Chapter 1: Number Systems
Q1. Is 0.9999... equal to 1? Prove it.
SolutionLet x = 0.9999...
10x = 9.9999...
10x - x = 9
9x = 9, x = 1
Therefore 0.9999... = 1
Q2. Simplify: (√5 + √2)/(√5 - √2).
SolutionMultiply by conjugate:
= (√5 + √2)²/((√5)² - (√2)²)
= (5 + 2 + 2√10)/(5 - 2)
= (7 + 2√10)/3
Chapter 2: Polynomials
Q1. Factorise: x² - 5x + 6.
Solutionx² - 5x + 6 = (x - 2)(x - 3)
Q2. Find the zero of polynomial p(x) = 2x + 5.
Solution2x + 5 = 0 ⇒ x = -5/2
Q3. Verify that x = 2 is a zero of x² - 3x + 2.
Solutionp(2) = 4 - 6 + 2 = 0 ✓
Yes, x = 2 is a zero.
Chapter 3: Coordinate Geometry
Q1. Find the distance between (2, 3) and (6, 7).
Solutiond = √[(6-2)² + (7-3)²] = √(16 + 16) = √32 = 4√2
Q2. Find the midpoint of (4, 6) and (8, 10).
SolutionMidpoint = ((4+8)/2, (6+10)/2) = (6, 8)
Chapter 4: Linear Equations in Two Variables
Q1. Find two solutions of 2x + 3y = 12.
SolutionWhen x = 0: y = 4 → (0, 4)
When y = 0: x = 6 → (6, 0)
Q2. Find the value of k if x = 2, y = 3 is a solution of 2x + ky = 13.
Solution2(2) + 3k = 13
3k = 9
k = 3
Chapter 6: Lines and Angles
Q1. If two angles of a triangle are 60° and 80°, find the third angle.
SolutionThird angle = 180° - 60° - 80° = 40°
Q2. Prove that vertically opposite angles are equal.
SolutionLet lines AB and CD intersect at O.
∠AOC + ∠COB = 180° (linear pair)
∠BOD + ∠COB = 180° (linear pair)
∠AOC = ∠BOD ✓
Chapter 7: Triangles
Q1. In ▹ABC, AB = AC and ∠A = 50°. Find ∠B and ∠C.
SolutionAB = AC → ∠B = ∠C
50° + 2∠B = 180°
∠B = 65°
∠B = ∠C = 65°
Chapter 8: Quadrilaterals
Q1. In a parallelogram ABCD, if ∠A = 70°, find ∠B, ∠C, ∠D.
Solution∠A + ∠B = 180° (adjacent angles supplementary)
∠B = 110°
∠C = ∠A = 70°
∠D = ∠B = 110°
Chapter 10: Circles
Q1. In a circle, the angle at the centre is twice the angle at the circumference subtended by the same arc. Prove this.
SolutionLet arc AB subtend ∠AOB at centre and ∠ACB at circumference.
Draw diameter through O and C.
In ▹OAC: OA = OC (radii)
∠OAC = ∠OCA = x
Exterior angle = 2x
Similarly for ▹OBC
Total: ∠AOB = 2∠ACB ✓
Chapter 12: Heron's Formula
Heron's Formula:
s = (a+b+c)/2
Area = √[s(s-a)(s-b)(s-c)]
Q1. Find the area of a triangle with sides 5, 12, 13.
Solutions = (5+12+13)/2 = 15
Area = √[15 × 10 × 3 × 2] = √900 = 30 sq. units
Chapter 13: Surface Areas and Volumes
• Cylinder: CSA = 2πrh, TSA = 2πr(r+h), V = πr²h
• Cone: CSA = πrl, V = ⅔πr²h
• Sphere: TSA = 4πr², V = ⅖πr³
Q1. Find the volume of a cylinder with radius 7 cm and height 10 cm.
SolutionV = πr²h = (22/7) × 49 × 10 = 1540 cm³
Q2. Find the curved surface area of a cone with radius 5 cm and slant height 13 cm.
SolutionCSA = πrl = (22/7) × 5 × 13 = 204.29 cm²
Chapter 14: Statistics
Q1. Find the mean of: 4, 7, 9, 11, 14.
SolutionMean = (4+7+9+11+14)/5 = 45/5 = 9
Q2. Find the median of: 3, 5, 7, 9, 11.
Solutionn = 5 (odd)
Median = (n+1)/2 th term = 3rd term = 7
Chapter 15: Probability
Q1. A die is thrown once. Find the probability of getting a number greater than 4.
SolutionFavorable outcomes: {5, 6} = 2
P = 2/6 = 1/3
Q2. A coin is tossed twice. Find the probability of getting at least one head.
SolutionSample space: {HH, HT, TH, TT}
At least one head: {HH, HT, TH} = 3
P = 3/4