1) **Problem:** Given $a = 2^2 \times 3^2$, $b = 2 \times 3 \times 5$, and $c = 2^2 \times 3 \times 7$, find $x$ such that the LCM$(a,b,c) = 3780$. **Step 1:** Express $a,b,c$ in prime factorization: $$a = 2^2 \times 3^2, \quad b = 2^1 \times 3^1 \times 5^1, \quad c = 2^2 \times 3^1 \times 7^1.$$ **Step 2:** LCM is found by taking the highest power of each prime factor: $$\text{LCM} = 2^{\max(2,1,2)} \times 3^{\max(2,1,1)} \times 5^{\max(0,1,0)} \times 7^{\max(0,0,1)} = 2^2 \times 3^2 \times 5^1 \times 7^1.$$ **Step 3:** Calculate the LCM numerical value: $$2^2 = 4, \quad 3^2 = 9, \quad 5 = 5, \quad 7 = 7,$$ $$\text{LCM} = 4 \times 9 \times 5 \times 7 = 1260.$$ **Step 4:** Given LCM$(a,b,c) = 3780$, which is $3 \times 1260$. To get this, multiply by $x$: $$x \times 1260 = 3780$$ $$x = \frac{3780}{1260} = 3.$$ **Answer for 1:** $x = 3$ (Option c). 2) **Problem:** Find the shortest distance from point $(2,3)$ to the $y$-axis. **Step 1:** Shortest distance from a point to the $y$-axis is the absolute $x$-coordinate value. **Step 2:** The point is $(2, 3)$, so distance to $y$-axis is: $$|2| = 2.$$ **Answer for 2:** $2$ (Option a). 3) **Problem:** For lines $3x + 2ky =2$ and $2x + 15y +1=0$, find all values of $k$ such that the lines are *not* parallel. **Step 1:** Lines are parallel if their slopes are equal. **Step 2:** Find slope of first line: $$3x + 2ky = 2 \implies y = \frac{2 - 3x}{2k} = -\frac{3}{2k}x + \frac{2}{2k}.$$ Slope = $m_1 = -\frac{3}{2k}$. **Step 3:** Find slope of second line: $$2x + 15y + 1 = 0 \implies 15y = -2x -1 \implies y = -\frac{2}{15}x - \frac{1}{15}.$$ Slope = $m_2 = -\frac{2}{15}$. **Step 4:** For lines to not be parallel, slopes must not be equal: $$-\frac{3}{2k} \ne -\frac{2}{15} \implies \frac{3}{2k} \ne \frac{2}{15}.$$ **Step 5:** Solve equality condition to find excluded value: $$\frac{3}{2k} = \frac{2}{15} \implies 3 \times 15 = 2k \times 2 \implies 45 = 4k \implies k = \frac{45}{4} = \frac{15}{4}.$$ **Step 6:** So, lines are parallel if $k=\frac{15}{4}$. **Answer for 3:** $k \ne \frac{15}{4}$ (Option b). 4) **Problem:** Quadrilateral $ABCD$ circumscribes a circle. Given $BC=7$ cm, $CD=4$ cm, $AD=3$ cm, find $AB$. **Step 1:** In a tangential quadrilateral (one circumscribing a circle), sum of lengths of opposite sides are equal: $$AB + CD = BC + AD.$$ **Step 2:** Substitute known values: $$AB + 4 = 7 + 3 = 10.$$ **Step 3:** Solve for $AB$: $$AB = 10 - 4 = 6 \, \text{cm}.$$ **Answer for 4:** $6$ cm (Option c).