## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4. Practice more so you score more!!

## Exercise 1.4

Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or non-terminating repeating decimal expansion:
(i) $$\frac { 13 }{ 3125 }$$
(ii) $$\frac { 17 }{ 8 }$$
(iii) $$\frac { 64 }{ 455 }$$
(iv) $$\frac{15}{1600}$$
(v) $$\frac { 29 }{ 343 }$$
(vi) $$\frac{23}{2^{3} 5^{2}}$$
(vii) $$\frac{129}{2^{2} 5^{7} 7^{5}}$$
(viii) $$\frac { 6 }{ 15 }$$
(ix) $$\frac { 35 }{ 50 }$$
(x) $$\frac { 77 }{ 210 }$$
Solution:
(i) $$\frac { 13 }{ 3125 }$$ = $$\frac{17}{2 \times 2 \times 2}$$ = $$\frac{17}{2^{3}}$$
Because the denominator can be in the form 2n 5n, hence it will have terminating decimal expansion.

(ii) $$\frac { 17 }{ 8 }$$ = $$\frac { 17 }{ 8 }$$
It will have terminating decimal expansion.

(iii) $$\frac { 64 }{ 455 }$$ = $$\frac{64}{5 \times 7 \times 13}$$
Non terminating repeating decimal expansion.

(iv) $$\frac{15}{1600}$$ = $$\frac{15}{2^{2} \times 5^{2}}$$
It will have terminating decimal expansion.

(v) $$\frac { 29 }{ 343 }$$ = $$\frac{29}{7^{3}}$$
Non terminating repeating decimal expansion.

(vi) $$\frac{23}{2^{3} 5^{2}}$$
It will have terminating decimal expansion.

(vii) $$\frac{129}{2^{2} 5^{7} 7^{5}}$$
Non terminating repeating decimal expansion.

(viii) $$\frac { 6 }{ 15 }$$ = $$\frac{6}{3 \times 5}$$
It will have terminating decimal expansion.

(ix) $$\frac { 35 }{ 50 }$$ = $$\frac{35}{2 \times 5^{2}}$$
It will have terminating decimal expansion.

(x) $$\frac { 77 }{ 210 }$$ = $$\frac{17}{2 \times 3 \times 5 \times 7}$$
Non terminating repeating decimal expansion.

Question 2.
Write down the decimal expansion of those rational numbers in Question 1 above which terminating decimal expansions.
Solution:
(i) $$\frac { 13 }{ 3125 }$$ = 0.00146
(ii) $$\frac { 17 }{ 8 }$$ = 2.125
(iii) $$\frac { 15 }{ 1600 }$$ = 0.009375
(iv) $$\frac{23}{2^{3} 5^{2}}$$ = $$\frac { 23 }{ 200 }$$ = 0.115
(v) $$\frac { 6 }{ 15 }$$ = 0.4
(vi) $$\frac { 35 }{ 50 }$$ = 0.7

Question 3.
The following real numbers have decimal expansions as given below. In each case decide whether they are rational or not. If they are rational and of the form $$\frac { p }{ q }$$, what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000
(iii) $$43 . \overline{123456789}$$
Solution:
(i) 43.123456789
$$=\frac{43123456789}{1000000000}=\frac{43123456789}{2^{9} 5^{9}}$$
it is a rational, number The prime factors of q are 2959

(ii) 0.120120012000120000 …….
\begin{aligned} &=\frac{120120012000012}{100000000000000 \ldots \ldots . .} \\ &=\frac{120120012000012}{\left(2^{1} \times 2^{2} \times 2^{3} \ldots . .\right) \times\left(5^{1} \times 5^{2} \times 5^{3} . \ldots .\right)} \end{aligned}
it is a rational number The prime factors of q are (21 x 2² x 2³ ….) x (51 x 5² x 5³ ….)

(iii) $$43 . \overline{123456789}$$
It is not-terminating decimal expansion.
But it is a rational number, whose prime factors of q will also have a factor other than 2 or 5.

We hope you found these NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 helped you with finding the solutions easily.

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## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3. Practice more so you score more!!

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Question 1.
Solve the following pair of linear equations by the substitution method, Solution:
(i) The pair of linear equation formed are
x + y = 14 … (i)
x – y = 14 … (ii)
We express x in the term of y from equation (ii) to get
x = y + 4
Now we substitute this value of x in equation (i) we get
(y + 4) + y = 14
2y = 14 – 4
y = 5
Putting this value of equation (ii)
x – 5 = 4
x = 9
∴ x = 9, y = 5

(ii) We have
s – t = 3 … (i)
$$\frac { s }{ 3 }$$ + $$\frac { t }{ 2 }$$ = 6 … (ii)
We express s in the terms of t from equation (i)
s = 3 + t
2t + 6 + 3t = 36
5t = 30
t = 6
Putting this value in equation (i)
s – 6 = 3
s = 9
∴ t = 6 and s = 9

(iii) We have
3x – y = 3 … (i)
9x – 3y = 9 … (ii)
From equation (i)
y = 3x – 3
Substituting this value in equation (ii) we get
9x – 3(3x – 3) = 9
9x – 9x + 9 = 0
9 = 9
The statement is true for all value of y
So y = 3x – 3
Where x can take any value i.e., infinite many solution.

