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By Karan Singh Bisht
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Updated on 6 Sep 2025, 12:29 IST
Infinity Learn extra Questions for Class 9 Maths are designed for focused practice. We’ve carefully graded them for stepwise difficulty. Try solving without checking solutions first; if you get stuck, then refer to the answers.
Here is the list of Extra Questions for Class 9 Maths with Answers based on latest CBSE syllabus prescribed by CBSE board.
Question 1. Simplify: (√5 + √2)2.
Solution: Here, (√5 + √22 = (√52 + 2√5√2 + (√2)2
= 5 + 2√10 + 2 = 7 + 2√10
Question 3. Identify a rational number among the following numbers : 2 + √2, 2√2, 0 and π
Solution: O is a rational number.
Question 4. Express 1.8181… in the form pq where p and q are integers and q ≠ 0.
Solution: Let x =1.8181… …(i)
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
100x = 181.8181… …(ii) [multiplying eqn. (i) by 100]
99x = 180 [subtracting (i) from (ii)]
x = 18099
Hence, 1.8181…
= 18099
= 2011
Question 5. Simplify : √45 – 3√20 + 4√5
Solution: √45 – 3√20 + 4√5 = 3√5 – 6√5 + 4√5 = √5.
Question 1. Evaluate : (√5 + √22 + (√8 – √5)2
Solution: (√5 + √2)2 + (√8 – √52 = 5 + 2 + 2√10 + 8 + 5 – 2√40
= 20 + 2√10 – 4√10 = 20 – 2√10
Question 2.
Express 23.43¯¯¯¯¯ in pq form, where p, q are integers and q ≠ 0.
Solution: Let x = 23.43¯¯¯¯¯
or x = 23.4343… ….(i)
100x = 2343.4343… …(ii) [Multiplying eqn. (i) by 100] 99x = 2320 [Subtracting (i) from (ii)
x = 2320
Question 3. Let ‘a’ be a non-zero rational number and ‘b’ be an irrational number. Is ‘ab’ necessarily an irrational ? Justify your answer with example.
Solution: Yes, ‘ab’ is necessarily an irrational.
For example, let a = 2 (a rational number) and b = √2 (an irrational number)
If possible let ab = 2√2 is a rational number.
Now, aba = 22√2 = √2 is a rational number.
[The quotient of two non-zero rational number is a rational] But this contradicts the fact that √2 is an irrational number.
Thus, our supposition is wrong.
Hence, ab is an irrational number.
CBSE Class 9 Important Questions |
CBSE Class 9 Science Important Questions |
CBSE Class 9 Mathematics Important Questions |
CBSE Class 9 Social Science Important Questions |
Solve these class 9 extra questions with Solutions and check the level of your performance.
Question 1. Express 1.32 + 0.35 as a fraction in the simplest form.
Solution: Let . x = 1.32 = 1.3222…..(i)
Multiplying eq. (i) by 10, we have
10x = 13.222…
Again, multiplying eq. (i) by 100, we have
100x = 132.222… …(iii)
Subtracting eq. (ii) from (iii), we have
100x – 10x = (132.222…) – (13.222…)
90x = 119
⇒ x = 11990
Again, y = 0.35 = 0.353535……
Multiply (iv) by 100, we have …(iv)
100y = 35.353535… (v)
Subtracting (iv) from (u), we have
100y – y = (35.353535…) – (0.353535…)
99y = 35
y = 3599
Question 8. Find ‘x’, if 2x-7 × 5x-4 = 1250.
Solution: We have 2x-7 × 5x-4 = 1250
2x-7 × 5x-4 = 2 5 × 5 × 5 × 5
2x-7 × 5x-4 = 21 × 54
Equating the powers of 2 and 5 from both sides, we have
x – 7 = 1 and x – 4 = 4
x = 8 and x = 8
Hence, x = 8 is the required value.
Question 1. Sudhir and Ashok participated in a long jump competition along a straight line marked as a number line. Both start the jumps one by one but in opposite directions. From ‘O’ Ashok jumps one unit towards the positive side while Sudhir jumps double in units as Ashok jumps, along negative side. After jumping 4 jumps each, at which point Ashok and Sudhir reached. What is the distance between their final positions ? Ashok argue that he is the winner since Sudhir is at negative side. Who do you think is winner and why? What is the value of the competition ?
