2023 - JAMB Mathematics Past Questions and Answers - page 4

The number of ways to choose a committee with 3 women and 2 men is given by \(^4C_3 \times ^6C_2\). Similarly, the number of ways to choose a committee with 4 women and 1 man is given by \(^4C_4 \times ^6C_1\).

\[ ^4C_3 \times ^6C_2 + ^4C_4 \times ^6C_1 = 4 \times 15 + 1 \times 6 = 66 \]

Let's denote the number of kids' tickets as \( k \) and the number of adult tickets as \( 5k \).

The total sales can be expressed as the sum of the product of the number of tickets and their respective prices:

\[ 520 \times 5k + 250 \times k = 171,000 \]

Now, solve for \( k \):

\[ 2600k + 250k = 171,000 \]

\[ 2850k = 171,000 \]

\[ k = \frac{171,000}{2850} \]

\[ k = 60 \]

So, the number of kids' tickets sold is 60.

Let's solve the quadratic inequality:

\[ x^2 - x - 4 \leq 2 \]

First, bring all terms to one side to form a quadratic expression:

\[ x^2 - x - 6 \leq 0 \]

Now, factor the quadratic expression:

\[ (x - 3)(x + 2) \leq 0 \]

Now, identify the intervals where this expression is less than or equal to zero by considering the signs of the factors:

1. When \( x - 3 \leq 0 \) and \( x + 2 \geq 0 \):
   - \( x \leq 3 \)
   - \( x \geq -2 \)

2. When \( x - 3 \geq 0 \) and \( x + 2 \leq 0 \):
   - \( x \geq 3 \)
   - \( x \leq -2 \)

Combine the intervals:

\[ -2 \leq x \leq 3 \]

The locus of points equidistant from two intersecting lines is the pair of bisectors of the angles between the two lines. This is based on the property that points on these bisectors are equidistant from the intersecting lines. 

To find the value of \(k\), substitute the coordinates of point A(-2, \(k\)) into the equation \(3y + 6x = 48\):

\[3k + 6(-2) = 48\]

Simplify the equation:

\[3k - 12 = 48\]

Add 12 to both sides:

\[3k = 60\]

Divide both sides by 3:

\[k = 20\]

The differentiation of the given function \(y = \sqrt[3]{x^2(2x - x^2)}\) involves the product and chain rules. Let's break down the solution:

\[y = \sqrt[3]{x^2(2x - x^2)}\]

We can rewrite this as:

\[y = x^{2/3}(2x - x^2)^{1/3}\]

Now, applying the product rule \((uv)' = u'v + uv'\) and chain rule \((g(h(x)))' = g'(h(x)) \cdot h'(x)\), we get:

\[y' = \frac{2}{3}x^{-1/3}(2x - x^2)^{1/3} + x^{2/3} \cdot \frac{1}{3}(2x - x^2)^{-2/3}(2 - 2x)\]

Simplify the expression:

\[y' = \frac{2}{3x^{1/3}}(2x - x^2)^{1/3} - \frac{2x^{2/3}(2 - 2x)}{3(2x - x^2)^{2/3}}\]

Now, let's factor out common terms:

\[y' = \frac{2x^{2/3}}{3(2x - x^2)^{2/3}} \left(3 - 3x - x^{1/3}(2x - x^2)^{1/3}\right)\]

The simplified expression for the derivative is:

\[y' = \frac{2x^{2/3}}{3(2x - x^2)^{2/3}} \left(3 - 3x - x^{1/3}(2x - x^2)^{1/3}\right)\]

Comparing with the given options, it seems there might be a mistake in the assumed correct option. The correct differentiation seems to be:

\[y' = \frac{2x^{2/3}}{3(2x - x^2)^{2/3}} \left(3 - x(2x - x^2)^{1/3}\right)\]

To find the value of \(y\) if \(402_y = 102_{ten}\), we need to interpret the given expression. The subscript \(y\) indicates that the number is written in base \(y\). The number \(402_y\) can be expressed as:

\[4 \times y^2 + 0 \times y^1 + 2 \times y^0 = 102_{ten}\]

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

\[4y^2 + 2 = 102\]

Subtract 2 from both sides:

\[4y^2 = 100\]

Divide by 4:

\[y^2 = 25\]

Take the square root of both sides:

\[y = 5\]

To find the probability of getting at least one head in three coin tosses, we can use the complement rule. The complement of "at least one head" is "no heads," i.e., getting all tails.

The probability of getting tails in a single toss is \( \frac{1}{2} \), so the probability of getting tails in all three tosses is \( \left(\frac{1}{2}\right)^3 \).

Now, the probability of getting at least one head is the complement of getting all tails:

\[ P(\text{at least one head}) = 1 - P(\text{no heads}) \]
\[ P(\text{at least one head}) = 1 - \left(\frac{1}{2}\right)^3 \]
\[ P(\text{at least one head}) = 1 - \frac{1}{8} \]
\[ P(\text{at least one head}) = \frac{7}{8} \]

To rationalize the given expression \(\frac {3 - \sqrt 3}{2 + \sqrt 3}\), we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac {3 - \sqrt 3}{2 + \sqrt 3} \times \frac {2 - \sqrt 3}{2 - \sqrt 3} \]

This simplifies to:

\[ \frac {(3 - \sqrt 3)(2 - \sqrt 3)}{(2 + \sqrt 3)(2 - \sqrt 3)} \]

Expanding the numerator and denominator:

Numerator:
\[ (3 - \sqrt 3)(2 - \sqrt 3) = 6 - 3\sqrt 3 - 2\sqrt 3 + 3 = 9 - 5\sqrt 3 \]

Denominator:
\[ (2 + \sqrt 3)(2 - \sqrt 3) = 4 - 3 = 1 \]

Now, the expression becomes:

\[ \frac {9 - 5\sqrt 3}{1} = 9 - 5\sqrt 3 \]

Comparing this with \(a + b\sqrt 3\), we get:
\[ a = 9 \]
\[ b = -5 \]

Therefore, the values are \(a = 9\) and \(b = -5\). 

To solve for \(x\), let's start with the given formula:

\[ y = \frac{3x - 9c}{4x + 5d} \]

Cross-multiplying to eliminate the fraction:

\[ y(4x + 5d) = 3x - 9c \]

Expanding and simplifying:

\[ 4xy + 5dy = 3x - 9c \]

Bringing all \(x\)-related terms to one side:

\[ 4xy - 3x = -9c - 5dy \]

Factoring out \(x\):

\[ x(4y - 3) = -9c - 5dy \]

Now, solving for \(x\):

\[ x = \frac{-9c - 5dy}{4y - 3} \]

This matches Option A:

\[ x = \frac{-9c - 5dy}{4y - 3} \]