(iv) We have
0.2x + 0.3y = 1.3 …(i)
0.4x + 0.5y = 2.3 …(ii)
From equation (i) we get
x = $$\frac{1.3-0.3 y}{0.2}$$
Substituting this value in equation (ii)
0.4$$\frac{(1.3-0.3 y)}{0.2}$$ + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
– 0.1y = – 0.3
y = 3
Putting this value in equation (i) we get
∴ 0.2x + 0.3 x 3 = 1.3
0.2x = 1.3-0.9 0.2x = 0.4 x= 2
∴ x = 2 and y = 3

(v) We have
$$\sqrt{2x}$$ + $$\sqrt{3y}$$y = 0 … (i)
$$\sqrt{3x}$$ + $$\sqrt{8y}$$y = 0 … (ii)
From equation (i) we can get
x = $$\frac{-\sqrt{3} y}{\sqrt{2}}$$
Substituting this value in equation (ii) Putting this value in equation (i) we get Substituting this value in equation (i) we get Putting this value if y if equation (i) we get

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
We have the two equations
2x + 3y = 11 … (i)
2x – 4y = – 24 … (ii)
From equation (ii)
2x = 4y – 24
x = 2y – 12
Substituting this value in equation (ii)
2(2y – 12) + 3y = 11
4y – 24 + 3y = 11
7y = 35
y = 5
Putting this value in equation (i)
2x + 15 – 11 x = – 2
Putting these value in
y = mx + 3
we get
5 = – 2m + 3
2 = – 2m
m = – 1
Hence the value of m = – 1 and x = – 2, y = 5

Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes 5/6, Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let 1st number be x and 2nd number be y.
Let x > y
1st condition :
x – y = 26
2nd condition :
x = 3y
Putting x = 3y in equation (i)
3y – y = 26 ⇒ 2y = 26 ⇒ y = 13
From (ii)
x = 3 x 13 = 39
∴ One number is 13 and the other number is 39.

(ii) Let one angle be x and its supplementary angle = y
Let x > y
1st Condition :
x + y = 180°
2nd Condition :
x – y = 18° ⇒ X = 18° + y
From equation (ii), putting the value ofx in equation (i),
18° + y + y = 180° ⇒ 18° + 2y = 180°
2y = 162° ⇒ y = 81°
From (ii) x = 18° + 81° = 99° ⇒ x = 99°
∴ One angle is 81° and another angle is 99°.

(iii) Let the cost of each bat be x
and the cost of each ball by y.
According to question
7x + 6y = 3800 … (i)
3x + 5y = 1750 … (ii)
From equation (i)
3$$\frac{(3800-6 y)}{7}$$ + 5y = 1750
3(3800 – 6y) + 35y = 7 x 1750
17y = 12250 – 11400
y = 50
Putting this value in equation (i)
7x + 300 = 3800
7x = 3500
x = 500
Hence the rate of each but in ₹ 500 and the rate of each ball is ₹ 50

(iv) Let the fixed charge be x
and let the charge per km be y.
Then according to question
x + 10y = 105 … (i)
x + 15 y = 155 … (ii)
From equation (i)
x = 105 – 10y
Substituting this value in equation (ii)
105 – 10y + 15y = 155
5y = 50
y = ₹ 10 per km.
Putting this value in equation (i)
x + 100 = 105
x = ₹ 5
A person travelling distance of 25 km the charges will be.
= x + 25y
= 5 + 25 x 10
= ₹ 255

(v) Let the numerator of the fraction be x
and the denominator of the fraction be y.
Therefore according to question (vi) Let the present age of Jacob be x
and the present age of his son be y
According to question
(x + 5) = 3(y + 5)
Again according to question five years ago
(x – 5) = 7(y – 5)
x – 7y = – 30
From equation (i) we get
x= 3y + 10
Substituting this value in equation (ii)
3y + 10 – 7y = – 30
– 4y = – 40
y = 10
Putting this value in equation (i)
x – 30 = 10
x = 40
Hence the present age of Jacob is 40 years and the present age of his son is 10 years.

We hope you found these NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 helped you with finding the solutions easily.

Make sure to practice the remaining parts of the Class 10 Maths NCERT Solutions English Medium Chapter 3.

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4. Practice more so you score more!!

