1. Problem 28: A total of Rs 35000 is invested at three interest rates: 4%, 5%, and 6%. Interest for the first year was Rs 1780. Second year, the amount invested at 6% earned 7% interest; others remained the same. Total second year interest was Rs 1910. Find amount invested at 6%. Let the investments at 4%, 5%, and 6% be $x$, $y$, and $z$. Then: $$x + y + z = 35000$$ Year 1 interest: $$0.04x + 0.05y + 0.06z = 1780$$ Year 2 interest (6% replaced by 7% on $z$): $$0.04x + 0.05y + 0.07z = 1910$$ 2. Subtract year 1 from year 2 interest equations: $$(0.04x - 0.04x) + (0.05y - 0.05y) + (0.07z - 0.06z) = 1910 - 1780$$ $$0.01z = 130$$ 3. Solve for $z$: $$z = \frac{130}{0.01} = 13000$$ So the amount invested at 6% is Rs 13000. --- 4. Problem 29: Wages of 8 men and 6 boys total Rs 3300. Four men earn Rs 450 more than five boys. Find wage of each man. Let wage of a man be $m$ and a boy be $b$. Equations: $$8m + 6b = 3300$$ $$4m = 5b + 450$$ 5. Solve the second equation for $b$: $$4m - 450 = 5b \Rightarrow b = \frac{4m - 450}{5}$$ 6. Substitute $b$ into the first equation: $$8m + 6 \times \frac{4m - 450}{5} = 3300$$ $$8m + \frac{24m - 2700}{5} = 3300$$ 7. Multiply both sides by 5: $$40m + 24m - 2700 = 16500$$ $$64m = 16500 + 2700 = 19200$$ 8. Solve for $m$: $$m = \frac{19200}{64} = 300$$ Each man earns Rs 300. --- 9. Problem 30: 56 is divided into two parts such that 3 times first part exceeds one-third of second part by 48. Find the larger part. Let first part = $x$, second part = $56 - x$. Given: $$3x - \frac{1}{3}(56 - x) = 48$$ 10. Multiply both sides by 3 to clear fraction: $$9x - (56 - x) = 144$$ $$9x - 56 + x = 144$$ $$10x = 200$$ 11. Solve for $x$: $$x = 20$$ Second part is: $$56 - 20 = 36$$ Largest part is 36. --- 12. Problem 31: If $\frac{x}{4} = \frac{y}{3} = \frac{z}{2} = k$ and $7x + 8y + 5z = 31$, find $x$. Express variables in terms of $k$: $$x = 4k, \quad y = 3k, \quad z = 2k$$ 13. Substitute into given equation: $$7(4k) + 8(3k) + 5(2k) = 31$$ $$28k + 24k + 10k = 31$$ $$62k = 31$$ 14. Solve for $k$: $$k = \frac{31}{62} = \frac{1}{2}$$ Calculate $x$: $$x = 4k = 4 \times \frac{1}{2} = 2$$ --- 15. Problem 32: To obtain 700 gallons of 24% acid by mixing 20% and 30% solutions, find gallons of 30% solution. Let $x$ = gallons of 30% solution, then $700 - x$ gallons of 20% solution. Concentration equation: $$0.20(700 - x) + 0.30x = 0.24 \times 700$$ 16. Simplify: $$140 - 0.20x + 0.30x = 168$$ $$140 + 0.10x = 168$$ 17. Solve for $x$: $$0.10x = 28 \Rightarrow x = 280$$ --- 18. Problem 33: Mix Rs 4/kg (A) and Rs 7/kg (B) nuts to get a 42 kg mixture selling at Rs 5/kg. Find kg of type B. Let $x$ kg of B, then $42 - x$ kg of A. Cost equation: $$4(42 - x) + 7x = 5 \times 42$$ 19. Simplify: $$168 - 4x + 7x = 210$$ $$3x = 42$$ 20. Solve for $x$: $$x = 14$$ --- 21. Problem 34: Length = 5 times width, height = 4 times width, width = 7 inches. Find volume. Length = $5 \times 7 = 35$ inches Height = $4 \times 7 = 28$ inches Volume: $$35 \times 7 \times 28 = 6860$$ cubic inches --- 22. Problem 35: Pair of shoes sells at 37.75 after 15% discount. Find original price. Let original price = $P$. Price after discount: $$P - 0.15P = 37.75 \Rightarrow 0.85P = 37.75$$ Solve: $$P = \frac{37.75}{0.85} = 44.41$$ --- 23. Problem 36: Zeros of $f(x) = 2x^2 - 4x - 6$. Use quadratic formula: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 2 \times (-6)}}{2 \times 2}$$ Calculate discriminant: $$16 + 48 = 64$$ Roots: $$x = \frac{4 \pm 8}{4}$$ First root: $$x = \frac{4 + 8}{4} = 3$$ Second root: $$x = \frac{4 - 8}{4} = -1$$ Zeros are 3 and -1. --- 24. Problem 37: Patio 10 ft by 12 ft, increase area by 50% by adding $x$ to each dimension. Original area = $10 \times 12 = 120$ sq ft Increased area = $120 + 0.5 \times 120 = 180$ sq ft Equation for new dimensions: $$(10 + x)(12 + x) = 180$$ --- Final Answers: 28: Rs 13000 29: Rs 300 30: 36 31: 2 32: 280 gallons 33: 14 kg 34: 6860 cubic inches 35: 44.41 36: 3 and -1 37: Equation is $(10 + x)(12 + x) = 180$