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.
Represent √3 on the number line.
Solution:


On the number line, take OA = 1 unit. Draw AB = 1 unit perpendicular to OA. Join OB.
Again, on OB, draw BC = 1 unit perpendicular to OB. Join OC.
By Pythagoras Theorem, we obtain OC = √3. Using
compasses, with centre O and radius OC, draw an arc, which intersects the number line at point
D. Thus, OD = √3 and D corresponds to √3.

Question 2.
Represent √3.2 on the number line.
Solution:
First of all draw a line of length 3.2 units such that AB = 3.2 units. Now, from point B, mark a distance of 1 unit. Let this point be ‘C’. Let ‘O’ be the mid-point of the distance AC. Now, draw a semicircle with centre ‘O’ and radius OC. Let us draw a line perpendicular to AC passing through the point ‘B’ and intersecting the semicircle at point ‘D’.
∴ The distance BD = √3.2

Now, to represent √3.2 on the number line. Let us take the line BC as number line and point ‘B’ as zero, point ‘C’ as ‘1’ and so on. Draw an arc with centre B and radius BD, which intersects the number line at point ‘E’.
Then, the point ‘E’ represents √3.2.

Question 3.
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

Also Check: 5 Toughest and Easiest Chapters in CBSE Class 9 Maths

Question 4.
Find the square root of 10 + √24 + √60 + √40.
Solution:

Question 5.
If x = 9 + 4√5, find the value of √x – 1x√.
Solution:
Here,
x = 9 + 4√5
x = 5 + 4 + 2 x 2√5
x = (√5² + (2² + 2 x 2x √5).
x = (√5 + 2)²
√x = √5 + 2
Now, 1x√ = 15√+2

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Question 6.

If x = 15√−2 , find the value of x3 – 32 – 5x + 3
Solution:

∴ x – 2 = √5
Squaring both sides, we have
x² – 4x + 4 = 5
x² – 4x – 1 = 0 …(i)
Now, x³ – 3² – 5x + 3 = (x² – 4x – 1) (x + 1) + 4
= 0 (x + 1) + 4 = 4 [using (i)]

Question 7.
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.

Question 8.
Evaluate:

Solution:

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 Long Answer Type

Question 1.

If x = p+q√+p−q√p+q√−p−q√, then prove that q2 – 2px + 9 = 0.
Solution:


Squaring both sides, we have
⇒ q²x² + p² – 2pqx = p² – q²
⇒ q²x² – 2pqx + q² = 0
⇒ q(q² – 2px + q) = 0
⇒ qx² – 2px + q = 0 (∵ q ≠ 0)

Question 2.

If a = 13−11√ and b = 1a, then find a2 – b2
Solution:

Question 3.
Simplify:

Solution:

Question 4.
Prove that:

Solution:

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Question 5.
Find a and b, if

Solution:

Number Systems Class 9 Extra Questions HOTS

Question 1.
If x^a = y, y^b = z and z^c = x, then prove that abc = 1.
Solution:
We have x^a = y, y^b = z and z^c = x

Question 2.
Prove that:

Solution:
Taking L.H.S., we have

Question 3.
Show that:

Solution:

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.

Question 2.
Manu went to his mathematics teacher and asked him “Sir, I want some chocolates to distribute among my classmates for my birthday but I have no money. Can you provide me some chocolates”. Teacher told Manu, I am giving you two numbers 13+22√ and 13−22√ and if you can find the value of sum of their squares, then I will provide you as many chocolates as the resulting value of sum of squares of given numbers. Find the number of chocolates. What value is depicted from this action?
Solution:

= (3 – 2√2)² + (3 + 2√2²
= 9 + 8 – 2 × 3 × (2√2) + 9 + 8 + 2 × 3 × 2√2 = 34.
Hence, resulting value of sum of squares of numbers = number of chocolates = 34. By doing this, teacher motivates the students to use their knowledge and apply it in day to day life with caring and kindness.

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 remainder when x3+ x2 + x + 1 is divided by x – 12 using remainder theorem.
Solution:
Let p(x) = x3+ x2 + x + 1 and q(x) = x – 12
Here, p(x) is divided by q(x)
∴ By using remainder theorem, we have

Question 6.
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 2.
If x + = 1x = 7, then find the value of x3 + 1×3
Solution:
We have x + 1x = 7
Cubing both sides, we have

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 1.
Factorise :

Solution:

Question 2.
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 3.
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 4.
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 5.
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 6.
Simplify:

Solution:

Question 7.
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 6.
Show that 13and 43 are zeroes of the polynomial 9×3 – 6×2 – 11x + 4. Also, find the third zero of the polynomial.
Solution:
Let p(x) = 9×3– 6×2 – 11x + 4
Put x = 13, we have

Thus, x = 43 is another zero of the polynomial p(x). Since x = 13 and x = 43 are the zeroes of p(x), therefore, (x−13) (x−43) (3x – 1) (3x – 4) or 9×2 – 15x + 4 exactly divides p(x).
⇒ 9×3 – 6×2 – 11x + 4 = (9×2 – 15x + 4) (x + 1)
Hence, x = -1 is its third zero.

Question 7.
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
p(1) = 6(1)³ – 5(1)² – √3(1) + 12 = 6 – 5 – √3 + 12 = 0
∴ (x – 1) is a factor of p(x).
Now, by long division, we have

Thus,
p(x) = (x – 1) (6x² + x – 12)
p(x) = (x – 1) (6x² + 9x – 8x – 12)
p(x) = (x – 1) {3x (2x + 3) – 4(2x + 3)}
p(x) = (x – 1) (3x – 4) (2x + 3).

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

Polynomials Class 9 Extra Questions HOTS

Question 1.
What must be added to polynomial f(x) = x⁴ + 2x² – 2x² + x – 1 so that resulting polynomial is exactly divisible by x² + 2x – 3?
Solution:

Here, remainder = -x + 2
To make remainder = 0, we must add -(remainder) in the polynomial
i.e., -(-x + 2) i.e., x – 2
Hence, x4 + 2x³ – 2x² + x – 1 + (x – 2)
Here, polynomial = x⁴ + 2x³– 2x² + 2x – 3 and required addition is (x – 2).

Question 2.
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 3.
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.
Cube.
Apply knowledge and use of example for clarity of subject, student friendly.

Question 2.
In a camp organised by the students of class-9 to donate amount collected to flood victims of Kashmir. At the time of payment of a juice glass at one stall of juice, stall holder asked the students to pay the remainder of x3+ 3×2 + 3x + 1 divided by (x−12) What is the price of the juice at the stall ? Which value is depicted by class-9 students by organising such camps ?
Solution:
Let
p(x) = x2 + 3×2 + 3x + 1 and

By long division method, we have

Remainder = 278 or 3 38
Thus, price of the juice glass is ₹ 3 38
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