Solution:
After jumping four jumps each, Ashok reached at 4 in positive direction and Sudhir reached at -8 i.e., in negative direction. Distance between their final positions is 12 units. Here, distance covered by Sudhir is 8 units and distance covered by Ashok is 4 units. Thus, Sudhir is the winner. Competition inculcate spirit of performance.
Question 1. Factorise : 125x3 – 64y3
Solution: 125x3– 6443 = (5x)3 – (4y)3
By using a3 – b3 = (a – b) (a2 + ab + b2), we obtain
125x3– 64y3 = (5x – 4y) (25x2 + 20xy + 16y2)
Question 2. Find the value of (x + y)2 + (x – y)2.
Solution:
(x + y)2 + (x – y)2 = x2 + y2 + 2xy + x2 + y2 – 2xy
= 2x2 + 2y2 = 21x2 + y2)
It is a free online tool that shows the division of two polynomials with Polynomial long division method.
Question 3. If p(x)= x2 – 2√2x+1, then find the value of p(2√2)
Solution:
Put x = 2√2 in p(x), we obtain
p(2√2) = (2√2)2 – 2√2(2√2) + 1 = (2√2)2 – (2√2)2 + 1 = 1
Question 4. Find the value of m, if x + 4 is a factor of the polynomial x2 + 3x + m.
Solution:
Let p(x) = x2 + 3x + m
Since (x + 4) or (x – (-4)} is a factor of p(x).
∴ p(-4) = 0
⇒ (-4)2 + 3(-4) + m = 0
⇒ 16 – 12 + m = 0
⇒ m = -4
Question 5. Find the common factor in the quadratic polynomials x2 + 8x + 15 and x2 + 3x – 10.
Solution:
x2 + 8x + 15 = x2 + 5x + 3x + 15 = (x + 3) (x + 5)
x2 + 3x – 10 = x2 + 5x – 2x – 10 = (x – 2) (x + 5)
Clearly, the common factor is x + 5.
More resources for class 9
Question 1. Expand :
(i) (y – √3)2
(ii) (x – 2y – 3z)2
Solution: (i)
(y – √3)2 = y2 -2 × y × √3 + (√3)2 = y2 – 2√3 y + 3 (x – 2y – 3z)2
= x2 + 1 – 2y)2 + (-3z)2 + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x
= x2 + 4y2 + 9z2 – 4xy + 12yz – 6zx
Question 3. Show that p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.
Solution:
Let f(p) = p10 + p8 + p6 – p4 – p2 – 1
Put p = 1, we obtain
f(1) = 110 + 18 + 16 – 14 – 12 – 1
= 1 + 1 + 1 – 1 – 1 – 1 = 0
Hence, p – 1 is a factor of p10 + p8 + p6 – p4 – p2 – 1.
Question 4. If 3x + 2y = 12 and xy = 6, find the value of 27x3 + 8y3
Solution:
We have 3x + 2y = 12
On cubing both sides, we have
⇒ (3x + 2y)3 = 123
⇒ (3x)3 +(2y)3 + 3 × 3x × 2y(3x + 2y) = √728
⇒ 27x3+ 8y3 + 18xy(3x + 2y) = √728
⇒ 27x3+ 8y3 + 18 × 6 × 12 = √728
⇒ 27x3+ 8y3 + 1296 = √728
⇒ 27x3+ 8y3 = √728 – 1296
⇒ 27x3+ 8y3 = 432
Question 5. Factorize : 4x2 + 9y2 + 16z22 + 12xy – 24 yz – 16xz.
Solution: 4x2 + 9y2 + 16z22 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(= 42) + 2(- 42)(2x)
By using a2 + b2 + 2ab + 2bc + 2ca = (a + b + c)2, we obtain
= (2x + 3y – 4z)2 = (2x + 3y – 4z) (2x + 3y – 4z)
Question 6. Factorize : 1 – 2ab – (a2 + b2).
Solution: 1 – 2ab – (a2 + b2) = 1 – (a2 + b2 + 2ab)
= 12 – (a + b)2
= (1 + a + b) (1 – a – b) [∵ x2 – y2 = (x + y)(x – y)]
Question . Factorize 64a3 – 27b3 – 144a2b + 108ab2.
