Half Yearly Class 9 Mathematics Set 1 (Solutions) | Mathematics (Maths) Class 9 PDF Download

Time: 3 Hours
Maximum Marks: 80
General Instructions:
(i) The question paper comprises four sections: A, B, C, and D. 
(ii) All questions are compulsory. However, internal choices are provided in some questions. 
(iii) Section A has 10 Questions carrying 1 mark each.
(iv) Section B has 5 Questions carrying 2 marks each.
(v) Section C has 10 Questions carrying 3 marks each.
(vi) Section D has 5 Questions carrying 6 marks each.
(vii) Use of calculators is not permitted.

Section A

Q1. Express as a fraction.    (1 Mark) 

Ans: Let Then, 10x =

Subtract:  10x − x =
⇒  9x = 6
⇒ x = 6/9 = 2/3.

Q2.  Find the degree of the polynomial 3x⁴ − 5x² + 2.   (1 Mark) 

Ans: Degree of 3x⁴ − 5x²  + 2 is 4.

Q3. Write the coordinates of the point where the x-axis and y-axis intersect.   (1 Mark) 

Ans: The x-axis and y-axis intersect at the origin, (0, 0).

Q4.  In ax + by + c = 0, what is b if the line is parallel to the x-axis?   (1 Mark) 

Ans: For a line parallel to the x-axis, y = k, so in ax + by + c = 0, b ≠ 0, a = 0.

Q5. State Euclid’s second postulate.   (1 Mark) 

Ans: Euclid’s second postulate: Any straight line segment can be extended indefinitely in a straight line.

Q6. If one angle of a triangle is 80and the other two are equal, find each equal angle.   (1 Mark) 

Ans: Let each equal angle be x. Then, x + x + 80° = 180°
⇒ 2 = 100°
⇒ x = 50°

Q7. Simplify: √18.   (1 Mark) 

Ans:

Q8.  If (x − 3) is a factor of x² − 5x + k, find k.   (1 Mark) 

Ans: If (x − 3) is a factor, p(3) = 0.
So, 3² − 5 · 3 + k = 0
⇒ 9 − 15 + k = 0
⇒ k = 6.

Q9.  In which quadrant does the point (2, −5) lie?   (1 Mark) 

Ans: Point (2, −5) has x > 0, y < 0, so it lies in the fourth quadrant.

Q10. If two lines intersect, what is the measure of a pair of vertically opposite angles?   (1 Mark) 

Ans: Vertically opposite angles are equal.

Section B

Q1. Rationalize the denominator:    (2 Marks) 

Ans:

Q2. Find the remainder when p(x) = x³ − 4x² + 6x − 1 is divided by x − 2.   (2 Marks) 

Ans: For p(x) = x³ − 4x² + 6x − 1, remainder is p(2) = 2³ − 4·2² + 6·2−1 = 8 − 16 + 12 − 1 = 3.

Q3. Find the distance between points P (1, 4) and Q(−3, −2).   (2 Marks) 

Ans: Distance:

Q4.  Solve 3x + 2y = 10 for y when x = 2.   (2 Marks) 

Ans: For 3x + 2y = 10, when x = 2: 3 · 2 + 2y = 10
⇒ 6 + 2y = 10
⇒ 2y = 4
⇒ y = 2.

Q5. If two angles on a straight line are (2x + 10)° and (x − 10)°, find x.   (2 Marks) 

Ans: (2x + 10)°  + (x − 10)° = 180° 
⇒ 3x = 180
⇒ x = 60. 

Section C

Q1.  Prove that √3 is irrational.   (3 Marks) 

Ans: Assume √3 = a/b, a, b coprime, b ≠ 0.
Then, 3 = a²/b² ⇒ a² = 3b²
So, 3 divides a², hence a.
Let a = 3k. Then, 9k² = 3b²
⇒ b² = 3k², so 3 divides b. Contradiction.

Q2.  Using factor theorem, show that (x + 1) is a factor of x³ + x² − 2x − 2.   (3 Marks) 

Ans: For p(x) = x³ + x² − 2x − 2, p(−1) = (−1)³ + (−1)² − 2(−1) − 2 = −1 + 1 + 2 − 2 = 0.
So, (x + 1) is a factor.

