1. Problem: Two cars A and B take 30 minutes and $p$ minutes respectively to complete 1 round. They meet first time after 90 minutes at the starting point. Highest Common Factor (HCF) of 30 and $p$ is 15. Find the value of $p$. Step 1: The first time they meet at starting point is the Least Common Multiple (LCM) of their times. Given time for A = 30 minutes, so LCM(30, $p$) = 90 minutes. Step 2: Use the relation for LCM and HCF: $$\text{LCM}(30,p) \times \text{HCF}(30,p) = 30 \times p$$ Substituting the known values: $$90 \times 15 = 30 \times p$$ Step 3: Calculate $p$: $$1350 = 30p$$ $$p = \frac{1350}{30} = 45$$ Answer: $p = 45$ minutes, which corresponds to option (a). 2. Problem: If the LCM of two positive integers $a$ and $b$ is equal to the product $ab$, find their HCF. Step 1: Recall the relation: $$\text{LCM}(a,b) \times \text{HCF}(a,b) = a \times b$$ Given $$\text{LCM}(a,b) = ab$$ Step 2: Substitute: $$ab \times \text{HCF}(a,b) = ab$$ Step 3: Divide both sides by $ab$: $$\text{HCF}(a,b) = 1$$ Answer: The HCF is 1, corresponding to option (c). 3. Problem: Zeroes of polynomial $$x^2 + (a+1) x + b$$ are 2 and -3. Find values of $a$ and $b$. Step 1: Sum of zeroes is given by $$-\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -(a+1)$$ Sum of roots: $$2 + (-3) = -1$$ Step 2: Set sum equations equal: $$- (a + 1) = -1$$ $$a + 1 = 1$$ $$a = 0$$ Step 3: Product of roots is $$\frac{b}{1} = 2 \times (-3) = -6$$ $$b = -6$$ Answer: $a = 0$, $b = -6$, which is option (d). 4. Problem: Find the ratio of sum and product of roots of quadratic equation $$5x^{2} - 6x + 21 = 0$$ Step 1: Sum of roots = $$-\frac{-6}{5} = \frac{6}{5}$$ Step 2: Product of roots = $$\frac{21}{5}$$ Step 3: Ratio of sum to product: $$\frac{6/5}{21/5} = \frac{6}{5} \times \frac{5}{21} = \frac{6}{21} = \frac{2}{7}$$ Answer: The ratio is $2 : 7$, which is option (b).