🎩 Math Magic: Complete Solutions 🎩
Simplification Problems
1(i). Simplify: (x/(x+1) + x/(x-1)) / (2x²/(x²-1))
1
Numerator: x/(x+1) + x/(x-1)
Find common denominator: (x+1)(x-1) = x²-1
2
= [x(x-1) + x(x+1)] / (x²-1)
Combine the fractions
3
= [x² - x + x² + x] / (x²-1) = 2x²/(x²-1)
Simplify the numerator
4
Now divide by denominator: [2x²/(x²-1)] / [2x²/(x²-1)]
Numerator and denominator are identical
Final Answer: 1
1(ii). Simplify: (x+2)/(x+3) + (x-1)/(x-2)
1
Find common denominator: (x+3)(x-2)
2
= [(x+2)(x-2) + (x-1)(x+3)] / [(x+3)(x-2)]
Rewrite each term with common denominator
3
= [x²-4 + x²+2x-3] / [x²+x-6]
Expand the numerators
4
= (2x² + 2x - 7) / (x² + x - 6)
Combine like terms
Final Answer: (2x² + 2x - 7)/(x² + x - 6)
Subtraction Problem
3. Subtract 1/(x²+2) from (2x³ + 3x² + 2)/(x² + 2)²
1
(2x³ + 3x² + 2)/(x² + 2)² - 1/(x² + 2)
Rewrite the problem
2
Find common denominator: (x² + 2)²
3
= [2x³ + 3x² + 2 - (x² + 2)] / (x² + 2)²
Adjust second term to common denominator
4
= (2x³ + 2x²) / (x² + 2)²
Simplify numerator
5
= 2x²(x + 1) / (x² + 2)²
Factor numerator
Final Answer: 2x²(x + 1)/(x² + 2)²
Find Expression Problem
4. What should be subtracted from (x² + 6x + 8)/(x³ + 8) to get 3/(x² - 2x + 4)?
1
Let expression to subtract be E
(x²+6x+8)/(x³+8) - E = 3/(x²-2x+4)
2
Factor expressions:
(x+2)(x+4)/[(x+2)(x²-2x+4)] - E = 3/(x²-2x+4)
3
Simplify first term: (x+4)/(x²-2x+4) - E = 3/(x²-2x+4)
4
Solve for E: E = (x+4-3)/(x²-2x+4)
5
= (x+1)/(x²-2x+4)
Final Answer: (x + 1)/(x² - 2x + 4)
Word Problems
7. Pari needs 4 hours to complete a work. Yuvan needs 6 hours. How long if they work together?
1
Pari's work rate: 1/4 work per hour
Yuvan's work rate: 1/6 work per hour
2
Combined rate: 1/4 + 1/6 = 3/12 + 2/12 = 5/12 work per hour
3
Time taken: 1 ÷ (5/12) = 12/5 hours
4
Convert to minutes: 0.4 hours × 60 = 24 minutes
Final Answer: 2 hours and 24 minutes
8. Iniya bought 50 kg of apples and bananas. Apples cost twice as much as bananas. She spent ₹1800 on apples and ₹600 on bananas. How many kg of each?
1
Let price of bananas = ₹p/kg → apples = ₹2p/kg
Let bananas = b kg, apples = a kg → a + b = 50
2
Cost equations: 2p×a = 1800 and p×b = 600
3
From second equation: p = 600/b
4
Substitute into first: 2×(600/b)×a = 1800 → 1200a/b = 1800
5
Simplify: a/b = 1800/1200 = 3/2 → a = (3/2)b
6
Substitute into a + b = 50: (3/2)b + b = 50 → (5/2)b = 50 → b = 20
7
Then a = 50 - 20 = 30
Final Answer: 30 kg apples and 20 kg bananas