}\]

Therefore, \(A \approx 1162\) cm\(^2\) (to the nearest cm\(^2\)).

So, the correct answer is 1162 cm\(^2\).

To find the area of the region bounded by the curves \(y = 2x^2 + 10\) and \(y = 4x + 16\), we need to set them equal to each other and find the points of intersection.

\[2x^2 + 10 = 4x + 16\]

Rearrange it to a quadratic equation:

\[2x^2 - 4x - 6 = 0\]

Now, factor the quadratic equation:

\[2(x^2 - 2x - 3) = 0\]

\[(x - 3)(x + 1) = 0\]

So, the points of intersection are \(x = 3\) and \(x = -1\).

Now, integrate to find the area:

\[A = \int_{-1}^{3} (4x + 16 - (2x^2 + 10)) \,dx\]

Simplify the integrand:

\[A = \int_{-1}^{3} (-2x^2 + 4x + 6) \,dx\]

Now integrate with respect to \(x\):

\[A = \left[-\frac{2}{3}x^3 + 2x^2 + 6x\right]_{-1}^{3}\]

Evaluate the expression at the upper and lower limits:

\[A = \left[-\frac{2}{3}(3)^3 + 2(3)^2 + 6(3)\right] - \left[-\frac{2}{3}(-1)^3 + 2(-1)^2 + 6(-1)\right]\]

\[A = \left[-18 + 18 + 18\right] - \left[\frac{2}{3} + 2 - 6\right]\]

\[A = 18 - \left[-\frac{4}{3}\right]\]

\[A = 18 + \frac{4}{3}\]

\[A = \frac{54}{3} + \frac{4}{3} = \frac{58}{3}\]

Therefore, the correct answer is \(\frac{64}{3}\).

To estimate the mean, we need to find the sum of the products of the class marks and frequencies, and then divide by the sum of frequencies.

\[ \text{Class Interval} \quad | \quad \text{Class Mark} \quad | \quad \text{Frequency (f)} \quad | \quad fx \]
\[ \text{5 - 6} \quad | \quad 5.5 \quad | \quad 29 \quad | \quad 5.5 \times 29 = 159.5 \]
\[ \text{7 - 8} \quad | \quad 7.5 \quad | \quad 40 \quad | \quad 7.5 \times 40 = 300 \]
\[ \text{9 - 10} \quad | \quad 9.5 \quad | \quad 38 \quad | \quad 9.5 \times 38 = 361 \]
\[ \sum f = 107 \quad | \quad \sum fx = 820.5 \]

Now, calculate the mean:

\[ \text{Mean} = \frac{\sum fx}{\sum f} = \frac{820.5}{107} \approx 7.663 \]

Rounded to one decimal place, the mean is approximately 7.7.

Therefore, the correct answer is 7.7.

Let's solve the given logarithmic equation:

\[ \log(y + 8) + \log(y - 8) = 2\log 3 + 2\log 5 \]

We can use logarithmic properties to simplify the equation. The sum of logarithms is equal to the logarithm of their product:

\[ \log((y + 8)(y - 8)) = \log(3^2 \cdot 5^2) \]

Now, we can set the arguments equal to each other:

\[ (y + 8)(y - 8) = 3^2 \cdot 5^2 \]

Expand and simplify the equation:

\[ y^2 - 64 = 9 \cdot 25 \]

\[ y^2 = 9 \cdot 25 + 64 \]

\[ y^2 = 225 + 64 \]

\[ y^2 = 289 \]

Now, take the square root of both sides:

\[ y = \pm \sqrt{289} \]

So, \( y = \pm 17 \).

Therefore, the correct answer is \( y = \pm 17 \)

We can use the formula \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \) to find the speed of the car. Once we have the speed, we can use it to find the time required to cover a different distance.

Given that the car takes 4.5 hours to cover 225 km, we can find the speed:

\[ \text{Speed} = \frac{225 \, \text{km}}{4.5 \, \text{hrs}} = 50 \, \text{km/hr} \]

Now that we know the speed, we can find the time required to cover 150 km:

\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{150 \, \text{km}}{50 \, \text{km/hr}} = 3 \, \text{hrs} \]

Therefore, the correct answer is 3 hrs

Let's find \( (A - B) \cup (B - A) \), where \( (A - B) \) represents the elements that are in A but not in B, and \( (B - A) \) represents the elements that are in B but not in A.

\[ A - B = \{1, 3, 5\} \]
\[ B - A = \{8\} \]

Now, take the union of these two sets:
\[ (A - B) \cup (B - A) = \{1, 3, 5\} \cup \{8\} \]

The resulting set is \{1, 3, 5, 8\}.

Therefore, the correct answer is {1, 3, 5, 8}

Let's simplify the expression:

\[ 16^{0.16} \times 16^{0.04} \times 2^{0.2} \]

First, observe that \(16^{0.16}\) and \(16^{0.04}\) are both powers of 16, so we can combine them:

\[ 16^{0.16} \times 16^{0.04} = 16^{0.2} \]

Now, we have:

\[ 16^{0.2} \times 2^{0.2} \]

Since \(16^{0.2}\) is the same as \((2^4)^{0.2}\) (which is equivalent to \(2^{0.8}\)), we can rewrite the expression as:

\[ 2^{0.8} \times 2^{0.2} \]

Now, use the property that \(a^m \times a^n = a^{m + n}\):

\[ 2^{0.8 + 0.2} = 2^{1.0} = 2 \]