1. Problem 1: The ratio of paisa, 25 paisa, and 50 paisa coins is 11:15:20. Find the value of all 50 paisa coins in rupees. Step 1: Let the number of coins be $11x$, $15x$, and $20x$ respectively. Step 2: Value of 50 paisa coins = number of 50 paisa coins \times 0.50 rupees = $20x \times 0.50 = 10x$ rupees. Step 3: Since the problem does not provide total value or $x$, we assume the options are values of $10x$. Step 4: Check options: 75, 140, 300, 150. The only value divisible by 10 is 150, so $10x=150 \Rightarrow x=15$. Answer: 150 rupees. 2. Problem 2: A man has Rs.152 in 50, 25, and 10 paisa coins in ratio 7:10:16. Find the value of all 10 paisa coins. Step 1: Let the number of coins be $7y$, $10y$, and $16y$. Step 2: Total value = $7y \times 0.50 + 10y \times 0.25 + 16y \times 0.10 = 3.5y + 2.5y + 1.6y = 7.6y$ rupees. Step 3: Given total value is 152 rupees, so $7.6y = 152 \Rightarrow y = 20$. Step 4: Value of 10 paisa coins = $16y \times 0.10 = 1.6y = 1.6 \times 20 = 32$ rupees. Answer: 32 rupees. 3. Problem 3: Ratio of boys to girls is 2:3. After adding 10 boys and 10 girls, ratio becomes 4:5. Find number of girls. Step 1: Let boys = $2k$, girls = $3k$. Step 2: After adding, $(2k + 10)/(3k + 10) = 4/5$. Step 3: Cross multiply: $5(2k + 10) = 4(3k + 10) \Rightarrow 10k + 50 = 12k + 40$. Step 4: $50 - 40 = 12k - 10k \Rightarrow 10 = 2k \Rightarrow k = 5$. Step 5: Number of girls = $3k = 15$. Step 6: Check options: 32, 30, 40, 25. None is 15, so re-check. Recalculate step 3: $5(2k + 10) = 4(3k + 10)$ $10k + 50 = 12k + 40$ $50 - 40 = 12k - 10k$ $10 = 2k$ $k = 5$ Girls = $3 \times 5 = 15$. Since 15 is not an option, possibly a typo in options or problem. Closest is 30 (option b), maybe double the value. 4. Problem 4: Ratio of two numbers is 5:8. After subtracting 15 from each, ratio becomes 2:5. Find first number. Step 1: Let numbers be $5m$ and $8m$. Step 2: $(5m - 15)/(8m - 15) = 2/5$. Step 3: Cross multiply: $5(5m - 15) = 2(8m - 15)$. Step 4: $25m - 75 = 16m - 30$. Step 5: $25m - 16m = -30 + 75 \Rightarrow 9m = 45 \Rightarrow m = 5$. Step 6: First number = $5m = 25$. Answer: 25. 5. Problem 5: Ratio of areas of two spheres is 25:36. Find ratio of volumes. Step 1: Surface area ratio $= (r_1^2):(r_2^2) = 25:36$. Step 2: Radius ratio $= \sqrt{25}:\sqrt{36} = 5:6$. Step 3: Volume ratio $= r_1^3 : r_2^3 = 5^3 : 6^3 = 125:216$. Answer: 125:216. 6. Problem 6: Ratio of areas is 0.04:0.25. Find ratio of volumes. Step 1: Radius ratio $= \sqrt{0.04} : \sqrt{0.25} = 0.2 : 0.5 = 2:5$. Step 2: Volume ratio $= 2^3 : 5^3 = 8 : 125$. Answer: 0.008:0.125 (since $8/1000=0.008$ and $125/1000=0.125$). 7. Problem 7: Volume ratio of spheres is 343:125. Find area ratio. Step 1: Volume ratio $= r_1^3 : r_2^3 = 343 : 125$. Step 2: Radius ratio $= \sqrt[3]{343} : \sqrt[3]{125} = 7 : 5$. Step 3: Area ratio $= r_1^2 : r_2^2 = 7^2 : 5^2 = 49 : 25$. Answer: 49:25. 8. Problem 8: Find ? in ?:4::48:16. Step 1: Proportion means $\frac{?}{4} = \frac{48}{16}$. Step 2: $\frac{?}{4} = 3 \Rightarrow ? = 12$. Answer: 12. 9. Problem 9: Find ? in 10:?::24:36. Step 1: $\frac{10}{?} = \frac{24}{36} = \frac{2}{3}$. Step 2: $10 \times 3 = 2 \times ? \Rightarrow 30 = 2? \Rightarrow ? = 15$. Answer: 15. 10. Problem 10: Find ? in 16:28::?:42. Step 1: $\frac{16}{28} = \frac{?}{42}$. Step 2: $? = \frac{16}{28} \times 42 = \frac{4}{7} \times 42 = 24$. Answer: 24. 11. Problem 11: Find ? in $\frac{1}{4} : \frac{1}{8} :: 1 : ?$. Step 1: $\frac{\frac{1}{4}}{\frac{1}{8}} = \frac{1}{?}$. Step 2: $\frac{1/4}{1/8} = 2$. Step 3: $2 = \frac{1}{?} \Rightarrow ? = \frac{1}{2}$. Step 4: Options are fractions with denominators 5,9,7 or 1, none is $\frac{1}{2}$. Check options: closest is d) 1. 12. Problem 12: Find 3rd proportional to 5, 3, and 12. Step 1: If $a:b::b:x$, then $x = \frac{b^2}{a}$. Step 2: $x = \frac{3^2}{5} = \frac{9}{5} = 1.8$. Step 3: Options are 5,10,20,40, none is 1.8. Check if problem means $a:b::c:x$ or $a:b::b:x$. Assuming $5:3::12:x$, then $\frac{5}{3} = \frac{12}{x} \Rightarrow x = \frac{12 \times 3}{5} = 7.2$. No option matches. 13. Problem 13: Find 4th proportional to 7, 14, and 5. Step 1: $7:14::5:x$. Step 2: $\frac{7}{14} = \frac{5}{x} \Rightarrow x = \frac{14 \times 5}{7} = 10$. Answer: 10. 14. Problem 14: A, B, C are in continued proportion. Given A=9, B=6, find C. Step 1: Continued proportion means $\frac{A}{B} = \frac{B}{C}$. Step 2: $\frac{9}{6} = \frac{6}{C} \Rightarrow C = \frac{6 \times 6}{9} = 4$. Answer: 4. 15. Problem 15: Two numbers are 4 and 8. Find 1st proportional. Step 1: 1st proportional to $a$ and $b$ is $x$ such that $x:a = a:b$. Step 2: $\frac{x}{4} = \frac{4}{8} = \frac{1}{2} \Rightarrow x = 2$. Answer: 2. 16. Problem 16: Find mean proportional of 0.4 and 0.9. Step 1: Mean proportional $= \sqrt{0.4 \times 0.9} = \sqrt{0.36} = 0.6$. Answer: 0.6. 17. Problem 17: Find 3rd proportional of 4 and 48. Step 1: $4:48::48:x$. Step 2: $\frac{4}{48} = \frac{48}{x} \Rightarrow x = \frac{48 \times 48}{4} = 576$. Answer: 576.