Solution:
64a2 – 27b2 – 144a2b + 108ab2
= (4a)3 – (3b)3 – 36ab(4a – 3b)
= (40)2 – (3b)3 – 3 × 4a × 3b (4a – 3b)
= (4a – 3b)3 [∵ (x – y)3 = x3 – y3 – 3xy(x – y)] = (40 – 3b) (4a – 3b) (4a – 3b)
Question. What are the possible expressions for the dimensions of a cuboid whose volume is given below ?
Volume = 12ky2 + 8ky – 20k.
Solution:
We have, volume = 12ky2 + 8ky – 20k
= 4k(3y2 + 2y – 5) = 4k(3y2 + 5y – 3y – 5)
= 4k[y(3y + 5) – 1(3y + 5)] = 4k(3y + 5) (y – 1)
∴Possible expressions for the dimensions of cuboid are 4k units, (3y + 5) units and (y – 1) units.
Question. If p(x) = x3 + 3x2 – 2x + 4, then find the value of p(2) + p(-2) – P(0).
Solution:
Here, p(x) = x3+ 3x2 – 2x + 4
Now, p(2) = 23 + 3(2)2 – 2(2) + 4
= 8 + 12 – 4 + 4 = 20
p(-2) = (-2)3 + 3(-2)2 – 21 – 2) + 4
= 8 + 12 + 4 + 4 = 12
and p(0) = 0 + 0 – 0 + 4 = 4
∴ p(2) + p(-2) – p(0) = 20 + 12 – 4 = 28.
Question. If one zero of the polynomial x2 – √3x + 40 is 5, which is the other zero ?
Solution:
Let
p(x) = x2 – √3x + 40
= x2 – 5x – 8x + 40 = x(x – 5) – 8(x – 5) = (x – 5) (x – 8)
Now, for zeroes of given polynomial, put p(x) = 0
∴ (x – 5) (x – 8) = 0
x = 5 or x = 8
⇒ Hence, other zero is 8.
Question 1. Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b + c) (c + a).
Solution:
L.H.S. = (a + b + c)3 – a3 – b3 – c3
= {(a + b + c)3 – 3} – {b3 + c3}
= (a + b + c – a) {(a + b + c)2 + a2 + a(a + b + c)} – (b + c) (b2 + c2 – bc)
= (b + c) {a2 + b2 + 2 + 2ab + 2bc + 2ca + a2 + a2 + ab + ac – b2 – a2 + bc)
= (b + c) (3a2 + 3ab + 3bc + 3ca}
= 3(b + c) {a2 + ab + bc + ca}
= 31b + c) {{a2 + ca) + (ab + bc)}
= 3(b + c) {a(a + c) + b(a + c)}
= 3(b + c)(a + c) (a + b)
= 3(a + b)(b + c) (c + a) = R.H.S.
Question 2. Factorize : (m + 2n)2 x2 – 22x (m + 2n) + 72.
Solution:
Let m + 2n = a
∴ (m + 2n)2 x2 – 22x (m + 2n) + 72 = a2x2 – 22ax + 72
= a2x2 – 18ax – 4ax + 72
= ax(ax – 18) – 4(ax – 18)
= (ax – 4) (ax – 18)
= {(m + 2n)x – 4)} {(m + 2n)x – 18)}
= (mx + 2nx – 4) (mx + 2nx – 18).
Question 3. If x – 3 is a factor of x2 – 6x + 12, then find the value of k. Also, find the other factor of the – polynomial for this value of k.
Solution:
Here, x – 3 is a factor of x2 – kx + 12
∴ By factor theorem, putting x = 3, we have remainder 0.
⇒ (3)2 – k(3) + 12 = 0
⇒ 9 – 3k + 12 = 0
⇒ 3k = 21
⇒ k = 7
Now, x2 – 7x + 12 = x2 – 3x – 4x + 12
= x(x – 3) – 4(x – 3)
= (x – 3) (x – 4)
Hence, the value of k is 7 and other factor is x – 4.