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
Solution:
(i) By Elimination Method:
Fquations are x + y = 5
and 2x – 3y = 4
Multiply equation (i) by 2 and subtract equation (ii) from it, we have (ii) By Elimination method:
Equations are 3x + 4y = 10
and 2x – 2y = 2
Multiplying equation (ii) by 2 and adding to equation (i), we (iii) By Elimination Method: (iv) By Elimination Method: Substracting equation (i) from (iii) we get
5y = – 15
⇒ y = – 3
Subtracting the value of y = – 3 in (ii), we get
3x – (- 3) = 9
3x + 3 = 9
3x = 6 x = 2
∴ x = 2 and y = 3
Substituting method
From equation (ii) we have
3x – y = 9
⇒ 3x = 9 + y
⇒ x = $$\frac { 9 + y }{ 3 }$$
Substituting the value of x = equation, (i) we get
3($$\frac { 9 + y }{ 3 }$$) + 4y = – 6
⇒ 9 + y + 4y = – 6
⇒ 5y = – 15
⇒ y = – 3
Again, substituting the value of y = -3 in equation (i) we get
3x – (-3) = 9
⇒ 3x + 3 = 9
⇒ 3x = 6
⇒ x = 2
∴ x = 2 and y = – 3

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes – if we only add 1 to the denominator. What is the fraction₹

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
(i) Let numerator be x and denominator be y.
$$\frac { x + 1 }{ y – 1 }$$ = 1
x – y = – 2 … (i)
According to question’s second condition
$$\frac { x }{ y + 1 }$$ = $$\frac { 1 }{ 2 }$$
2x – y = 1 … (i)
Substract equation (i) from equation we eliminate y
x = 3
Substitute this value in equation (i) we get
3 – y = – 2
y = 5
Hence the fraction is $$\frac { 3 }{ 5 }$$

(ii) Let the present age of Nuri be x
and let the present age of Sonu be y.
According to questions first condition
(x – 5) = 3 (y – 5)
x – 3y = – 10 … (i)
According to eq. question’s second condition
(x + 10) = 2 (y+ 10)
x – 2y = 10 … (ii)
Substracting equation (i) from equation (ii)
We eliminate for x.
y = 20
Putting this value in equation (i)
x – 60 = – 10
x = 50
Present age of Nuri is 50 years old and Sonu is 20 years old.

(iii) Let the ten’s digit of the number be x
and the unit’s digit of the number by y.
According to question’s first condition
x + y = 9 … (i)
According to question’s second condition
9(10 x + y) = 2(10y + x)
90x + 9y – 20y + 2x
88x = 11y
8x – y = 0 (dividing by 11)
8x – y = 0 … (ii)
Adding equation (i) and (ii) we eleminate for y, we get
9x = 9
x = 1
Putting the value in equation (i)
1 + y = 9
y = 8
Hence the units digit is 8 and the ten’s digit is 1. Then the number is 18.

(iv) Let the notes of ₹ 50 be x
and let the notes of ₹ 100 be y.
According to question’s first condition
50x + 100y = 2000
x + 2 y= 40 … (i)
According to question second condition
x + y = 25
Substract equation (i) from equation (ii) we eleminate for x.
– y = – 15
y = 15
Putting this value in equation (i)
x + 30 = 40
x = 40
Hence ₹ 50 notes is 10 and the notes of ₹ 100 is 15.

(v) Let the fixed charge be ₹ x and let the additional charge per day be ₹ y.
According to questions first condition
x + 4y = 27 … (i)
According to questions second condition
x + 2y = 21 … (ii)
Eliminating x for equation (i) and (ii)
We subtract equation (ii) from equation (1)
2y = 6
y = 3
Putting this value in equation (ii)
x + 6 = 21
x = 21 – 6
x = 15
Hence the fixed charge is ₹ 15 and the additional charge per day is ₹ 3.

We hope you found these NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 helped you with finding the solutions easily.

Make sure to practice the remaining parts of the Class 10 Maths NCERT Solutions English Medium Chapter 3.

## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4. Practice more so you score more!!

