Extra Questions for Class 9 Maths with Solutions Chapter Wise

Number Systems Class 9 Extra Questions Short Answer Type 2

Solve these Extra Questions for Class 9 Maths 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

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Question 8.
Find ‘x’, if 2^x-7 × 5^x-4 = 1250.
Solution:
We have 2^x-7 × 5^x-4 = 1250
⇒ 2^x-7 × 5^x-4 = 2 5 × 5 × 5 × 5
⇒ 2^x-7 × 5^x-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.

Related Links:

• Polynomials Class 9 Extra Questions Maths Chapter 2
• NCERT Books for Class 9 – Free PDF Download for 2024-25
• Lines and Angles Class 9 Extra Questions Maths Chapter 6
• Quadrilaterals Class 9 Extra Questions Maths Chapter 8
• NCERT Exemplar Solutions Class 9 Maths

Number Systems Class 9 Extra Questions Value Based (VBQs)

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.

Polynomials Class 9 Extra Questions Very Short Answer Type

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.

Polynomials Class 9 Extra Questions Short Answer Type 1

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. Factorise : 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. Factorise : 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)]

Polynomials Class 9 Extra Questions Short Answer Type 2

Question .
Factorise 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.

Polynomials Class 9 Extra Questions Long Answer Type

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.
Factorise : (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.

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.
Factorise : 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.
By putting x = 1, we have

Also check: NCERT Solutions Class 9 Maths Chapter 12 – Heron’s Formula

Polynomials Class 9 Extra Questions HOTS

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

Polynomials Class 9 Extra Questions Value Based (VBQs)

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, √olume = 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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