Class 9 Maths Questions

Class 9 maths questions with solutions are given here to help students with their exam preparations. Practice these questions for the last-minute revisions and brush up on all the important concepts and formulae for Class 9 maths.

Also Refer:

• CBSE Important Questions for Class 9
• Sample Papers for CBSE Class 9
• Class 9 Maths MCQs

The following questions almost cover all the topics of Class 9 and are designed as per the previous year’s asked questions of Class 9 maths.

Class 9 Maths Questions with Solutions

Let us solve the following Class 9 maths questions for a quick revision.

Question 1: Find three rational numbers between ½ and ⅔.

Solution:

If x and y are rational numbers, then ½(x + y) is the rational number between x and y.

Therefore, 7/12 is a rational number between ½ and ⅔.

Again, the rational number between ½ and 7/12

Now, 13/24 and 7/12 are the rational numbers between ½ and ⅔.

Again, the rational number between 7/12 and ⅔

Three rational numbers between ½ and ⅔ are 13/24, 7/12 and 15/24.

Question 2: Express 1.656565… as a fraction in simplest form.

Solution:

Let

Multiplying both sides by 100, we get:

Subtract (i) from (ii), we get:

99x = 164

⇒ x = 164/9

Question 3: Insert a rational and an irrational number between 1/7 and 2/7.

Solution:

Rational number between 1/7 and 2/7 is ½(1/7 + 2/7) = 3/14

Irrational number between 1/7 and 2/7 is √(1/7 × 2/7) = √2/7

Also Refer: Important Formulas List for Class 9

Question 4: Find the values of p and q, if [(3 + √2)/(3 – √2)] = p + q√2.

Solution:

∴ p = 11/7 and q = 6/7

Question 5: Find the remainder when x³ – 2x² + 5x – 4 is divided by (x – 5).

Solution:

Let p(x) = x³ – 2x + 5x – 4 and g(x) = x – 5, by remainder theorem, when p(x) is divided by x – 5, then p(5) is the remainder.

So, p(5) = 5³ – 2.5 + 5.5 – 4 = 125 – 10 + 25 – 4 = 136.

∴ 136 is the remainder.

Question 6: Without actual division, prove that (2x⁴ + 3x³ – 12x – 7x + 6) is divisible by (x² + x – 6).

Solution:

Let p(x) = 2x⁴ + 3x³ – 12x² – 7x + 6 and g(x) = x² + x – 6 = (x + 3)(x – 2)

If g(x) divides p(x), then p( –3) and p(2) will be zero.

p( –3) = 2 × ( –3)⁴ + 3 × ( –3)³ – 12 × ( –3)² – 7 × ( –3) + 6

= 162 – 81 – 108 + 21 + 6

= 189 – 189 = 0

And p(2) = 2 × 2⁴ + 3 × 2³ – 12 × 2² – 7 × 2 + 6

= 32 + 24 – 48 – 14 + 6

= 56 – 62 + 6 = 0

Hence, g(x) divides p(x).

Also Read:

• Lines and Angles Class 9 Notes
• Polynomials Class 9 Notes
• Coordinate Geometry
• Congruency of Triangles
• Surface Area and Volume Class 9

Question 7: Factorise: x(x – y)³ + 3x²y(x – y).

Solution:

x(x – y)³ + 3x²y(x – y) = x(x – y) [(x – y)² + 3xy]

= x(x – y)[x² – xy – y²]

Question 8: In the figure, ∠AOB = 180^o. Find the value of angles ∠AOC, ∠COD and ∠DOB.

Solution:

Given ∠AOB = 180^o, then:

∠AOC + ∠COD + ∠DOB = ∠AOB

⇒ (3x + 7) + (2x – 19) + x = 180^o

⇒ 6x – 12 = 180^o

⇒ x – 2 = 30^o

⇒ x = 32^o

∴ ∠AOC = 3 × 32 + 7 = 103^o.

∠COD = 2 × 32 – 19 = 45^o

and ∠DOB = 32

Question 9: Prove that opposite angles of a parallelogram are equal.

Solution:

Let ABCD be a parallelogram, such that AB || CD and BC || AD.

Now, when AB || CD, AD is the transversal.

Then ∠BAD + ∠CDA = 180^o (pair of consecutive interior angles are supplementary)

Again, BC || AD, DC is the transversal

∠BCD + ∠CDA = 180^o

∴ ∠BAD + ∠CDA = ∠BCD + ∠CDA

⇒ ∠BAD = ∠BCD

⇒ ∠A = ∠C

Similarly, ∠B = ∠D.