## Exercise 2.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5x + 2;  $$\frac { 1 }{ 2 }$$, 1, – 2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Solution:
(i) Comparing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 2, b – 1, c = -5 and d = 2.
Now, we get the zeroes Therefore, $$\frac { 1 }{ 2 }$$, 1, – 2 are the zeroes of 2x³ + x² – 5x + 2.
So we take α, ß, γ are the co-efficient of cubic polynomial.
Sum of zeroes (ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Compearing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 1, b = -4, c = 5 and d = – 2.
∴ p (x) = x3 – 4x2 + 5x – 2
⇒  p(2) = (2)3 – 4(2)2 + 5 x 2 – 2
= 8 – 16+ 10 – 2 = 0
p(1) = (1)3 – 4(1)2 + 5 x 1- 2
= 1 – 4 + 1 – 2
= 6 – 6 = 0
Therefore, 2, 1 and 1 are the zeroes of x3 – 4x2 + 5x – 2.
So, take α, ß, γ are the coefficient of cubic polynomial.
Comparing x² – 4x² + 5x – 2 = 0
Sum of zeroes Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:
We know that a cubic equation is ax³ + bx² + cx + d = 0
But given, α + β + γ = 2
αβ + βγ + γα = -7 and αβγ = – 14
Also, we know that a + b + g = $$\frac { -b }{ a }$$ = 2, hence, b = – 2
αβ + βγ + γα = $$\frac { c }{ a }$$ = – 7, hence c = – 7 a
αβγ = $$\frac { -d }{ a }$$ = – 14, hence d = – 14, and a = 1
Now, put the value of a, b, c, d in equation (1), we get
x³ – 2x² – 7x + 14 = 0

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b, find a and b.
Solution:
Zeroes of the polynomial are a-b, a, a + b
Sum of zeroes = a-b + a + a + b = 3a ⇒ a³ – b²a = 1
Put a = 1, in a² – b²a = – 1
(1)³ – b² = – 1 ⇒ b² = 2
b= ± $$\sqrt{2}$$
values of a and b are 1, ± $$\sqrt{2}$$

Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± $$\sqrt{3}$$, finnd other zeroes.
Solution:
Let two zeroes are 2 + $$\sqrt{3}$$ and 2 – $$\sqrt{3}$$, (x² – 4x +1) is a factor of the given polynomial.
Now, we divide the given polynomial by
So, x4 – 6x3 – 26x2 + 138x – 35 = (x2 – 4x + 1) (x2 – 2x – 35)
Now, by spilitting – 2x, we factorise x² – 2x – 35
= x3 – 7x + 5x – 35
= x(x – 7) (x + 5)
= (x – 7) (x + 5) px = 7, x = – 5
So, zeroes of the given polynomials are
2 + $$\sqrt{3}$$, 2 – $$\sqrt{3}$$, 7 and – 5.

Question 5.
If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
If p(x) and g(x) are any two polynomials of the form x² – 2x and x² + bx + c. We know by formula.
Dividend = Divisor x Quotient + Remainder
x4 – 6x² + 16x3 – 25x + 10 = (x² – 2x + k) (x² + bx + c) + (x + a)
x4 – 6x3 + 16x² – 25x + 10 = x4 + bx3 + x²c – 2x3 – 2bx² – 2cx + kx² + kbx + kc + x + a
x4 – 6x3 + 16x² – 25x + 10 = x4 + (b – 2)x3 + (c – 2b + k) x² + (- 2x + kb + 1)x + kc + a
Now comparing the co-efficients of both sides b – 2 = – 6 … (1)
[Comparing the co-efficient of x3]
c – 2b + k = 16 … (2)
[Comparing the co-efficient of x²]
– 2c + kb + 1 = – 25 … (3)
[Comparing the co-efficient of x]
kc + a = 10 … (4)
[Comparing the constant term]
From (1), b = – 4
Now, putting the vlaue of b in equation (2) and (3) we get,
⇒ c – 2(- 4) + k = 16
or c + k = 8 … (5)
and -2c – 4k + 1 = – 25
or – 2c – 4k = – 26
or – c – 2k = – 13 … (6)
Adding equation (5) and (6), we get, c + k = 8
– c – 2k = – 13 – k = – 5
So, k = 5,
From (5), c = 8 – 5 ⇒ c = 3,
Now put the value of k and c in equation (4), we get,
kc + a = 10
5(3) + a = 10
a = – 5
∴ Value of k and a is 5 and – 5 respectively.

We hope you found these NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 helped you with finding the solutions easily.

Make sure to practice the remaining parts of the Class 10 Maths NCERT Solutions English Medium Chapter 2.

## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.1

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.1. Practice more so you score more!!

## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1

Question 1.
The graphs of y = p(x) are given in the below given figures, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Solution: (i) No. of zeroes of p(x) is zero because the graph does not intersect x-axis point.
(ii) No. of zeroes of p(x) is 1 as the graph intersects the x-axis at one point only.
(iii) No. of zeroes p(x) is 3 as the graph intersects x-axis at three points.
(iv) No. of zeroes p(x) is as the graphs intersects the x-axis at two points.
(v) No. of zeroes p(x) is 4 as the graph intersects x-axis at four points.
(vi) No. of zeroes p(x) is 3 as the graph intersects the x-axis at three points.

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## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3. Practice more so you score more!!

