Self Assessment Test 7
Based on Chapter 7
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.
-
Question: Simplify \(\sqrt{49}\).
Solution: \(\sqrt{49} = 7\).
-
Question: Convert \(\frac{3}{4}\) to a decimal.
Solution: \(\frac{3}{4} = 0.75\).
-
Question: If \(f(x) = 2x + 5\), find \(f(3)\).
Solution: \(f(3) = 2(3) + 5 = 6 + 5 = 11\).
-
Question: Solve for \(x\): \(3x - 7 = 8\).
Solution: \(3x - 7 = 8 \Rightarrow 3x = 15 \Rightarrow x = 5\).
-
Question: Simplify: \(3(x + 4) - 2(2x - 1)\).
Solution: \(3(x + 4) - 2(2x - 1) = 3x + 12 - 4x + 2 = -x + 14\).
-
Question: Factorize: \(x^2 - 6x + 9\).
Solution: \(x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2\).
-
Question: Solve the system of equations for \(x\) and \(y\): \(x + y = 7\), \(x - y = 3\).
Solution: Adding the two equations, \(2x = 10 \Rightarrow x = 5\). Substituting into the first equation, \(5 + y = 7 \Rightarrow y = 2\). So, \(x = 5\) and \(y = 2\).
-
Question: Find the area of a triangle with sides of length 6, 8, and 10 units.
Solution: Using Heron's formula, \(s = \frac{6+8+10}{2} = 12\), \(Area = \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12 \cdot 6 \cdot 4 \cdot 2} = 24\).
-
Question: If the circumference of a circle is 44 cm, find its radius.
Solution: Circumference = \(2 \pi r = 44\), \(\Rightarrow r = \frac{44}{2 \pi} \approx 7\).
-
Question: Solve the inequality: \(3x + 5 > 14\).
Solution: \(3x + 5 > 14 \Rightarrow 3x > 9 \Rightarrow x > 3\).
No comments:
Post a Comment