Thus, opposite angles of a parallelogram are equal.

Question 10: In the given figure, OR and OQ are the angle bisectors of ∠R and ∠Q, respectively. Prove that ∠ROQ = 90^o + ½ ∠P.

Solution:

In ∆PRQ,

∠PRQ + ∠PQR + ∠RPQ = 180^o (angle sum property of triangle)

⇒ 2∠ORQ + 2∠OQR + ∠RPQ = 180^o ( ∵ OR and OQ are the angle bisectors)

⇒ ∠ORQ + ∠OQR = ½ (180^o – ∠RPQ)

⇒ ∠ORQ + ∠OQR = 90^o – ½ ∠RPQ ….(i)

In ∆ORQ,

∠ORQ + ∠OQR + ∠ROQ = 180^o (Angle sum property of triangle)

⇒ 90^o – ½ ∠RPQ + ∠ROQ = 180^o

⇒ ∠ROQ = 180^o – 90^o + ½ ∠RPQ

⇒ ∠ROQ = 90^o + ½ ∠RPQ

⇒ ∠ROQ = 90^o + ½ ∠P

Question 11: In the given figure, AB || CD and AO = OD. Prove that ∆ AOB ≅∆ DOC and OB = OC.

Solution:

Given AB || CD, then

∠OAB = ∠ODC (Alternate interior angles)

AO = OD (given)

∠AOB = ∠COD (Vertically opposite angles)

By ASA congruency criterion, ∆ AOB ≅∆ DOC also

OB = OC (by c.p.c.t)

Question 12: The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find the measure of each angle.

Solution:

Let the angles of the quadrilateral be 2x, 4x, 5x and 7x, then:

2x + 4x + 5x + 7x = 360^o

⇒ 18x = 360^o

⇒ x = 360^o/18 = 20^o

Therefore, the angles are 40^o, 80^o, 100^o and 140^o.

Question 13: In the given figure, S divides RQ in the ratio m : n. Prove that area (∆PRS) : area (∆PSQ) = m : n.

Solution:

Let RS = mx and SQ = nx. Now,

Area of ∆PRS = ½ × RS × PT

= ½ × mx × PT

Area of ∆PSQ = ½ × SQ × PT

= ½ × nx × PT

∴ Area(∆PRS) : Area(∆PSQ) = m : n.

For chapter-wise notes: Class 9 Maths (Chapterwise Notes and Study Materials)

Question 14: Find the length of the chord, which is at a distance of 8 cm from the centre of the circle, whose radius is 17 cm.

Solution:

Let O be the centre of the circle and AB be the chord, which at distance of 8 cm from the centre.

In the right triangle AOC,

OA² = OC² + AC²

⇒ 17² = 8² + AC²

⇒ AC = √(17² – 8²)

⇒ AC = √(289 – 64) = 15 cm.

Since the perpendicular from the centre bisects the chord, then AB = 2AC = 2 × 15 cm = 30 cm.

∴ length of chord AB is 30 cm.

Question 15: The sides of a triangle are in the ratio 5 : 12 : 13, and its perimeter is 150 m. Find the area of the triangle.

Solution:

Let the sides of the triangle be 5x, 12x and 13x. Then:

5x + 12x + 13 = 150

⇒ 30x = 150

⇒ x = 5

∴ sides of the triangle are 25 cm, 60 cm, and 65 cm.

Semi perimeter, s = 150/2 = 75 cm

Area of the triangle = √[s × (s – a) × (s – b) × (s – c)]

= √[75 × 50 × 15 × 10]

= √[15 × 5 × 5 × 10 × 15 × 10]

= 15 × 10 × 5 = 750 cm^2.

Practice Questions on Class 9 Maths

1. Express 89.999… in rational form.

2. Rationalise: (√3 – 2√2)/(√3 + √2).

3. In ∆PQR, ∠P = 70^o, ∠Q = 52^o, OQ and OR are angle bisectors of ∠Q and ∠R, respectively. Find the measures of ∠OQR and ∠ORQ.

4. The relation between two scales of temperature Fahrenheit (^oF) and Celcius (^oC) is given by F = (9/5)C + 32. Draw a linear graph for the given equation. Using the graph, fill in the blanks.

(i) –6^o C = ___^oF

(ii) 15^o F = ___^oC

(iii) 2^oC = ____^oF.

5. The depth and width of a river are 3 m and 40 m, respectively. If the water flowing speed is 2 km/hr, how much water will flow within a minute?

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