## Exercise 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder, in each of the following:
(i) p(x) = x3 – 3x2 + 5x -3, g(x) = x2-2
(ii) p(x) =x4 – 3x2 + 4x + 5, g(x) = x2 + 1 -x
(iii) p(x) = x4 – 5x + 6, g(x) = 2 -x2
Solution:
(i) Therefore,
quotient = x – 3 and remainder = 7x – 9

(ii) First we arrange the terms of the dividend and the divisor in the decreasing order of their degrees.
∴ p(x) = x4– 3x2 + 4x + 5 and g(x) = x2 – x + 1 Therefore,
quotient = x2 +x-3 and remainder = 8

(iii) First we arrange the terms of the dividend and the divisior in the decreasing order of their degrees.
∴ p(x) = x2 + x – 3 and g(x) = – x2 + 2 Therefore,
quotient = – x² – 2 and remainder = – 5x + 10

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t2 – 3, 2t4 + t3 – 2t2 – 9t – 12
(ii) x2 + 3x + 1, 3x4 + 5x3 -7x2 + 2x + 2
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + l
Solution:
We have,
P(t) = 2t4 + 3t3 – 2t2 – 9t – 12
q(x) = t2 – 3
By actual division, we have Here, remainder is zero.
Therefore, q(x) = t2 – 3 is the factor of p(x) = 2t4 + 3t3 – 2t2 – 9t – 12.

(ii) We have,
p(x) = 3x4 + 5x3 – 7x2 + 2x + 2
and q(x) = x² + 3x + 1
By actual division, we have Here, remainder is not zero.
Therefore, q(x) = x² – 3x + 1 is not a factor of p(x) = 3x4 + 5x3 – 7x2 + 2x + 2.

(iii) We have,
p(x) = x5 – x3 + x2 + 3x + 1
and q(x) = x³ – 3x + 1
By actual division, we have Here, remainder is not zero.
Therefore, q(x) = x² – 3x + 1 is not a factor of p(x) = x5 – x3 + x2 + 3x + 1.

Question 3.
Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are $$\sqrt { \frac { 5 }{ 3 } }$$ and – $$\sqrt { \frac { 5 }{ 3 } }$$
Solution:
We have given that two zeroes of polynomial p(x) = 3x4 + 6x3 – 2x2 – 10x – 5 are $$\sqrt { \frac { 5 }{ 3 } }$$ and – $$\sqrt { \frac { 5 }{ 3 } }$$
∴ (x – $$\sqrt { \frac { 5 }{ 3 } }$$)(x + $$\sqrt { \frac { 5 }{ 3 } }$$) = x² – $$\frac { 5 }{ 3 }$$ is a factor is given polynomial p(x).
Now apply the division algorithm to the given polynomial and x² – $$\frac { 5 }{ 3 }$$ So, 3x4 + 6x3 – 2x2 – 10x – 5 If 3x4 + 6x3 – 2x2 – 10x – 5 = 0
Then, (x + $$\sqrt { \frac { 5 }{ 3 } }$$)(x – $$\sqrt { \frac { 5 }{ 3 } }$$)(x + 1)(3x + 3) = 0
∴ x = $$\sqrt { \frac { 5 }{ 3 } }$$ or x = $$\sqrt { \frac { 5 }{ 3 } }$$ Therefore, the zeroes of polynomial p(x) = 3x4 + 6x3 – 2x2 – 10x – 5 are – $$\sqrt { \frac { 5 }{ 3 } }$$, $$\sqrt { \frac { 5 }{ 3 } }$$, -1 and -1.

Question 4.
On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x).
Solution:
We know that
Dividend = Divisor x Quotiet + Remainder
∴ x² – 3x² + x + 2 = g(x) x (x – 2) + (- 2x + 4)
or x3– 3x2 + x + 2 = g(x) (x – 2) + (-2x + 4)
or x3 – 3x2 + x + 2 + 2 x- 4 = g(x) x (x-2)
or x3 – 3x2 + 3x – 2 = g(x) x (x – 2) Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) deg p(x) = deg q(x)
Polynomial p(x) = 2x2– 2x + 14; g(x) = 2
q(x) = x2 – x + 7 r(x) = 0
Here, deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)
Polynomial p(x) = x3+ x2 + x + 1; g(x) = x2 – 1,
q(x) = x + 1, r(x) = 2x + 2

(iii) deg r(x) is 0.
Polynomial p(x) = x2+ 2x2 – x + 2; g(x) = x2 – 1, q(x) = x + 1, r(x) = 4

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## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3. Practice more so you score more!!

