Question (b) - Probability

Here are the solutions to the questions:

Question (b) - Probability A number is selected from each of the sets $\{2, 3, 4\}$ and $\{1, 3, 5\}$. Find the probability that the sum of the two numbers is greater than 3 and less than 7.

Step 1: List all possible sums. The first set has 3 numbers, and the second set has 3 numbers. The total number of possible sums is $3 \times 3 = 9$. The possible sums are: $2+1=3$ $2+3=5$ $2+5=7$ $3+1=4$ $3+3=6$ $3+5=8$ $4+1=5$ $4+3=7$ $4+5=9$

Step 2: Identify favorable outcomes. We need sums that are greater than 3 AND less than 7. These sums are 4, 5, 6. The pairs that result in these sums are: $(3,1) \to 4$ $(2,3) \to 5$ $(4,1) \to 5$ $(3,3) \to 6$ There are 4 favorable outcomes.

Step 3: Calculate the probability. $P = \frac{Number of favorable outcomes}{Total number of outcomes} = \frac{4}{9}$ The probability is $\boxed{\frac{4}{9}}$.

Question 2 (a) - Inequality Solve the inequality: $4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1$.

Step 1: Multiply by the least common multiple (LCM) of the denominators (4 and 8), which is 8, to clear fractions. $8 \left(4 + \frac{3}{4}(x + 2)\right) \leq 8 \left(\frac{3}{8}x + 1\right)$ $32 + 6(x + 2) \leq 3x + 8$

Step 2: Distribute and simplify both sides of the inequality. $32 + 6x + 12 \leq 3x + 8$ $44 + 6x \leq 3x + 8$

Step 3: Gather $x$ terms on one side and constant terms on the other. $6x - 3x \leq 8 - 44$ $3x \leq -36$

Step 4: Solve for $x$. $x \leq \frac{-36}{3}$ $x \leq -12$ The solution to the inequality is $\boxed{x \leq -12}$.

Question 2 (b) - Geometry The diagram shows a rectangle PQRS from which a square of side $x$ cm has been cut. The area of the shaded portion is $484 cm^2$. Find the values of $x$.

Step 1: Determine the dimensions of the large rectangle and the cut-out square. The total length of the rectangle PQRS is $10 cm + x cm + 10 cm = (20+x) \text{ cm}$. The width of the rectangle PQRS is $20 \text{ cm}$. The cut-out square has a side length of $x \text{ cm}$.

Step 2: Set up the equation for the shaded area. The area of the shaded portion is the area of the large rectangle minus the area of the cut-out square.