CBSE Class 9 Syllabus 2025-26 |
CBSE Class 9 Syllabus Maths |
CBSE Class 9 Syllabus Science |
CBSE Class 9 Syllabus Hindi |
CBSE Class 9 Syllabus Social Science |
CBSE Class 9 Syllabus AI |
CBSE Class 9 Syllabus IT |
Question 4. Find a and b so that the polynomial x3– 10x2 + ax + b is exactly divisible by the polynomials (x – 1) and (x – 2).
Solution:
Let p(x) = x3– 10x2 + ax + b
Since p(x) is exactly divisible by the polynomials (x – 1) and (x – 2).
∴ By putting x = 1, we obtain
(1)3 – 10(1)2 + a(1) + b = 0
⇒ a + b = 9
And by putting x = 2, we obtain
(2)3 – 10(2)2 + a(2) + b = 0
8 – 40 + 2a + b = 0
⇒ 2a + b = 32
Subtracting (i) from (ii), we have
a = 23
From (i), we have 23 + b = 9 = b = -14
Hence, the values of a and b are a = 23 and b = -14
Question 5. Factorize : x2 – 6x2 + 11x – 6.
Solution:
Let p(x) = x2 – 6x2 + 11x – 6
Here, constant term of p(x) is -6 and factors of -6 are ± 1, ± 2, ± 3 and ± 6
By putting x = 1, we have
p(1) = (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 -6 = 0
∴ (x – 1) is a factor of p(x)
By putting x = 2, we have
p(2) = (2)3 – 6(2)2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 0
∴ (x – 2) is a factor of p(x)
By putting x = 3, we have
p(3) = (3)3 – 6(3)2 + 11(3) – 6 = 27 – 54 + 33 – 6 = 0
∴ (x – 3) is a factor of p(x) Since p(x) is a polynomial of degree 3, so it cannot have more than three linear factors.
∴ x3 – 6x2 + 11x – 6 = k (x – 1) (x – 2) (x – 3)
By putting x = 0, we obtain
0 – 0 + 0 – 6 = k (-1) (-2) (3)
-6 = -6k
k = 1
Hence, x3 – 6x2 + 11x – 6 = (x – 1) (x – 2)(x – 3).
Question. Factorise : 6x2 – 5x2 – √3x + 12
Solution:
Let p(x) = 6x3– 5x2 – √3x + 12
Here, constant term of p(x) is 12 and factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12.
MCQ Questions for Class 9 Subject-wise | |
MCQ Questions for Class 9 Maths | MCQ Questions for Class 9 Science |
MCQ Questions for Class 9 Social Science | MCQ Questions for Class 9 Hindi |
Question. If x = 2 – √3, y = √3 – √7 and 2 = √7 – √4, find the value of x’ + 43 + 2?.
Solution:
Here, x + y + z = 2 – √3+ √3 – √7+√7 – 2 = 0
x3+ √3 + x3= 3(x)(y)(z)
= 3(2 – √3)(√3 – √7)(√7 – 2)
= 3(2√3 – 2√7 – 3 + √21)(√7 – 2)
= 3(2√21 – 14 – 3√7 + 7√3 – 4√3 + 4√7 + 6 – 2√21)
= 3(3√3 + √7 – 8)
Question. If (x – a) is a factor of the polynomials x2 + px – q and x2 + rx – t, then prove that a = t−qr−p
Solution:
Let f(x) = x + px -q and g(x) = x2 + x – t
Since x-a is factor of both f(x) and g(x)
⇒ f(a) = g(a) = 0
Now, here f(a) = a2 + pa – q and
g(a) = a2 + ra- t
⇒ a2 + pa – q = a + ra – t (considering f(a) = g(a)] ⇒ pa – q = ra – t
⇒ ra – pa = t – q
⇒ a(r – p) = t – q
a = t−qr−p
Question 1. If a teacher divides a material of volume 27x3 + 54x2 + 36x + 8 cubic units among three students. Is it possible to find the quantity of material ? Can you name the shape of the figure teacher obtained ? Which value is depicted by the teacher ?
Solution: We know that,
√volume = Length × Breadth × Height
Now, 27x3+ 54x2 + 36x + 8
= (3x)3 + 3(3x)2(2) + 3(3x)(2)2 + (2)3
= (3x + 2)2 = (3x + 2) (3x + 2) (3x + 2)
Thus, volume = (3x + 2) (3x + 2) (3x + 2)
Yes, it is possible to find the quantity of material. (3x + 2) units..
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