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

Question 1.
Prove that $$\sqrt{5}$$ is irrational.
Solution:
Let us assume, to the contrary, that $$\sqrt{5}$$ is irrational that is we can find integers a and b (b ≠ 0) such that $$\sqrt{5}$$ = $$\frac { a }{ b }$$. suppose a and b have a common factor other than 1 then we can divide by the common factor and assume that a and b are coprime.
So b$$\sqrt{5}$$ = a
Squaring on both sides, and rearranging we get 5b² = a³.
Thus for a² is divisible by 5, it follows that a is also divisible by 5.
So, we can write a 5c for some integer c.
Substituting for a, we get
5b² = 25c²
b² = 5c²
This means that b2 is divisible by 5 and so b is also divisible by 5.
Therefore, a and b have at least 5 as common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that $$\sqrt{5}$$ is irrational.
So we conclude that $$\sqrt{5}$$ is irrational.

Question 2.
Prove that 3 + 2A$$\sqrt{5}$$ is irrational.
Solution:
Let us assume, to contrary, that 3 + 2$$\sqrt{5}$$ is rational.
That is, we can find coprime a and b (b ≠ 0)
such that 3 + 2$$\sqrt{5}$$ = $$\frac { a }{ b }$$
Therefore 3 – $$\frac { a }{ b }$$ = – 2$$\sqrt{5}$$
Rearranging this equation we get 2$$\sqrt{5}$$ = $$\frac { a }{ b }$$ – 3 = $$\frac { a – 3b }{ b }$$
Since a and b are integers, $$\frac { a }{ b }$$ – 3 we get is
rational and so 2$$\sqrt{5}$$ is rational and so $$\sqrt{5}$$ is rational.
But this contradicts the fact $$\sqrt{5}$$ is irrational.
This contadiction has arise because of our incorrect assumption 3 + 2$$\sqrt{5}$$ is rational.
Thus, we conclude that 3 + 2$$\sqrt{5}$$ is irrational.

Question 3.
Prove that the following are irrationals:
(i) $$\frac{1}{\sqrt{2}}$$
(ii) 7$$\sqrt{5}$$
(iii) 6 + $$\sqrt{2}$$
Solution:
Let us assume to the contrary, that $$\frac{1}{\sqrt{2}}$$ is rational that is, we can find coprime a and b(b ≠ 0) such that = $$\frac{1}{\sqrt{2}}$$ = $$\frac { a }{ b }$$
Since a and b are integers so $$\frac { a }{ b }$$ is rational and so $$\sqrt{2}$$ is rational.
But this contradicts the fact that $$\sqrt{2}$$ is irrational.
Thus, we conclude that $$\frac{1}{\sqrt{2}}$$ is irrational

(ii) 7$$\sqrt{5}$$
Let us assume, to the contrary that 7$$\sqrt{5}$$ is rational that is, we can find coprime a and b (≠ 0)
such that 7$$\sqrt{5}$$ = $$\frac { a }{ b }$$ Rearranging, we get $$\sqrt{5}$$ = $$\frac { a }{ b }$$
Since 7, a and b are integers, $$\frac { a }{ 7b }$$ is rational and so $$\sqrt{5}$$ is rational
But this contradicts the fact that $$\sqrt{5}$$ is irrational. So, we conclude that 7$$\sqrt{5}$$ is irrational.

(iii) 6 + $$\sqrt{2}$$
Let us assume, to the contrary, that 6 + $$\sqrt{2}$$ is irrational.
That is, we can find co prime a and b (* 0) such that 6 + $$\sqrt{2}$$ = 7 b
Rearranging, we get $$\sqrt{2}$$ = $$\frac { a-6b }{ b }$$
Since a, b and 6 are integers, so $$\frac { a-6b }{ b }$$ is rational and so is rational.
But this contradicts the fact that $$\sqrt{2}$$ is
irrational. So we conclude that 6 + $$\sqrt{2}$$ is irrational.

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## NCERT Solutions for Class 10 Hindi Kshitij क्षितिज भाग 2

We hope you all started your preparation for the upcoming tests. Well, these NCERT Solutions for class 10 Hindi क्षितिज भाग 2 can surely help you with the preparation and scoring better than your classmates. Make sure to check these NCERT Solutions for Class 10 Hindi and also to check the NCERT Solutions for Class 10 Hindi Kshitij PDF for reference.

NCERT Solutions for Class 10 Hindi Kshitij Bhag 2 क्षितिज भाग 2

काव्य – खंड

गद्य – खंड

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1. Practice more so you score more!!