Q3. Find the coordinates of the midpoint of the line segment joining A(−2, 3) and B(4, −1).   (3 Marks) 

Ans: Midpoint of A(−2, 3), B(4, −1): 

Q4. Solve: x + y = 5 and 2x − y = 4 using substitution method.   (3 Marks) 

Ans: From x + y = 5, y = 5 − x.
Substitute in 2x − y = 4: 2x − (5 − x) = 4
⇒ 3x − 5 = 4
⇒ 3x = 9
⇒  x = 3, y = 5 − 3 = 2.

Q5. Using Euclid’s postulates, prove that a line segment can be drawn joining any two points.   (3 Marks) 

Ans: Euclid’s first postulate: A straight line can be drawn joining any two points. Proof follows directly.

Q6. In △PQR, ∠P = 50°, ∠Q = 70°. Find ∠R. If ST ∥ QR, find ∠P ST.   (3 Marks) 

Ans: ∠R = 180° - (50° + 70°) = 60°
Since ST ∥ QR, ∠PST = ∠PQR = 70° (corresponding angles).

Q7. Find k such that x²  + kx + 4 has equal roots.   (3 Marks) 

Ans: Discriminant: k² − 4 · 1 · 4 = 0
⇒ k² = 16
⇒ k = ±4.

Q8. Express as a fraction.   (3 Marks) 

Ans: Let x =
Then, 10x =
100x =

Subtract: 100x−10x =
⇒ 90x = 43
⇒ x = 43/90.

Q9. Simplify:    (3 Marks) 

Ans:
Similarly,

Sum:

Q10. Solve: 4x − y = 8 and 2x + y = 7 using substitution method.   (3 Marks) 

Ans: From 4x − y = 8, y = 4x − 8.
Substitute in 2x + y = 7: 2x + (4x − 8) = 7
⇒ 6x = 15
⇒ x = 5/2, y = 4· 5/2 - 8 = 2.

Section D

Q1. Factorize x³  − 7x  + 14x − 8 completely.   (6 Marks) 

Ans: For p(x) = x³ −7x² + 14x − 8, p(1) = 1−7+14−8 = 0, so (x−1) is a factor.
Synthetic division: 1, −7, 14, −8 with root 1: 1, 1 − 6, −6 + 8, 2 − 8
⇒ x² − 6x + 8. Factorize: (x − 2)(x − 4).
Thus, p(x) = (x − 1)(x − 2)(x − 4).

Q2. In the coordinate plane, find the area of △ABC  with vertices A(0, 0), B(3, 0), C (1, 4). Is it a right-angled triangle?   (6 Marks) 

Ans: Area: 1/2 |0(0 − 4) + 3(4 − 0) + 1(0 − 0)| = 1/2|12| = 6 sq.
units.  Sides: AB = 3, 

Check: 3² + (√17)² ≠ (√20)² not right-angled.

Q3. Prove that vertically opposite angles are equal when two lines intersect. Use this to show that the sum of angles in a triangle is 180°.   (6 Marks) 

Ans: Let lines AB and C D intersect at O.
∠AOC = ∠BOD (same arcs).
For triangle, extend BC to D, draw C E ∥ AB.
∠BAC = ∠ACE, ∠ABC = ∠BCE (corresponding).
∠ACE + ∠ACB, ∠BCE = 180°, so ∠BAC + ∠ABC + ∠ACB = 180°

Q4. Solve graphically: x + y = 6 and 2x − y = 3. Find the area of the triangle formed by these lines and the y-axis.   (6 Marks) 

Ans: For x + y = 6:  (6, 0), (0, 6).
For 2x − y = 3:  (3, 3), (0, −3).
Intersection:  x = 3, y = 3
Triangle vertices:  (0, −3), (0, 6), (3, 3).
Area: 1/2 · 3 · 9 = 13.5 sq.  units.

Q5. Solve: 2x + 3y = 11 and x − 2y = −2 using elimination method. Verify the solution.   (6 Marks) 

Ans: Multiply second equation by 2: 2x − 4y = −4.
Add to first: 2x + 3y + 2x − 4y = 11 − 4
⇒ 4x−y = 7.
Solve with x−2y = −2: x = 3, y = 5/2
Verify: 2·3 + 3· 5/2 = 11, 3−2·5/2 = - 2.