Test

Self Assessment Test 6

Based on Chapter 6

Time: 1 hour

M.M. 25

Note: Q.1-2 carry 1 mark each, Q.3-5 carry 2 marks each, Q.6-8 carry 3 marks each and Q.9-10 carry 4 marks each.

1. Question: Express \( \frac{5}{12} \) as a percentage.

   Solution: \( \frac{5}{12} = \frac{5}{12} \times 100\% = 41.67\% \).

2. Question: Simplify \( \sqrt[3]{64} \).

   Solution: \( \sqrt[3]{64} = 4 \).

3. Question: Simplify \( \frac{4x^2 - 9}{2x + 3} \).

   Solution: \( \frac{4x^2 - 9}{2x + 3} = \frac{(2x - 3)(2x + 3)}{2x + 3} = 2x - 3 \).

4. Question: If \( f(x) = x^2 + 2x - 3 \), find \( f(-1) \).

   Solution: Substituting \( x = -1 \) into \( f(x) \), we get \( f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \).

5. Question: Solve the inequality \( 2x - 3 < 7 \).

   Solution: Adding 3 to both sides, we get \( 2x < 10 \). Dividing by 2, we get \( x < 5 \).

6. Question: Find the slope of the line passing through the points (3, 4) and (6, -2).

   Solution: The slope is \( \frac{-2 - 4}{6 - 3} = \frac{-6}{3} = -2 \).

7. Question: Evaluate \( \frac{2^3 \times 3^2}{6} \).

   Solution: \( \frac{2^3 \times 3^2}{6} = \frac{8 \times 9}{6} = 12 \).

8. Question: Solve for \( x \) in the equation \( 4x - 7 = 3x + 8 \).

   Solution: Subtracting \( 3x \) from both sides, we get \( x - 7 = 8 \). Adding 7 to both sides, we get \( x = 15 \).

9. Question: Factorize \( x^2 - 6x + 9 \).

   Solution: \( x^2 - 6x + 9 = (x-3)^2 \).

10. Question: Find the area of a triangle with sides of length 5, 12, and 13 units.

    Solution: Using Heron's formula, let \( s = \frac{5 + 12 + 13}{2} = 15 \). The area is \( \sqrt{15(15-5)(15-12)(15-13)} = 30 \) square units.