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

Question 1.
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting)? Represent this situation algebraically and graphically.
Solution:
Let Aftab’s present age be x and his daughter’s present age by y.
∴ Seven years ago,
Aftab’s age was x – 7
and his daughter’s age was y – 7
According to question
(x – 7) = 7 (y – 7)
or, x – 7 = 7y – 49
or, x – 7y + 42 = 0 … (i)
In case II
Three vears hence
Aftab’s age will be = x + 3
and his daughter’s age will be = y + 3
Again, according to question,
(x + 3) = 3(y + 3)
or, x + 3 = 3y + 9
or, x – 3y – 6 = 0 … (ii)
From equation (i)
x – 7y + 42 = 0
x = 7y – 42 Now from equation (ii)
x – 3y – 6 = 0
x = 3y + 6 x – 7y + 42=0
x – 3y – 6 = 0
Subtracting both equations
4y = 48
y = 12
Putting this value in equation (ii)
x – 3(12) – 6 = 0
x – 36 – 6 = 0
x = 2
Hence the age of Aftab is 42 years and his daughter is 12 years old.

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Solution:
Let the cost of each bat = x
and the cost of each ball = y
Then the algebric representation is given by the following equation. Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
Solution:
Let cost of one kg of apples = ₹ x
and the cost of one kg of grapes = ₹ y
Then the algebric representation is given by the following equations : We hope you found these NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 helped you with finding the solutions easily. Make sure to practice the remaining parts of the Class 10 Maths NCERT Solutions English Medium Chapter 3.

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2. Practice more so you score more!!

## Exercise 1.2

Question 1.
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) 140
140 = 2 x 2 x 7 x 5
140 = 22 x 7 x 5

(ii) 156 = 2 x 2 x 3 x 13 = 22 x 3 x 13

(iii) 3825 = 5 x 5 x 3 x 3 x 17
= 32 x 52 x 17

(iv) 5005 = 5 x 7 x 11 x 13

(v) 7429 = 17 x 19 x 23

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
The Prime factorisation of 26 and 91 is
26 = 2 x 13
91 = 7 x 13
Therefore, the LCM is 2 x 7 x 13 = 182.
Also HCF is 13.
By using the formula LCM x HCF = Product of two numbers
182 x 13 = 26 x 91
2366 = 2366
∴ LMC x HCF = Product of two numbers. [Verified]

(ii) 510 and 92
The prime factorisation of 510 and 92 is
510 = 2 x 3 x 5 x 17 92 = 2 x 2 x 23
Therefore, the LCM is 2 x 2 x 3 x 5 x 17 x 23 = 23,460 and HCF is 2.
Now, LCM x HCF = Product of two numbers
L.H.S. = LCM x HCF = 23640 x 2 = 46920
R.H.S. = Product of two numbers
= 510 x 92 = 46920
L.H.S. = R.H.S. [Verified]

(iii) 336 and 54
The prime factorisation of 336 and 54 is
336 = 24 x 3 x 7 54
= 2 x 3³
Therefore, the LCM is 24 x 3³ x 7 = 3024
and HCF is 2 x 3 = 6
Now, LCM x HCF = Product of two numbers
L.H.S. = LCM x HCF = 3024 x 6 = 18144
R.H.S. = Product of two numbers = 336 x 54 = 18144
Now, L.H.S.= R.H.S. [Verified]

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12,15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) 12,15 and 21
12 = 2 x 2 x 3
15 = 3 x 5
21 = 3 x 7
Therefore the LCM is 2² x 3 x 7 = 420
And the HCF is 3! = 3

(ii) 17, 23 and 29
17 = 1 x 17
23 = 1 x 23
29 = 1 x 29
Therefore tht LCM is 1 x 17 x 23 x 29 = 11339
And the HCF = 1

(iii) 8,9 and 25
8 = 2 x 2 x 2 x 1
9= 3 x 3 x 1
25 = 5 x 5 x 1
Therefore the LCM is 1 x 2³ x 3² x 5² = 1800
And HCF is = 1

Question 4.
Given that HCF (306,657) = 9, find LCM (306, 657).
Solution:
HCF (306, 657) = 9 We know that
HCF x LCM = Product of two numbers
9 x LCM = 306 x 657
LCM = $$\frac { 201042 }{ 9 }$$
LCM = 22338

Question 5.
Check wether 6n can end with the digit 0 for any natural number n.
Solution:
If the number 6n for any n, were to end with digit zero, then it will be divisible by 5. That is, the prime factorisation of 6n would contain the prime 5. This is not possible because 6n = (2 x 3)n = 2n. 3n, so the only prime in the factorisation of 6n is 2. So the uniqnessess of the fundamental theorem of arithemetic gurantees that, there are no other primes in the factorisation of 6n.

Question 6.
Explain why 7 x 11 x 13 +13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
Solution:
We know by fundamental theorem of arithmetic, “every composite number can be factorised as a product of primes”. So we can explain 7 x 11 x 13 + 13 and 7 x 3 x 2 x 5 x 2 x 2 x 3 x 2 x 1 + 5 are composite numbers or 7 x 11 x 13 + 13 and 7 x 32 x 24 + 5 are composite numbers.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to derive one round of the field, while Ravi takes 12 minutes for same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point.
Solution:
L.C.M (12,18)
12 = 2 x 2 x 3
18 = 2 x 3 x 3
So the L.C.M of 12 and 18 is 2 x 3 x 2 x 3 = 36 After 36 minutes they will will meet at the starting point.

