1. **Problem:** Find the perimeter of a room whose length is 5 cm longer than its width and whose area is 50 cm\(^2\). 2. **Step 1:** Let the width be $x$ cm. Then the length is $x + 5$ cm. 3. **Step 2:** The area of a rectangle is given by $\text{Area} = \text{length} \times \text{width}$. 4. **Step 3:** Using the given area, we have: $$x(x + 5) = 50$$ 5. **Step 4:** Expand and form the quadratic equation: $$x^2 + 5x - 50 = 0$$ 6. **Step 5:** Solve the quadratic equation using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=5$, and $c=-50$. 7. **Step 6:** Calculate the discriminant: $$\Delta = 5^2 - 4(1)(-50) = 25 + 200 = 225$$ 8. **Step 7:** Find the roots: $$x = \frac{-5 \pm \sqrt{225}}{2} = \frac{-5 \pm 15}{2}$$ 9. **Step 8:** Possible values for $x$: - $x = \frac{-5 + 15}{2} = 5$ - $x = \frac{-5 - 15}{2} = -10$ (not possible since width cannot be negative) 10. **Step 9:** Width $x = 5$ cm, length $x + 5 = 10$ cm. 11. **Step 10:** Perimeter of a rectangle is: $$P = 2(\text{length} + \text{width}) = 2(10 + 5) = 30 \text{ cm}$$ --- **Answer for 1:** 30 (Option C) --- **Problem 2:** Form a quadratic equation for a rectangle with area 50 and length 5 units more than width. **Answer:** From above, the quadratic equation is: $$x^2 + 5x - 50 = 0$$ which corresponds to option B. --- **Problem 3:** Two books are selected from 6 mathematics and 12 accounts books. Find the probability both are mathematics or both are accounts. 1. Total books = 6 + 12 = 18 2. Total ways to select 2 books: $$\binom{18}{2} = \frac{18 \times 17}{2} = 153$$ 3. Ways to select 2 mathematics books: $$\binom{6}{2} = \frac{6 \times 5}{2} = 15$$ 4. Ways to select 2 accounts books: $$\binom{12}{2} = \frac{12 \times 11}{2} = 66$$ 5. Total favorable ways: $$15 + 66 = 81$$ 6. Probability: $$P = \frac{81}{153} = \frac{27}{51} = \frac{9}{17} \approx 0.5294$$ 7. Closest option is D: 0.53 --- **Problem 4:** A line passes through (2,4) with gradient twice its y-intercept. Find the gradient. 1. Let the y-intercept be $c$. 2. Gradient $m = 2c$. 3. Equation of line: $$y = mx + c$$ 4. Substitute point (2,4): $$4 = m(2) + c = 2m + c$$ 5. Substitute $m = 2c$: $$4 = 2(2c) + c = 4c + c = 5c$$ 6. Solve for $c$: $$c = \frac{4}{5} = 0.8$$ 7. Gradient: $$m = 2c = 2 \times 0.8 = 1.6$$ **Answer:** 1.6 (Option B) --- **Problem 5:** Lines joining (-1,1) & (2,-2) and (1,2) & (2,k) are parallel. Find $k$. 1. Gradient of first line: $$m_1 = \frac{-2 - 1}{2 - (-1)} = \frac{-3}{3} = -1$$ 2. Gradient of second line: $$m_2 = \frac{k - 2}{2 - 1} = k - 2$$ 3. Since lines are parallel: $$m_1 = m_2$$ $$-1 = k - 2$$ 4. Solve for $k$: $$k = 1$$ **Answer:** 1 (Option A)