Quiz
1. Algebra – Logarithms: If \(\log_{2}(x) + \log_{2}(x-2) = 3\), find the value of \(x\).
Solution: \(\log_{2}(x) + \log_{2}(x-2) = 3\) simplifies to \(\log_{2}[x(x-2)] = 3\). Solving \(x(x-2) = 8\), we get \(x^2 – 2x – 8 = 0\), factoring to \((x-4)(x+2) = 0\). Since \(x > 2\), \(x = 4\).
2. Bar Graph: A company’s annual revenue (in millions) over 5 years is represented as follows:
- 2018: 120
- 2019: 150
- 2020: 200
- 2021: 180
- 2022: 220
If the percentage increase in revenue from 2021 to 2022 matches the average percentage increase for the entire period, what was the average annual percentage increase over these five years?
Solution: The total increase is \(220 – 120 = 100\). Average annual percentage increase: \(\frac{100}{120 \times 5} \times 100 = 12.5\%\).
3. Numbers (Factors): Find the total number of factors of \(3600\), and how many of them are perfect squares.
Solution: Prime factorization of \(3600 = 2^4 \times 3^2 \times 5^2\). Total factors: \((4+1)(2+1)(2+1) = 45\). Perfect squares are formed by even exponents: \((2+1)(2+1)(2+1) = 12\).
4. Numbers – HCF (Word Problem): Three ropes of lengths 140 m, 224 m, and 308 m need to be cut into equal pieces of maximum possible length. What is the length of each piece, and how many pieces can be obtained from all three ropes?
Solution: HCF of \(140, 224, 308\) is \(28\). Pieces: \(\frac{140}{28} + \frac{224}{28} + \frac{308}{28} = 24\).
5. Averages: The average age of 20 students in a class is 15 years. If five new students join, and their average age is 18 years, what is the new average age of the class?
Solution: Total age of first group: \(20 \times 15 = 300\). New group: \(5 \times 18 = 90\). Average: \(\frac{300 + 90}{25} = 15.6\).
6. Numbers – LCM (Word Problem): Two buses start from the same point. The first bus stops every 15 minutes, while the second bus stops every 20 minutes. If they start together at 8:00 AM, at what time will they both stop at the same time next?
Solution: The LCM of 15 and 20 is 60 minutes. Thus, both buses will stop together 60 minutes after 8:00 AM, which is 9:00 AM.
7. Algebra – Linear Equations: Solve for \(x\) and \(y\):
\(3x + 4y = 20 \quad \text{and} \quad 2x – y = 4\)
Solution: From the second equation, \(y = 2x – 4\). Substituting into the first equation: \(3x + 4(2x – 4) = 20\) gives \(3x + 8x – 16 = 20\), simplifying to \(11x = 36\), so \(x = 4\) and \(y = 2\).
8. Algebra – Exponents/Square Root: Simplify and find the value of \(x\) if \(4^{x} + 2 \cdot 2^{x} = 20\).
Solution: Let \(2^x = y\), so \(4^x = y^2\). The equation becomes \(y^2 + 2y = 20\), or \(y^2 + 2y – 20 = 0\). Factoring: \((y – 4)(y + 5) = 0\), so \(y = 4\) (since \(y > 0\)). Hence, \(2^x = 4\), and \(x = 2\).
9. Geometry – Trapezium: In a trapezium, the parallel sides are 12 cm and 28 cm, and the distance between them is 10 cm. Find the area of the trapezium.
Solution: The area of a trapezium is \(\frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}\). Here: \(\frac{1}{2} \times (12 + 28) \times 10 = 200 \text{ cm}^2\).
10. Probability: A bag contains 5 red balls, 4 green balls, and 3 blue balls. Two balls are drawn at random. What is the probability that both balls are of different colors?
Solution: Total balls = 12. Total ways to draw 2 balls: \(\binom{12}{2} = 66\). Favorable outcomes: \(\binom{5}{1}\binom{7}{1} + \(\binom{4}{1}\binom{8}{1} + \binom{3}{1}\binom{9}{1} = 47\). Probability: \(\frac{47}{66} = \frac{8}{11}\).
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