We hope you found these NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 helped you with finding the solutions easily.

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## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2. Practice more so you score more!!

## Exercise 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 -15
(vi) 3x2 – x – 4
Solution:
(i) x² – 2x – 8
⇒ x² – 4x + 2x – 8
⇒ x(x – 4) + 2(x – 4)
⇒ (x – 4) (x + 2)
⇒ x – 4 = 0 or x + 2 = 0
⇒ x = 4 or x= -2
Verification: (ii) 4s2 – 4s + 1
⇒ 4s² – 2s – 2s + 1
⇒ 2s(2s – 1) – 1(2s – 1)
⇒ (2s – 1) (2s – 1)
2s – 1 = 0 or 2s – 1 = 0
⇒ s = $$\frac { 1 }{ 2 }$$ or s = $$\frac { 1 }{ 2 }$$
Verification: (iii) 6x2 – 3 – 7x
⇒ 6x² – 7x + 2x – 3
⇒ 3x(2x – 3) + 1(2x – 3)
⇒ (2x – 3) (3x + 1)
2x – 3 = 0 or 3x + 1 = 0
⇒ x = $$\frac { 3 }{ 2 }$$ or x = $$\frac { -1 }{ 3 }$$
Verification: (iv) 4u2 + 8u
⇒ 4u(u + 2)
⇒ u – 4 = 0 or u + 2 = 0
⇒ u = 4 or u= – 2
Verification: (v) t2 -15
⇒ t² = 15
⇒ t = ± $$\sqrt{15}$$
t = $$\sqrt{15}$$ or t = – $$\sqrt{15}$$
Verification: (vi) 3x2 – x – 4
⇒ 3x² – 4x + 3x – 4
⇒ x(3x – 4) + 1(3x – 4)
⇒ (3x – 4) (x + 1)
3x – 4 = 0 or x + 1 = 0
⇒ x = $$\frac { 4 }{ 3 }$$ or x = – 1
Verification: Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $$\frac { 1 }{ 4 }$$, – 1
(ii) $$\sqrt{2}$$, $$\frac { 1 }{ 3 }$$
(iii) 0, $$\sqrt{5}$$
(iv) 1, 1
(v) $$\frac { -1 }{ 4 }$$, $$\frac { 1 }{ 4 }$$
(vi) 4, 1
Solution:
(i) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = $$\frac { 1 }{ 4 }$$ = $$\frac { -b }{ a }$$
and α.β = – 1 = $$\frac { c }{ a }$$
If a = 4, then b = – 1 and c = – 4
So, one quadratic polynomial which fits the given condition is 4x² – x – 4.

(ii) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = $$\sqrt{2}$$ = $$\frac { -b }{ a }$$
and α.β = $$\frac { 1 }{ 3 }$$ = $$\frac { c }{ a }$$
If a = 4, then b = – $$\sqrt{2}$$ and c = $$\frac { 1 }{ 3 }$$
So, one quadratic polynomial which fits the given condition is x² – $$\sqrt{2x}$$ + $$\frac { 1 }{ 3 }$$ or 3x² – 3$$\sqrt{2x}$$ + 1

(iii) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 0 = $$\frac { -b }{ a }$$
and α.β = $$\sqrt{5}$$ = $$\frac { c }{ a }$$
If a = 1, then b = 0 and c = $$\sqrt{5}$$
So, one quadratic polynomial which fits the given condition is x² – $$\sqrt{5}$$

(iv) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 1 = $$\frac { -b }{ a }$$
and α.β = 1 = $$\frac { c }{ a }$$
If a = 1, then b = – 1 and c = 1
Now, put the values of a, b and c in equation ax² + bx + c
So, one quadratic polynomial which fits the given condition is x² – $$\sqrt{5}$$
1x² – 1x 1 = 0
or x² – x + 1 = 0

(v) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = –$$\frac { 1 }{ 4 }$$ = $$\frac { -b }{ a }$$
and α.β = $$\frac { 1 }{ 4 }$$ = $$\frac { c }{ a }$$
If a = 1, then b = 1 and c = 1
Now, put the values of a, b and c in equation ax² + bx + c
So, one quadratic polynomial which fits the given condition is 4x² + x + 1 = 0

(vi) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 4 = $$\frac { -b }{ a }$$
and α.β = 1 = $$\frac { c }{ a }$$
If a = 1, then b = – 4 and c = 1
So, one quadratic polynomial which fits the given condition is x² – 4x + 1 = 0

We hope you found these NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 helped you with finding the solutions easily.

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