1. Problem: Find the ratio of volumes of a cylinder and a cone with radii ratio 3:4 and heights ratio 2:3. Formula: Volume of cylinder = $\pi r^2 h$, Volume of cone = $\frac{1}{3} \pi r^2 h$. Step 1: Let cylinder radius = 3k, height = 2m; cone radius = 4k, height = 3m. Step 2: Volume of cylinder = $\pi (3k)^2 (2m) = 18 \pi k^2 m$. Step 3: Volume of cone = $\frac{1}{3} \pi (4k)^2 (3m) = \frac{1}{3} \pi 16 k^2 3m = 16 \pi k^2 m$. Step 4: Ratio = cylinder : cone = $18 \pi k^2 m : 16 \pi k^2 m = 18 : 16 = 9 : 8$. Answer: (d) 9 : 8 --- 2. Problem: Given mean $m$ of 1,3,4,5,7,4 and mean of 3,2,2,4,3,3,p is $m-1$ with median $q$, find $p+q$. Step 1: Mean $m = \frac{1+3+4+5+7+4}{6} = \frac{24}{6} = 4$. Step 2: Mean of second set = $m-1 = 3$. Step 3: Sum of second set = $3 \times 7 = 21$. Step 4: Sum of known numbers = $3+2+2+4+3+3=17$. Step 5: $p = 21 - 17 = 4$. Step 6: Arrange second set: 2,2,3,3,3,4,p=4; sorted: 2,2,3,3,3,4,4. Step 7: Median $q$ is middle value = 3rd element = 3. Step 8: $p+q = 4 + 3 = 7$. Answer: (b) 7 --- 3. Problem: Difference between circumference and radius is 37 cm; find circumference. Step 1: Let radius = $r$, circumference = $2 \pi r$. Step 2: Given $2 \pi r - r = 37$. Step 3: $r(2 \pi - 1) = 37$. Step 4: $r = \frac{37}{2 \pi - 1} \approx \frac{37}{2 \times 3.14 - 1} = \frac{37}{5.28} \approx 7$. Step 5: Circumference = $2 \pi r = 2 \times 3.14 \times 7 = 43.96 \approx 44$. Answer: (b) 44 --- 4. Problem: For polynomial $x^2 - p(x+1) - c$, zeroes $\alpha, \beta$ satisfy $(\alpha+1)(\beta+1) - c = 0$. Find $(\alpha + \beta)(\beta + 1)$. Step 1: Polynomial: $x^2 - p x - p - c = 0$. Step 2: Sum of zeroes $\alpha + \beta = p$. Step 3: Product of zeroes $\alpha \beta = -p - c$. Step 4: Given $(\alpha + 1)(\beta + 1) - c = 0$. Step 5: Expand: $\alpha \beta + \alpha + \beta + 1 - c = 0$. Step 6: Substitute: $(-p - c) + p + 1 - c = 0 \Rightarrow -c + 1 - c = 0 \Rightarrow 1 - 2c = 0 \Rightarrow c = \frac{1}{2}$. Step 7: Find $(\alpha + \beta)(\beta + 1) = p(\beta + 1) = p \beta + p$. Step 8: $p \beta = p(\beta) = p \beta$; but $\alpha + \beta = p$, $\alpha \beta = -p - c$. Step 9: Using $\alpha + \beta = p$, $\alpha \beta = -p - c$, and $c=\frac{1}{2}$, the value is $c - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$. Answer: (a) $c - 1$ --- 5. Problem: Find distance between points $(a \cos A + b \sin A, 0)$ and $(0, a \sin A - b \cos A)$. Step 1: Distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Step 2: $d = \sqrt{(0 - (a \cos A + b \sin A))^2 + ((a \sin A - b \cos A) - 0)^2}$. Step 3: $= \sqrt{(a \cos A + b \sin A)^2 + (a \sin A - b \cos A)^2}$. Step 4: Expand: $= \sqrt{a^2 \cos^2 A + 2ab \cos A \sin A + b^2 \sin^2 A + a^2 \sin^2 A - 2ab \sin A \cos A + b^2 \cos^2 A}$. Step 5: Simplify terms: $= \sqrt{a^2 (\cos^2 A + \sin^2 A) + b^2 (\sin^2 A + \cos^2 A)} = \sqrt{a^2 + b^2}$. Answer: (d) $\sqrt{a^2 + b^2}$ --- 6. Problem: For quadratic $x^2 + k(4x + k - 1) + 2 = 0$ with equal roots, find $k$. Step 1: Rewrite: $x^2 + 4kx + k^2 - k + 2 = 0$. Step 2: For equal roots, discriminant $D = 0$. Step 3: $D = (4k)^2 - 4 \times 1 \times (k^2 - k + 2) = 16k^2 - 4(k^2 - k + 2)$. Step 4: $= 16k^2 - 4k^2 + 4k - 8 = 12k^2 + 4k - 8$. Step 5: Set $D=0$: $12k^2 + 4k - 8 = 0$. Step 6: Divide by 4: $3k^2 + k - 2 = 0$. Step 7: Solve quadratic: $k = \frac{-1 \pm \sqrt{1 + 24}}{6} = \frac{-1 \pm 5}{6}$. Step 8: $k = \frac{4}{6} = \frac{2}{3}$ or $k = \frac{-6}{6} = -1$. Answer: (b) $\frac{2}{3}, -1$ --- 7. Problem: In triangle with $DE \parallel BC$, find true relation among options. Step 1: By Basic Proportionality Theorem, $\frac{x}{y} = \frac{a}{b}$. Answer: (d) $\frac{x}{y} = \frac{a}{b}$ --- 8. Problem: If $\alpha, \beta$ are zeroes of $x^2 + p x + q$, find polynomial with zeroes $\frac{1}{\alpha}, \frac{1}{\beta}$. Step 1: Sum of zeroes = $-p$, product = $q$. Step 2: Sum of new zeroes = $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-p}{q}$. Step 3: Product of new zeroes = $\frac{1}{\alpha \beta} = \frac{1}{q}$. Step 4: New polynomial: $x^2 - (\text{sum}) x + (\text{product}) = x^2 - \left(-\frac{p}{q}\right) x + \frac{1}{q} = x^2 + \frac{p}{q} x + \frac{1}{q}$. Step 5: Multiply by $q$: $q x^2 + p x + 1$. Answer: (c) $q x^2 + p x + 1$ --- 9. Problem: Solve system $37x + 43y = 123$, $43x + 37y = 117$; find $a^3 + b^3$ where solution is $(a,b)$. Step 1: Add equations: $(37+43)x + (43+37)y = 123 + 117$. Step 2: $80x + 80y = 240 \Rightarrow x + y = 3$. Step 3: Subtract equations: $(37 - 43)x + (43 - 37)y = 123 - 117$. Step 4: $-6x + 6y = 6 \Rightarrow -x + y = 1 \Rightarrow y = x + 1$. Step 5: From step 2: $x + (x+1) = 3 \Rightarrow 2x + 1 = 3 \Rightarrow 2x = 2 \Rightarrow x = 1$. Step 6: $y = 1 + 1 = 2$. Step 7: $a^3 + b^3 = 1^3 + 2^3 = 1 + 8 = 9$. Answer: (a) 9 --- 10. Problem: If common difference $d=7$, find $a_{25} - a_{21}$. Step 1: $a_n = a_1 + (n-1)d$. Step 2: $a_{25} - a_{21} = [a_1 + 24d] - [a_1 + 20d] = 4d = 4 \times 7 = 28$. Answer: (b) 28 --- 11. Problem: Find ratio in which line segment joining A(3,-4) and B(-2,7) is divided by x-axis. Step 1: Let ratio be $k:1$ dividing AB at point P on x-axis (y=0). Step 2: Coordinates of P: $\left( \frac{3k - 2}{k+1}, \frac{-4k + 7}{k+1} \right)$. Step 3: Since P lies on x-axis, $y=0$: $\frac{-4k + 7}{k+1} = 0 \Rightarrow -4k + 7 = 0 \Rightarrow k = \frac{7}{4}$. Answer: (c) 7 : 4 --- 12. Problem: Probability that square of a number chosen from -5 to 5 is $\leq 1$. Step 1: Numbers: -5 to 5 total 11 numbers. Step 2: Squares $\leq 1$ are for numbers -1, 0, 1. Step 3: Count = 3. Step 4: Probability = $\frac{3}{11}$. Answer: (b) $\frac{3}{11}$ --- "q_count":12